### In Gauss’S Law, To What Does Qencl Refer?

Strategy – From Gauss’s law, the flux through each surface is given by q enc / ε 0, q enc / ε 0, where q enc q enc is the charge enclosed by that surface.

## What is Q in Gauss’s law?

Gauss Law Formula –

• As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ 0 is electric constant, the Q enclosed by the surface is;
• Q = ϕ ϵ 0
• The Gauss law formula is expressed by;
• ϕ = Q/ϵ 0
• Where,
• Q = total charge within the given surface,
• ε 0 = the electric constant.

## What is E in Gauss’s law?

Gauss’s Law – The flux $$\Phi$$ of the electric field $$\vec$$ through any closed surface S (a Gaussian surface) is equal to the net charge enclosed $$(q_ )$$ divided by the permittivity of free space $$(\epsilon_0)$$: \ To use Gauss’s law effectively, you must have a clear understanding of what each term in the equation represents.

The field $$\vec$$ is the total electric field at every point on the Gaussian surface. This total field includes contributions from charges both inside and outside the Gaussian surface. However, $$q_$$ is just the charge inside the Gaussian surface. Finally, the Gaussian surface is any closed surface in space.

That surface can coincide with the actual surface of a conductor, or it can be an imaginary geometric surface. The only requirement imposed on a Gaussian surface is that it be closed (Figure $$\PageIndex$$). Figure $$\PageIndex$$: A Klein bottle partially filled with a liquid. Could the Klein bottle be used as a Gaussian surface?

#### What does Q enclosed equal?

Maxwells Electromagnetic Field Equation No.1 Maxwells Electromagnetic Field Equation No.1 By 1.0 Statement of Equation The following Electrostatic Field equations will be developed in this section:

 Integral form Differential forms

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• Maxwells first equation is based on Gauss law of electrostatics published in 1832, wherein Gauss established the relationship between static electric charges and their accompanying static fields.
• The above integral equation states that the electric flux through a closed surface area is equal to the total charge enclosed.
• The differential form of the equation states that the divergence or outward flow of electric flux from a point is equal to the volume charge density at that point.
• 1.1. Maxwells Equation No.1; Area Integral We will derive the integral equation by considering the summation of electric flux density on a surface area, and then as a summation of volume containing electric charge. The two integrals are shown to be equal when they are based on the same charge.

Two examples using the equations are shown.1.1.1 Gauss Law Gauss electrostatics law states that lines of electric flux, f E, emanate from a positive charge, q, and terminate, if they terminate, on a negative charge. The space within which the charges exert their influence is called the electrostatic field.

The sketch in Figure 1.1 represents the charges and the three dimensional field. The field is visualized as being made up of lines of flux. For an isolated charge, the lines of flux do not terminate and are considered to extend to infinity. To obtain the equation relating an electric charge q, and its flux f E, assume that the charge is centered in a sphere of radius r meters. The electric flux density, D, is then equal to the electric flux emanating from the charge, q, divided by the area of the sphere.

coulombs per square meter; where the area is perpendicular to the lines of flux. (One coulomb is equal to the magnitude of charge of 6.25 X 10 18 electrons.) The charge enclosed in the sphere is then equal to the electric flux density on its surface times the area enclosing the charge. q (coulombs enclosed) = D x 4 r 2,

The lines of flux contributing to the flux density are those that leave the sphere perpendicular to the surface of the sphere. This leads to the integral statement of this portion of Gauss law; The integral sign indicates the summation of infinitesimal areas, d a, in order to obtain the entire surface area. The circle on the integral sign indicates that the integral or summation of area is taken of a closed continuous surface. Bold face letters indicate that the letter represents a vector, i.e., this quantity has magnitude and direction.

Distance, velocity, acceleration and force are common examples of vectors. D is the electric flux density vector in coulombs per square meter. The (said dot) following D shows that the vector dot product must be used when multiplying the two vectors, D and d a, The dot product (discussed below) indicates that the magnitudes of the two vectors are multiplied together and then that product is multiplied by the cosine of the angle between the two vectors.

The dot product here enables us to determine the effective lines of flux flowing through the surface.1.1.2 Vector Dot Product A vector dot product application can be illustrated by calculating work in the following physics problem. Recall that work = force times distance. A force of 20 pounds is applied through the mop handle. As shown in the diagram, only that component of the force in the direction parallel to the floor is used in the calculation of work. We see that the force parallel to the floor is; 20 pounds x cosine 60 0 = 10 pounds. What is the work done when pushing the mop 8 feet across the floor? 10 lbs x 8 ft = 80 ft. lbs.

1. Using the dot product, the equation for work is:
2. Work = F orce d istance.
3. The dot product indicates that work equals the magnitude of the force times the magnitude of the distance moved, times the cosine of the angle between the two vectors.
4. Or, Work = Force on the mop handle, times the distance the force moves, times the cosine of the angle between the force and the floor.
5. Work = 20 pounds times 8 feet times 1/2 = 80 foot pounds.
6. Only that component of the total force in the direction parallel to the floor, as obtained through the dot product, is used in the calculation of work.

This shows that the dot product is defined as the method of vector multiplication in which the vector magnitudes are multiplied together and then that product is multiplied by the cosine of the vectors included angle. There must always be a vector on each side of the dot in the dot product.

Therefore, when the dot product is used in Gauss Law, only that component of flux parallel to the vector representing area will contribute to the total enclosed charge.1.2 Gauss Law; Area Integral Examples The method of determining charge by using the dot product of is similar to finding Work as the dot product between applied Force and Distance.

Through this method, only those components of the vector lines of flux in the same direction as the vector representing area will be summed in the calculation of charge. Or, stated in another way: Only those flux lines perpendicular to the surface are incorporated in the result of the dot product to obtain the charge enclosed.

A scalar value is always the resultant of a dot product. In this case, the result is a number of coulombs. Examples of other scalar quantities are temperature, mass and power. A scalar quantity, in contrast to a vector, does not have direction. The differential element of area is d a, A vector representing an area is pointed normal, i.e.

perpendicular, to that area. Using the dot product between the vector representing area d a and the flux density D, results in obtaining the effective flux through the area. The summation of the entire area is in square meters. The preliminary equation (Gauss Law) in our procedure to obtain Maxwells first equation is now; This integral equation states that the amount of electric flux density normal to a surface is caused by a specific amount of charge, q, enclosed by the surface. Consider the following examples of finding the electric flux density on a spherical surface and on a cylindrical surface.1.2.1 Determine Flux Density on a Sphere Assume a charge of one coulomb is centered in a sphere of radius r meters as in Figure 1.3. A vector representing an area is directed normal to that area. The vector representing the small area, d a, is then directly in line with a line of electric flux leaving the sphere. D represents the density of those lines of electric flux leaving the sphere. one coulomb = D x 4 p r 2, coulomb per square meter on the surface of the sphere.1.2.2 Determine Flux Density on a Cylinder Assume a long line of stationary charges of q coulombs per meter as shown in Figure 1.4. There is a cylinder of length “L” and radius “r” centered on the charges.

What is the electric flux density on the surface of the cylinder? Gauss equation is: Q T, the total charge enclosed, = q coulombs per meter x L meters. = Q T = D x 2 p r x L, D in coulombs per square meter = 1.3 Maxwells Equation No.1; Volume Integral Gauss electrostatics law is also written as a volume integral: This equation states that the charge enclosed in a volume is equal to the volume charge density, r, (rho) summed for the entire volume.

q is the charge enclosed in the volume. r is the volume charge density in coulombs per cubic meter. is an infinitesimal element of volume. The entire volume is in cubic meters. The total charge enclosed in the volume is the volume in cubic meters times the charge density in coulombs per cubic meter.

The average volume charge density summed for the entire volume is the charge enclosed. A discussion of r is found in Section 1.5.3.1.4 Maxwells Equation No.1; Integral Form Completed To obtain the integral form of Maxwells Equation No.1, assume that an experiment is set up so that the same charge of q coulombs is contained in each of Gauss law equations.

Then the integrals due to the same charge must be equal. Then, Thus we have obtained the integral form of Maxwells Equation No.1. This equation states that the effective electric field through a surface enclosing a volume is equal to the total charge within the volume.

The equation shows that the area enclosed by the left hand integral must enclose the volume of the right integral. This is similar to stating that the surface area of a ball or box encloses the volume of the ball or box. The area and volume indicated by the equations need not be observable physical surfaces, often they will be mathematical limits.

To remember the integral form of Maxwells Equation No.1, consider that a charge q, enclosed in a volume, must be equal to the volume charge density, r, times the volume. Also, the same charge q, will cause a certain area flux density, D, times a certain area.

The area must enclose the volume. The integrals (summations) must be equal since the same number of coulombs must be obtained on each side of the equal sign.1.5 Maxwells Equation No.1; Differential form The differential form of Maxwells Equation No.1 is: is a differential operator read “del” (discussed below).

is read “divergence”. D is the electric flux density in coulombs per square meter. r (rho) is the volume charge density in coulombs per cubic meter.1.5.1 Discussion of (del) is the mathematical extension of the ordinary single dimension calculus derivative into three dimensions.

• Velocity (v) =.
• As the change in time is made very small the differential calculus symbol is used for velocity.
1. Now consider the ordinary single dimension derivative for acceleration.
2. Recall the equation for obtaining the velocity of an object when it is dropped from a height. The velocity that the object attains is found by;
3. Velocity, v, = acceleration due to gravity, g, times the time during which the object is falling.
4. Velocity = v = g t. Or

In the integral form of Gauss Law we summed the infinitesimal values of area and volume, da and dv. Here we are using the differential, ds, dv and dt to find instantaneous rates of change of distance and velocity with respect to time. There is a rule of differential calculus that we will point out here as we will use the rule below.

• Notice that velocity is equal to acceleration (g) times time(t).
• So we can take the derivative of velocity in this manner: The calculus rule is that the derivative of a variable times a constant is the constant.
• We will now extend that concept of ordinary derivatives to partial derivatives,This will allow us to obtain the rate of change of a volume in three dimensions, which in turn leads to the definition of Ñ (del).

To illustrate the rate of change in three dimensions, assume a box is placed at the origin of a rectangular coordinate system, as shown in Figure 1.5. Box volume, V, = length x width x height; = L x W x H. What is the rate of change of volume when only the length increases by a small amount but the width and height remain constant? This is where the symbol for the partial derivative is used.

• The symbol for partial derivatives is slightly different than the symbol for ordinary (one dimension) derivatives.
• The symbol indicates that only one of the independent variables is changing at the moment under consideration.
• The dependent variable, volume, v changes as determined by changes of the independent variables; L,W and H.

When only the length changes; The partial derivative symbol, shows that the change in volume is due to a change in length only. The width and height are being held constant. We also see that the derivative of a variable times a constant is the constant, W times H.

1. It follows, by symmetry, that the rate of change of volume as a function of either width or height is expressed as a partial derivative.
2. When the change in volume is due to a simultaneous change in length, width and height, the changes will occur in the x, y and z directions and the partial derivatives are added to find the resultant rate of change of volume.

This is accomplished in vector form by multiplying each partial derivative by unit vectors pointing in the x, y and z directions. Unit vectors are indicated here and discussed further in Section 1.5.2. Using unit vectors and simultaneous changes in three dimensions, the total change in volume is designated by: For the more general case of a volume, V changing in the x, y and z directions; This discussion of ordinary and partial derivatives was aimed at obtaining the group of three partial derivative terms in the above parenthesis.

• We will not use the gradient in this paper.
• We will have need for Ñ (del dot, discussed below) and Ñ 5 (del cross, discussed in Section 3).1.5.2 Discussion of Vector Components in Relation to Before continuing the part of the differential form of Maxwells equation No.1, we must consider the x, y and z components of a vector in rectangular coordinates.

In section 1.1.2 the dot product of two vectors, force and distance, was used to calculate work. We will here calculate the same work using the vector components of force and distance, and employ the dot product. Notice that the vector components of the force in the rectangular coordinates are placed at either at zero or 90 degrees to the distance movement.

The cosine of zero degrees is one and the cosine of 90 degrees is zero. From Figure 1.3; The vector components of force = 17.3 lbs. y + 10 lbs.x, The vector components of distance = 0 ft. y + 8 ft.x, The vector (dot) multiplication procedure now is to multiply the vector component magnitudes and the cosine of the angle between them, term by term.

The zero y distance times the two components of the force is zero. The 8 ft x distance times the 17.8 lb force in the y direction times the cosine of 90 degrees is zero. The 8 ft x directed distance times the 10 lb force in the x direction times the cosine of zero degrees is the same 80 ft lbs that we found in the previous example.

• This procedure of multiplying vector x, y and z components is followed in performing theproduct below.
• The multiplication of vector components, that are always at zero or 90 degrees apart, greatly simplifies vector mathematics.
• We determined the components of in Section 1.5.1.
• Above we discussed dot product multiplication of vectors using their components.

These two concepts are now used to calculate 1.5.3 Calculate The components of vector D are its projections on the x, y and z axis. The vector directions of D components are designated by the unit vectors x, y and z, In Figure 1.6, the vector D starts at the origin, points up and to the right and is indicated as coming out of the paper.

The magnitudes of projections of D along the axes are D x, D y, and D z, Figure 1.7 shows the unit vectors in the x, y and z directions that give the components of D their vector relationship. The same unit vectors are designated in The equation for vector D as projected on the three coordinate axes is, D = D x x + D y y + D z z.

We will now do the indicated dot product of, The dot product indicates that we must multiply the parentheses, term by term, times the cosine of the included angle between each pair of terms. This series of multiplications could result in nine terms, but notice that a unit vector dotted into the same unit vector:

The other six combinations of unit vector dot product multiplications contain the cosine of 90 degrees and are therefore, zero. The final result of the operation is a scalar of only three terms: This equation indicates the sum of a change in electric flux density, D, in each of the three orthogonal directions.

The change is due to a small, (approaching zero), distance change in the same orthogonal directions. The change of distance in the three orthogonal directions is a volume change, as shown in Section 1.5.1. Therefore, the electric density ( D ) change in the three directions, that we obtained by using the dot product with the unit vectors in del, is actually a per unit volume change.

Since charge is measured in coulombs, the sum of charge is in coulombs. The result of adding the three electric density changes is coulombs per cubic meter. This defines r, the volume charge density, as noted in Section 1.3.1.6 Equation No.1, Differential Form Completed By doing the indicated operation we obtained r, the volume charge density.

This is the differential statement of Maxwells equation No.1. The equation states that the divergence of the electric flux density at a point is equal to the charge per unit volume at that point. The dot product, as always, produces a scalar result. In this case, the result is r, the number of coulombs of charge per cubic meter.1.7 Divergence Theorem It is instructive at this point to continue using the integral and differential equations just developed for Maxwells Equation No.1 in order to illustrate a vector identity called, “Gauss Divergence Theorem”.

This identity equates a vector surface integral to a vector volume integral, and will be required later in Section 2.5,

1. From Section 1.4,
2. From Section 1.5;
3. By substituting for

r in the integral equation we obtain; This is a typical illustration of Gauss divergence theorem, using vector D as the example. The point here is that any time we have a vector surface integral of this type we can substitute the volume integral. If we have a vector volume integral of the above type, we can substitute the surface integral.

The integral of the divergence of a vector summed throughout the volume is equal to the integral of the product of the vector times its effective area summed over the area. This is analogous to stating that the volume of a ball is contained within its surface area. The circle on the integral sign indicates that the integral is taken over a continuous area.

If we had simply used Gauss Divergence Theorem from a textbook list of vector identities, we could have immediately written down the differential form of Maxwells equation No.1 from the integral form. This more detailed way of obtaining the identity will be helpful in later derivations.1.8 Relation of D, E and e The space in which electric charges exert their influence is called the field of the electric charge.

Surrounding an electric charge q, there is an electric field of field strength E, It is the electric field strength, E, that will cause an amount of flux density, D, depending on the permittivity, e of the surrounding medium. D is in coulombs per square meter. E is in newtons per coulomb or volts per meter.

e is in coulomb 2 per newton meter 2, E is the stress in space that causes D to be manifested. Due to this equation and for reasons discussed in Section 3.7, D is often designated as electric flux displacement density in addition to electric flux density.

Also, it will be shown in Section 4, that a magnetically induced electric field is also designated E, with a dimension of volts per meter. That induced electric field is the same field as the static field strength discussed here but it is generated by a changing magnetic field. The permittivity e, is the degree to which the surrounding medium will permit the electric flux density, D, to occur due to a given electric field strength, E,

In the medium of air or free space, e = 8.85×10 -12 coulomb 2 per newton meter 2, These concepts and definitions will be used in Sections 6 and 7.1.9 Coulombs Law In Section 1.2.1 we found that the electric flux density, D, due to a charge q, located within the sphere is: Then, using E and e as defined in Section 1.8; When another charge q 2, is placed r meters from q 1, a force is experienced by q 2,

• The force is E times q 2 Newtons.
• Or, where q 1 and q 2 designate individual charges.
• This equation is Coulombs law.
• Recall that a force of one newton will accelerate a mass of one kilogram at one meter per second 2,
• Coulombs law states that the force between two charges is proportional to the product of the two charges over the distance between the two charges squared.

The equation is the basis for experimentally determining the force between two charges and the permittivity of different mediums. This important electrical law is not included in Maxwells list as it is considered derivable from Gauss law, and is not used in these field equations.

## How do you find the Q enclosed of a sphere?

For example, if we are inside the sphere, we get qenc=ρV=qV1V2=q43πR3(43πr3)=qr3R3.

### What does q mean in electric field?

Q is the symbol used to represent charge, while n is a positive or negative integer, and e is the electronic charge, 1.60 x 10 – 19 Coulombs.

## What is q in electric flux?

Overview – An electric charge, such as a single electron in space, has an electric field surrounding it. In pictorial form, this electric field is shown as a dot, the charge, radiating “lines of flux”. These are called Gauss lines. Note that field lines are a graphic illustration of field strength and direction and have no physical meaning.

The density of these lines corresponds to the electric field strength, which could also be called the electric flux density: the number of “lines” per unit area. Electric flux is proportional to the total number of electric going through a surface. For simplicity in calculations, it is often convenient to consider a surface perpendicular to the flux lines.

If the electric field is uniform, the electric flux passing through a surface of S is Φ E = E ⋅ S = E S cos ⁡ θ, =\mathbf \cdot \mathbf =ES\cos \theta,} where E is the electric field (having units of V/m ), E is its magnitude, S is the area of the surface, and θ is the angle between the electric field lines and the normal (perpendicular) to S, For a non-uniform electric field, the electric flux dΦ E through a small surface area d S is given by d Φ E = E ⋅ d S }\Phi _ =\mathbf \cdot }\mathbf } (the electric field, E, multiplied by the component of area perpendicular to the field). The electric flux over a surface S is therefore given by the : Φ E = ∬ S E ⋅ d S =\iint _ \mathbf \cdot }\mathbf } where E is the electric field and d S is a differential area on the closed surface S with an outward facing defining its direction. For a closed, electric flux is given by: Φ E = =\,\!} S E ⋅ d S = Q ε 0 \cdot }\mathbf = }}\,\!} where

• E is the,
• S is any,
• Q is the total inside the surface S,
• ε 0 is the (a universal constant, also called the ” of free space”) ( ε 0 ≈ 8.854 187 817 × 10 −12 / )

This relation is known as for electric field in its form and it is one of, While the electric flux is not affected by charges that are not within the closed surface, the net electric field, E can be affected by charges that lie outside the closed surface.

While Gauss’s law holds for all situations, it is most useful for “by hand” calculations when high degrees of symmetry exist in the electric field. Examples include spherical and cylindrical symmetry. The unit of electric flux is the volt-meter ( V·m ), or, equivalently, -meter squared per ( N·m 2 ·C −1 ).

Thus, the unit of electric flux expressed in terms of base units is kg·m 3 ·s −3 ·A −1, Its dimensional formula is L 3 M T − 3 I − 1 }^ }^ }^ },

### What is E in terms of charge?

Elementary charge
Definition: charge of a proton
Symbol: e
SI value: 1.602 176 634 × 10 −19 C

The elementary charge, usually denoted by e is the electric charge carried by a single proton or, equivalently, the magnitude of the negative electric charge carried by a single electron, which has charge −1 e, This elementary charge is a fundamental physical constant, In the SI system of units, the value of the elementary charge is exactly defined as = 1.602 176 634 × 10 −19 coulombs, or 160.2176634 zeptocoulombs (zC). Since the 2019 redefinition of SI base units, the seven SI base units are defined by seven fundamental physical constants, of which the elementary charge is one. In the centimetre–gram–second system of units (CGS), the corresponding quantity is 4.803 2047,

### What is E in flux equation?

What is Electric Flux? – Definition, Formula, Unit, Symbol In this section, we will discuss the concept of electric flux, its calculation, and the analogy between the flux of an electric field and that of water. Let us imagine the flow of water with a velocity v in a pipe in a fixed direction, say to the right.

1. If we take the cross-sectional plane of the pipe and consider a small unit area given by ds from that plane, the volumetric flow of the liquid crossing that plane normal to the flow can be given as vds,
2. When the plane is not normal to the flow of the fluid but is inclined at an angle θ, the total volume of liquid crossing the plane per unit time is given as vds,cos θ,

Here, ds cos θ is the projected area in the plane perpendicular to the flow of the liquid. The electric field is analogous to the liquid flow in the case shown above. The quantity we will deal with here is not an observable quantity as the liquid we considered above. Here, we see that the of magnitude E pass through a plane of area A that is kept at an angle θ to the direction of the electric field. The total number of electric field lines passing a given area in a unit of time is defined as the electric flux. Similar to the example above, if the plane is normal to the flow of the electric field, the total flux is given as:

 $$\begin \phi _ =EA\end$$

When the same plane is tilted at an angle θ, the projected area is given as Acosθ, and the total flux through this surface is given as:

 $$\begin \phi=EAcos\theta\end$$

Where,

E is the magnitude of the electric field A is the area of the surface through which the electric flux is to be calculated θ is the angle made by the plane and the axis parallel to the direction of flow of the electric field

Watch this enticing video on Electric Flux and reimagine the concept like never before. To learn more about electric flux, the, and other related topics, download “BYJU’S – The Learning App. See the video below to learn problems based on electric flux, electric charges and fields (Gauss’s law). An electric field is a physical field that surrounds electrically activated particles or bodies. It exerts a force on every other charged particle or body in the field (repelling or attracting). In other words, it can be defined as the physical field for a body of charged particles.

• The total number of electric field lines passing a given area in a unit time is defined as the electric flux.
• An electric field line originates on a positive electric charge and terminates on a negative charge.
• The mathematical relationship between enclosed charge and electric flux is called Gauss’s law (case of an electric field).

The SI base unit of electric flux is voltmeters (V m). Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz Visit BYJU’S for all Physics related queries and study materials

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View Quiz Answers and Analysis : What is Electric Flux? – Definition, Formula, Unit, Symbol

### What is E in potential energy?

What is Electric Potential Difference? –

• In an electrical circuit, the potential between two points (E) is defined as the amount of work done (W) by an external agent in moving a unit charge (Q) from one point to another.
• Mathematically we can say that,
• E = W/Q
• Where,
• E = electrical potential difference between two points
• W = Work done in moving a charge from one point to another
• Q = Quantity of charge in coulombs

#### What is q equal to in physics?

The quantity of charge (Q) on an object is equal to the number of elementary charges on the object (N) multiplied by the elementary charge (e).

## What does the q variable represent in coulomb’s law?

Answer and Explanation: In Coulomb’s law q1 represents the signed magnitudes of the charge in object one and q2 represents the magnitude of the charge in the second object. In the equation the distance of separation between the charges can be represented by r or d.

#### What is q coulomb?

The symbol Q represents the amount of charge in coulombs, and the symbol n refers to the number of electrons or protons. If both sides of the equation are divided by e, the number of particles in a coulomb can be calculated as the following: Q = n ⋅ e. Q / e = (n ⋅ e) / e. n = Q / e (after reversing Q / e = n)

### How do you find the Q electric field?

Teacher Support – Ask students whether they have seen movies that use the concept of fields as in force fields, Have them describe how such fields work. Describe how gravity can be thought of as a field that surrounds a mass and with which other masses interact.

1. Explain that electric fields are very similar to gravitational fields.
2. You may have heard of a force field in science fiction movies, where such fields apply forces at particular positions in space to keep a villain trapped or to protect a spaceship from enemy fire.
3. The concept of a field is very useful in physics, although it differs somewhat from what you see in movies.

A field is a way of conceptualizing and mapping the force that surrounds any object and acts on another object at a distance without apparent physical connection. For example, the gravitational field surrounding Earth and all other masses represents the gravitational force that would be experienced if another mass were placed at a given point within the field.

Michael Faraday, an English physicist of the nineteenth century, proposed the concept of an electric field, If you know the electric field, then you can easily calculate the force (magnitude and direction) applied to any electric charge that you place in the field. An electric field is generated by electric charge and tells us the force per unit charge at all locations in space around a charge distribution.

The charge distribution could be a single point charge ; a distribution of charge over, say, a flat plate; or a more complex distribution of charge. The electric field extends into space around the charge distribution. Now consider placing a test charge in the field.

• A test charge is a positive electric charge whose charge is so small that it does not significantly disturb the charges that create the electric field.
• The electric field exerts a force on the test charge in a given direction.
• The force exerted is proportional to the charge of the test charge.
• For example, if we double the charge of the test charge, the force exerted on it doubles.

Mathematically, saying that electric field is the force per unit charge is written as E → = F → q test E → = F → q test 18.15 where we are considering only electric forces. Note that the electric field is a vector field that points in the same direction as the force on the positive test charge.

The units of electric field are N/C. If the electric field is created by a point charge or a sphere of uniform charge, then the magnitude of the force between this point charge Q and the test charge is given by Coulomb’s law F = k | Q q test | r 2 F = k | Q q test | r 2 where the absolute value is used, because we only consider the magnitude of the force.

The magnitude of the electric field is then E = F q test = k | Q | r 2, E = F q test = k | Q | r 2,18.16 This equation gives the magnitude of the electric field created by a point charge Q, The distance r in the denominator is the distance from the point charge, Q, or from the center of a spherical charge, to the point of interest.

1. If the test charge is removed from the electric field, the electric field still exists.
2. To create a three-dimensional map of the electric field, imagine placing the test charge in various locations in the field.
3. At each location, measure the force on the charge, and use the vector equation E → = F → / q test E → = F → / q test to calculate the electric field.

Draw an arrow at each point where you place the test charge to represent the strength and the direction of the electric field. The length of the arrows should be proportional to the strength of the electric field. If you join together these arrows, you obtain lines. Figure 18.17 Three-dimensional representation of the electric field generated by a positive charge.

#### What is Q in e Fe Q?

Assignment SE10: Electric Field – Objectives:

The student should be able to define electric field and recognize the variables which effect (and do not effect) the electric field intensity at a given location. The student should be able to use electric field equations and relatively simple numbers to determine the magnitude and the direction of the electric field intensity at a given location.

Reading: The Physics Classroom – Static Electricity Unit, Lesson 4, Part a

 A charge Q creates an electric field. A test charge q is used to measure the strength of the electric field at a distance d from Q. The force F is experienced by the test charge q. The electric field strength at this location is given by the expression _. List all that apply,, Definition of Electric Field Strength : Any source of charge Q will create an electric field in the space surrounding it. The strength of an electric field ( E ) at any given location in this space can be determined by placing a test charge q in the space and measuring the force ( F ) which is exerted upon it. The electric field strength is defined as the amount of force per unit of charge on the test charge. E = F / q The electric field strength ( E ) is defined as the amount of force exerted upon a test charge per unit of charge on the test charge ( q ). That is, E = F / q, The electric force ( F ) depends upon a number of variables as described by Coulomb’s law. F elect = k • Q 1 • Q 2 / d 2 In the above equation, Q 1 might be the source charge Q and Q 2 might be the test charge q, If the expression for force as given by the Coulomb’s law equation is substituted in for F in the electric field strength equation, then the equation for electric field becomes E = k • Q / d 2 What variables effect the strength of a charge’s electric field? How can the force on a test charge be used to determine the strength of another charge’s electric field?

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A charge Q creates an electric field. A test charge q is used to measure the strength of the electric field at a distance d from Q. The electric field strength is defined as _. Definition of Electric Field Strength : Any source of charge Q will create an electric field in the space surrounding it. The strength of an electric field ( E ) at any given location in this space can be determined by placing a test charge q in the space and measuring the force ( F ) which is exerted upon it. The electric field strength is defined as the amount of force per unit of charge on the test charge. E = F / q It is easy to become confused by the mathematics of electric field strength. It is important to bear in mind that there are always two charges involved in any electrical interaction. In this case, the charges are Q and q, Big Q represents the source charge which creates the electric field. Little q represents the test charge which is used to measure the strength of the electric field at a given location surrounding the source charge. Give considerable attention to the charge quantity – Q or q – being used in each equation. How can the force on a test charge be used to determine the strength of another charge’s electric field?

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The standard metric unit on electric field strength is the _. The electric field strength ( E ) at any location surrounding a source charge can be determined by measuring the force ( F ) exerted upon some test charge ( q ) which is placed at that location. E = F / q The standard metric units of a quantity can be understood by thinking about its formula. Electric field strength is the ratio of force to charge (see Formula Fix section above). So the units on electric field strength are units of force divided by units of charge. The standard metric unit of force is the Newton; the standard metric unit of charge is the Coulomb. So the standard metric unit of electric field strength is Newton/Coulomb, abbreviated N/C. What units are used to express electric field strength?

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A 4 microCoulomb charge exerts a force of 9.0 x 10^-6 Newton on a 3.0 x 10^-9 Coulomb test charge. The electric field strength created by the 4 microCoulomb charge is _ N/C. Definition of Electric Field Strength : Any source of charge Q will create an electric field in the space surrounding it. The strength of an electric field ( E ) at any given location in this space can be determined by placing a test charge q in the space and measuring the force ( F ) which is exerted upon it. The electric field strength is defined as the amount of force per unit of charge on the test charge. E = F / q The electric field strength ( E ) is defined as the amount of force exerted upon a test charge per unit of charge on the test charge ( q ). That is, E = F / q, The electric force ( F ) depends upon a number of variables as described by Coulomb’s law. F elect = k • Q 1 • Q 2 / d 2 In the above equation, Q 1 might be the source charge Q and Q 2 might be the test charge q, If the expression for force as given by the Coulomb’s law equation is substituted in for F in the electric field strength equation, then the equation for electric field becomes E = k • Q / d 2 It is easy to become confused by the mathematics of electric field strength. It is important to bear in mind that there are always two charges involved in any electrical interaction. In this case, the charges are Q and q, Big Q represents the source charge which creates the electric field. Little q represents the test charge which is used to measure the strength of the electric field at a given location surrounding the source charge. Give considerable attention to the charge quantity – Q or q – being used in each equation. What variables effect the strength of a charge’s electric field? How can the force on a test charge be used to determine the strength of another charge’s electric field?

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TRUE or FALSE : The quantity electric field is a vector quantity. (Note: Your actual True-False statement is picked at random from a collection of choices and may vary from the one listed here.) Electric Field as a Vector: The electric field at a given location about a source charge ( Q ) is a vector quantity. That is, it has a direction. The force upon a test charge ( q ) could be an attractive force (towards the source charge) or a repulsive force (away from the source charge) depending upon whether the Q and q are like-charged or oppositely charged. In defining the direction of the electric field, a convention is used so that the direction is always in the same direction regardless of the type of charge on q, According to the convention, the direction of the electric field is in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding Q, Is electric field a scalar or a vector quantity?

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By convention, the direction of the electric field is _. Electric Field as a Vector: The electric field ( E ) at a given location about a source charge ( Q ) is a vector quantity. That is, it has a direction. The force upon a test charge ( q ) could be an attractive force (towards the source charge) or a repulsive force (away from the source charge) depending upon whether the Q and q are like-charged or oppositely charged. In defining the direction of the electric field, a convention is used so that the direction is dependent upon the type of charge on the source charge Q, According to the convention, the direction of the electric field is in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding Q, How can one determine the direction of an electric field?

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A positive charge creates an electric field. The direction of the electric field would be _. Electric Field as a Vector: The electric field ( E ) at a given location about a source charge ( Q ) is a vector quantity. That is, it has a direction. The force upon a test charge ( q ) could be an attractive force (towards the source charge) or a repulsive force (away from the source charge) depending upon whether the Q and q are like-charged or oppositely charged. In defining the direction of the electric field, a convention is used so that the direction is dependent upon the type of charge on the source charge Q, According to the convention, the direction of the electric field is in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding Q, By convention, the direction of the electric field vector at any given location is the direction which a positive test charge would be pushed or pulled if placed at that location. Combining this convention with the rule that like charged objects repel, one can determine the direction of the electric field in the space surrounding a positive source charge. A positive source charge and a positive test charge will repel each other. That is, the positive test charge will be pushed away from the positive source charge at all locations in the space surrounding the source charge. How can one determine the direction of an electric field?

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A negative charge creates an electric field. The direction of the electric field would be _. Electric Field as a Vector: The electric field ( E ) at a given location about a source charge ( Q ) is a vector quantity. That is, it has a direction. The force upon a test charge ( q ) could be an attractive force (towards the source charge) or a repulsive force (away from the source charge) depending upon whether the Q and q are like-charged or oppositely charged. In defining the direction of the electric field, a convention is used so that the direction is dependent upon the type of charge on the source charge Q, According to the convention, the direction of the electric field is in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding Q, By convention, the direction of the electric field vector at any given location is the direction which a positive test charge would be pushed or pulled if placed at that location. Combining this convention with the rule that oppositely charged objects attract, one can determine the direction of the electric field in the space surrounding a negative source charge. A negative source charge and a positive test charge will attract each other. That is, the positive test charge will be pulled towards the negative source charge at all locations in the space surrounding the source charge. How can one determine the direction of an electric field?

#### What is the charge enclosed by Gaussian surface?

Uniformly Charged Cylindrical Shell – A very long non-conducting cylindrical shell of radius R has a uniform surface charge density $$\sigma_0$$ Find the electric field (a) at a point outside the shell and (b) at a point inside the shell. Strategy Apply the Gauss’s law strategy given earlier, where we treat the cases inside and outside the shell separately. Solution a. Electric field at a point outside the shell. For a point outside the cylindrical shell, the Gaussian surface is the surface of a cylinder of radius $$r > R$$ and length L, as shown in Figure $$\PageIndex$$. The charge enclosed by the Gaussian cylinder is equal to the charge on the cylindrical shell of length L, Therefore, $$\lambda_$$ is given by \ Figure $$\PageIndex$$: A Gaussian surface surrounding a cylindrical shell. Hence, the electric field at a point P outside the shell at a distance r away from the axis is \ where $$\hat$$ is a unit vector, perpendicular to the axis and pointing away from it, as shown in the figure. The electric field at P points in the direction of $$\hat$$ given in Figure $$\PageIndex$$ if $$\sigma_0 > 0$$ and in the opposite direction to $$\hat$$ if $$\sigma_0 <0$$.b. Electric field at a point inside the shell. For a point inside the cylindrical shell, the Gaussian surface is a cylinder whose radius r is less than R (Figure $$\PageIndex$$). This means no charges are included inside the Gaussian surface: \ Figure $$\PageIndex$$: A Gaussian surface within a cylindrical shell. This gives the following equation for the magnitude of the electric field $$E_$$ at a point whose r is less than R of the shell of charges. \ This gives us \ Significance Notice that the result inside the shell is exactly what we should expect: No enclosed charge means zero electric field.

### What is q in magnetic field?

The magnetic field of naturally occurring magnetite is too weak to be used in devices such as modern motors and generators; these magnetic fields must come from electric currents. Magnetic fields affect moving charges, and moving charges produce magnetic fields; therefore, the concepts of magnetism and electricity are closely intertwined.

1. Magnetic fields and lines of force A bar magnet attracts iron objects to its ends, called poles,
2. One end is the north pole, and the other is the south pole,
3. If the bar is suspended so that it is free to move, the magnet will align itself so that its north pole points to the geographic north of the earth.

The suspended bar magnet acts like a compass in the earth’s magnetic field. If two bar magnets are brought close together, the like poles will repel each other, and the unlike poles attract each other. ( Note: By this definition, the magnetic pole under the earth’s north geographical pole is the south pole of the earth’s magnetic field.) This magnetic attraction or repulsion can be explained as the effect of one magnet on the other, or it can be said that one magnet sets up a magnetic field in the region around it that affects the other magnet.

1. The magnetic field at any point is a vector.
2. The direction of the magnetic field ( B ) at a specified point is the direction that the north end of a compass needle points at that position.
3. Magnetic field lines, analogous to electric field lines, describe the force on magnetic particles placed within the field.

Iron filings will align to indicate the patterns of magnetic field lines. Force on a moving charge If a charge moves through a magnetic field at an angle, it will experience a force. The equation is given by F = q v × B or F = qvB sin θ, where q is the charge, B is the magnetic field, v is the velocity, and θ is the angle between the directions of the magnetic field and the velocity; thus, using the definition of the cross product, the definition for the magnetic field is Magnetic field is expressed in SI units as a tesla (T), which is also called a weber per square meter: The direction of F is found from the right‐hand rule, shown in Figure 1.

 Figure 1 Using the right-hand rule to find the direction of magnetic force on a moving charge.

/td>

To find the direction of the force on the charge, with a flat hand point your thumb in the direction of the velocity of the positive charge and your fingers in the direction of the magnetic field. The direction of the force is out of the palm of your hand.

• If the moving charge is negative, point your thumb opposite to its direction of motion.) Mathematically, this force is the cross product of the velocity vector and the magnetic field vector.
• If the velocity of the charged particle is perpendicular to the uniform magnetic field, the force will always be directed toward the center of a circle of radius r, as shown in Figure 2.

The x symbolizes a magnetic field into the plane of the paper—the tail of the arrow. (A dot symbolizes a vector out of the plane of the paper—the tip of the arrow.)

 Figure 2 The force on a charge moving perpendicular to a magnetic field is toward the center of a circle.

/td>

The magnetic force provides centripetal acceleration: or The radius of the path is proportional to the mass of the charge. This equation underlies the operation of a mass spectrometer, which can separate equally ionized atoms of slightly different masses. The singly ionized atoms are given equal velocities, and because their charges are the same and they travel through the same B, they will travel in slightly different paths and can then be separated.

• Force on a current-carrying conductor Charges confined to wires can also experience a force in a magnetic field.
• A current (I) in a magnetic field ( B ) experiences a force ( F ) given by the equation F = I l × B or F = IlB sin θ, where l is the length of the wire, represented by a vector pointing in the direction of the current.

The direction of the force may be found by a right‐hand rule similar to the one shown in Figure, In this case, point your thumb in the direction of the current—the direction of motion of positive charges. The current will experience no force if it is parallel to the magnetic field.

• Torque on a current loop A loop of current in a magnetic field can experience a torque if it is free to turn.
• Figure (a) depicts a square loop of wire in a magnetic field directed to the right.
• Imagine in Figure (b) that the axis of the wire is turned to an angle (θ) with the magnetic field and that the view is looking down on the top of the loop.

The x in a circle depicts the current traveling into the page away from the viewer, and the dot in a circle depicts the current out of the page toward the viewer. Figure 3 (a) Square current loop in a magnetic field B, (b) View from the top of the current loop. (c) If the loop is tilted with respect to B, a torque results. The right‐hand rule gives the direction of the forces. If the loop is pivoted, these forces produce a torque, turning the loop.

The magnitude of this torque is t = N I A × B, where N is the number of turns of the loop, B is the magnetic field, I is the current, and A is the area of the loop, represented by a vector perpendicular to the loop. Galvanometers, ammeters, and voltmeters The torque on a current loop in a magnetic field provides the basic principle of the galvanometer, a sensitive current‐measuring device.

A needle is affixed to a current coil—a set of loops. The torque gives a certain deflection of the needle, which is dependent upon the current, and the needle moves over a scale to allow a reading in amperes. An ammeter is a current‐measuring instrument constructed from a galvanometer movement in parallel with a resistor.

Ammeters are manufactured to measure different ranges of current. A voltmeter is constructed from a galvanometer movement in series with a resistor. The voltmeter samples a small portion of the current, and the scale provides a reading of potential difference—volts—between two points in the circuit. Magnetic field of a long, straight wire A current‐carrying wire generates a magnetic field of magnitude B in circles around the wire.

The equation for the magnetic field at a distance r from the wire is where I is the current in the wire and μ (the Greek letter mu) is the proportionality constant. The constant, called the permeability constant, has the value The direction of the field is given by a second right‐hand rule, shown in Figure 4.

 Figure 4 Using the second right-hand rule to determine the direction of the magnetic field resulting from a current.

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Grasp the wire so that your thumb points in the direction of the current. Your fingers will curl around the wire in the direction of the magnetic field. Ampere’s law Ampere’s law allows the calculation of magnetic fields. Consider the circular path around the current shown in Figure, Or in integral form, Somewhat analogous to the way Gauss’s law can be used to find the electric field for highly symmetric charge configurations, Ampere’s law can be used to find the magnetic fields for current configurations of high symmetry. For example, Ampere’s law can be used to derive the expression for the magnetic field generated by a long, straight wire: Magnetic fields of the loop, solenoid, and toroid A current generates a magnetic field, and the field differs as the current is shaped into (a) a loop, (b) a solenoid (a long coil of wire), or (c) a toroid (a donut‐shaped coil of wire). The equations for the magnitudes of these fields follow.

 Figure 5 Magnetic field resulting from (a) a current loop, (b) a solenoid, and (c) a toroid.

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a. The field at the center of a single loop is given by where r is the radius of the loop.b. The field due to a solenoid is given by B = μ 0 NI, where N is the number of turns per unit length.c. The field due to a toroid is given by where R is the radius to the center of the toroid.

## What does q mean in current?

Q=It – The definition of current is the electric charge transferred per unit time.

## What is the formula of q in physics?

What is an Electric Charge? – Protons, electrons, and neutrons are examples of the basic subatomic particles, in which protons possess positive electrical charge, electrons possess negative electric charge and neutrons have neutral charge. The symbol of charge is ‘q’ or ‘Q’.

The total charge of electrons present in an atom is the number of electrons multiplied by the charge of an electron. According to this definition, the formula for charge can be written as, Q = ne, Where Q is a charge, e is charge on one electron, and n is the number of electrons. It is possible to measure the charge of a body by comparing it to a standard value.

According to a study, the charge of electrons is 1.6 x 10-19C. The S.I. unit or Standard unit of electric charge is Coulomb and it is denoted by ‘C’.

#### What does q stand for in e QV?

E = energy transferred in joules (J) Q = quantity of electric charge in coulombs (C) V = potential difference (V)

### What is Q in e Fe Q?

Assignment SE10: Electric Field – Objectives:

The student should be able to define electric field and recognize the variables which effect (and do not effect) the electric field intensity at a given location. The student should be able to use electric field equations and relatively simple numbers to determine the magnitude and the direction of the electric field intensity at a given location.

Reading: The Physics Classroom – Static Electricity Unit, Lesson 4, Part a

 A charge Q creates an electric field. A test charge q is used to measure the strength of the electric field at a distance d from Q. The force F is experienced by the test charge q. The electric field strength at this location is given by the expression _. List all that apply,, Definition of Electric Field Strength : Any source of charge Q will create an electric field in the space surrounding it. The strength of an electric field ( E ) at any given location in this space can be determined by placing a test charge q in the space and measuring the force ( F ) which is exerted upon it. The electric field strength is defined as the amount of force per unit of charge on the test charge. E = F / q The electric field strength ( E ) is defined as the amount of force exerted upon a test charge per unit of charge on the test charge ( q ). That is, E = F / q, The electric force ( F ) depends upon a number of variables as described by Coulomb’s law. F elect = k • Q 1 • Q 2 / d 2 In the above equation, Q 1 might be the source charge Q and Q 2 might be the test charge q, If the expression for force as given by the Coulomb’s law equation is substituted in for F in the electric field strength equation, then the equation for electric field becomes E = k • Q / d 2 What variables effect the strength of a charge’s electric field? How can the force on a test charge be used to determine the strength of another charge’s electric field?

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A charge Q creates an electric field. A test charge q is used to measure the strength of the electric field at a distance d from Q. The electric field strength is defined as _. Definition of Electric Field Strength : Any source of charge Q will create an electric field in the space surrounding it. The strength of an electric field ( E ) at any given location in this space can be determined by placing a test charge q in the space and measuring the force ( F ) which is exerted upon it. The electric field strength is defined as the amount of force per unit of charge on the test charge. E = F / q It is easy to become confused by the mathematics of electric field strength. It is important to bear in mind that there are always two charges involved in any electrical interaction. In this case, the charges are Q and q, Big Q represents the source charge which creates the electric field. Little q represents the test charge which is used to measure the strength of the electric field at a given location surrounding the source charge. Give considerable attention to the charge quantity – Q or q – being used in each equation. How can the force on a test charge be used to determine the strength of another charge’s electric field?

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The standard metric unit on electric field strength is the _. The electric field strength ( E ) at any location surrounding a source charge can be determined by measuring the force ( F ) exerted upon some test charge ( q ) which is placed at that location. E = F / q The standard metric units of a quantity can be understood by thinking about its formula. Electric field strength is the ratio of force to charge (see Formula Fix section above). So the units on electric field strength are units of force divided by units of charge. The standard metric unit of force is the Newton; the standard metric unit of charge is the Coulomb. So the standard metric unit of electric field strength is Newton/Coulomb, abbreviated N/C. What units are used to express electric field strength?

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A 4 microCoulomb charge exerts a force of 9.0 x 10^-6 Newton on a 3.0 x 10^-9 Coulomb test charge. The electric field strength created by the 4 microCoulomb charge is _ N/C. Definition of Electric Field Strength : Any source of charge Q will create an electric field in the space surrounding it. The strength of an electric field ( E ) at any given location in this space can be determined by placing a test charge q in the space and measuring the force ( F ) which is exerted upon it. The electric field strength is defined as the amount of force per unit of charge on the test charge. E = F / q The electric field strength ( E ) is defined as the amount of force exerted upon a test charge per unit of charge on the test charge ( q ). That is, E = F / q, The electric force ( F ) depends upon a number of variables as described by Coulomb’s law. F elect = k • Q 1 • Q 2 / d 2 In the above equation, Q 1 might be the source charge Q and Q 2 might be the test charge q, If the expression for force as given by the Coulomb’s law equation is substituted in for F in the electric field strength equation, then the equation for electric field becomes E = k • Q / d 2 It is easy to become confused by the mathematics of electric field strength. It is important to bear in mind that there are always two charges involved in any electrical interaction. In this case, the charges are Q and q, Big Q represents the source charge which creates the electric field. Little q represents the test charge which is used to measure the strength of the electric field at a given location surrounding the source charge. Give considerable attention to the charge quantity – Q or q – being used in each equation. What variables effect the strength of a charge’s electric field? How can the force on a test charge be used to determine the strength of another charge’s electric field?

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TRUE or FALSE : The quantity electric field is a vector quantity. (Note: Your actual True-False statement is picked at random from a collection of choices and may vary from the one listed here.) Electric Field as a Vector: The electric field at a given location about a source charge ( Q ) is a vector quantity. That is, it has a direction. The force upon a test charge ( q ) could be an attractive force (towards the source charge) or a repulsive force (away from the source charge) depending upon whether the Q and q are like-charged or oppositely charged. In defining the direction of the electric field, a convention is used so that the direction is always in the same direction regardless of the type of charge on q, According to the convention, the direction of the electric field is in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding Q, Is electric field a scalar or a vector quantity?

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By convention, the direction of the electric field is _. Electric Field as a Vector: The electric field ( E ) at a given location about a source charge ( Q ) is a vector quantity. That is, it has a direction. The force upon a test charge ( q ) could be an attractive force (towards the source charge) or a repulsive force (away from the source charge) depending upon whether the Q and q are like-charged or oppositely charged. In defining the direction of the electric field, a convention is used so that the direction is dependent upon the type of charge on the source charge Q, According to the convention, the direction of the electric field is in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding Q, How can one determine the direction of an electric field?

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A positive charge creates an electric field. The direction of the electric field would be _. Electric Field as a Vector: The electric field ( E ) at a given location about a source charge ( Q ) is a vector quantity. That is, it has a direction. The force upon a test charge ( q ) could be an attractive force (towards the source charge) or a repulsive force (away from the source charge) depending upon whether the Q and q are like-charged or oppositely charged. In defining the direction of the electric field, a convention is used so that the direction is dependent upon the type of charge on the source charge Q, According to the convention, the direction of the electric field is in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding Q, By convention, the direction of the electric field vector at any given location is the direction which a positive test charge would be pushed or pulled if placed at that location. Combining this convention with the rule that like charged objects repel, one can determine the direction of the electric field in the space surrounding a positive source charge. A positive source charge and a positive test charge will repel each other. That is, the positive test charge will be pushed away from the positive source charge at all locations in the space surrounding the source charge. How can one determine the direction of an electric field?

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A negative charge creates an electric field. The direction of the electric field would be _. Electric Field as a Vector: The electric field ( E ) at a given location about a source charge ( Q ) is a vector quantity. That is, it has a direction. The force upon a test charge ( q ) could be an attractive force (towards the source charge) or a repulsive force (away from the source charge) depending upon whether the Q and q are like-charged or oppositely charged. In defining the direction of the electric field, a convention is used so that the direction is dependent upon the type of charge on the source charge Q, According to the convention, the direction of the electric field is in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding Q, By convention, the direction of the electric field vector at any given location is the direction which a positive test charge would be pushed or pulled if placed at that location. Combining this convention with the rule that oppositely charged objects attract, one can determine the direction of the electric field in the space surrounding a negative source charge. A negative source charge and a positive test charge will attract each other. That is, the positive test charge will be pulled towards the negative source charge at all locations in the space surrounding the source charge. How can one determine the direction of an electric field?

#### What is Q electric potential?

E = W/Q. Where, E = electrical potential difference between two points. W = Work done in moving a charge from one point to another. Q = Quantity of charge in coulombs.

#### What does Q stand for in Q CV?

Dynamics Track Inclined Plane Momentum Capacitor Plate Sep Plate Sep/Volt Dielectrics Circuits Ohms Law Series/Parallel Wave Tank Freq/Wavelength Two Pt Interf. Optical Bench Refraction Focal Length Parallel Plate Capacitor Capacitor Charge, Plate Separation, and Voltage A capacitor is used to store electric charge. The more voltage (electrical pressure) you apply to the capacitor, the more charge is forced into the capacitor. Also, the more capacitance the capacitor possesses, the more charge will be forced in by a given voltage. This relation is described by the formula q=CV, where q is the charge stored, C is the capacitance, and V is the voltage applied. Looking at this formula, one might ask what would happen if charge were kept constant and the capacitance were varied. The answer is, of course, that the voltage will change! That is what you will do in this lab. The Lab Capacitor A parallel plate capacitor is a device used to study capacitors. It reduces to barest form the function of a capacitor. Real-world capacitors are usually wrapped up in spirals in small packages, so the parallel-plate capacitor makes it much easier to relate the function to the device. This capacitor works by building up opposite charges on parallel plates when a voltage is applied from one plate to the other. The amount of charge that moves into the plates depends upon the capacitance and the applied voltage according to the formula Q=CV, where Q is the charge in Coulombs, C is the capacitance in Farads, and V is the potential difference between the plates in volts.

Capacitors store energy If a voltage is applied to a capacitor and then disconnected, the charge that is stored in the capacitor remains until the capacitor is discharged in some way. An electric field then exists between the plates, which allows the capacitor to store energy. This is one of the useful aspects of capacitors, the ability to store energy in an electric field so that can be utilized later on.

What determines the capacitance? The amount of charge which may be stored per volt applied is determined by the surface area of the plates and the spacing between them. The larger the plates and the more closely they are spaced, the more charge can be stored for every volt of potential difference between the plates.

The charge stored per volt applied is the capacitance, measured in Farads. Can altering the capacitance of a charged capacitor change its voltage? The lab capacitor is adjustable, so we can do an interesting experiment involving capacitance and voltage. If the capacitor has a constant charge, changing the capacitance should cause the voltage to vary.

Moving the plates apart will reduce the capacitance, so the voltage should increase. How can capacitance of our capacitor be mathematically determined? For a parallel plate capacitor, the capacitance is given by the following formula: C = ε 0A/d Where C is the capacitance in Farads, ε 0 is the constant for the permittivity of free space (8.85×10 -12), A is the area of the plates in square meters, and d is the spacing of the plates in meters.

A Farad is a very large quantities of capacitance, so we will use metric prefixes to produce more usable numbers. Capacitance is normally measured in microfarads (µF) which is 1.0×10 -6F or picofarads (pF) which is 1.0×10 -12F.1.0F = 1,000,000µF = 1,000,000,000,000pF! Be very careful with your calculations! This calculation will give you an approximation of the capacitance of the lab capacitor.

However, there are other factors that introduce errors into the real-world measurement of capacitance and voltage. You need to be careful to take these factors into account. Lab Equipment: To get good results, this lab activity requires some specialized equipment.

You need a good regulated power supply so that the voltage applied to the capacitor is the same in each trial. You also need a very accurate way of measuring the voltage between the plates without putting a resistive load on the capacitor. The amount of charge stored is very small, so a conventional voltmeter will not work.

The minute charge built up in the capacitor would simply discharge through the meter, rendering any measurement useless. You will use a special voltage-measuring device called an Electrometer that measures voltage without discharging the capacitor. One problem with the electrometer is that it has some capacitance of their own.

Since this capacitance is in parallel with that of the capacitor, the built-in capacitance of the leads must be added to that of the capacitor. Purpose: The purpose of this lab is to investigate the relationship between plate separation and voltage in a parallel plate capacitor kept at constant charge.

Equipment:

• Variable capacitor
• Electrometer
• Regulated Power Supply

Cautions: This equipment is delicate. Everything should go together with the lightest of touches. Do not force anything! Your first task is to predict what will happen to the capacitor’s voltage when you charge it with a 10V source and then spread the plates apart (which reduces the capacitance) You will do this in the next section.

1. Measure the diameter of the capacitor plates in centimeters. Your measurement should be near 17.8cm
2. Divide the diameter by 100 to put the measurement in meters. The result is 0.178m. Divide this by two to get the radius: 0.089m
3. The area of the plate is determined by the common formula A=πr 2. Plug in the numbers to get A = π(0.089) 2 = 0.0249m 2
4. Convert the plate spacing (1mm) to meters by dividing by 1000.1/1000 =,001m.
5. Use these number in the formula C = ε 0A/d to determine the calculated capacitance thus: C = 8.85×10 -12(.0249)/.001 = 2.20×10 -10. This is equal to 220×10 -12F or 220pF
6. Add the built-in capacitance of the electrometer (50pF) to the theoretical capacitance to get 270pF.
7. Write this result (270 pF ) in the “Calculated Capacitance” column and the 1mm row.
8. Repeat this process for the other plate spacings. Note that the plate area is the same for all, so all you need to do is repeat steps 5, 6, and 7, inserting the correct values for the spacing in each case.
9. Now you will calculate the theoretical voltage for each spacing. We will assume a voltage of 10V for the 1.0mm spacing, so you can just put that value into the table directly. First, you determine the amount of charge in the capacitor at this spacing and voltage. Use the formula Q=CV to determine the charge thus: Q=270×10 -12F(10V)=2700×10 -12C. This charge stays the same at all plate spacings, so you can fill the same value into the entire Calculated Charge column! Now use this charge value to determine the calculated voltage at all other spacings. For example, at the 5mm spacing, use the formula V=Q/C thus: V=2700×10 -12C/94.0×10 -12F=28.7V. Enter this value in the Calculated Voltage column at the 5mm row.
10. Repeat the same voltage calculation for the remaining plate spacings. Use the calculated capacitance and the constant charge for each spacing and enter the voltage value in the Calculated Voltage column of the table.
11. Congratulations! You have finished the preliminary calculations! All you have to do now is make the actual measurements!

In the next sections, you will perform the actual experiment to verify (or perhaps not verify!) your theoretical calculations. Procedure to set up the variable capacitor (if the lab is already set up, proceed to the next section!)

1. Place the variable capacitor in the middle of the lab table, with the 0cm mark to your left. Don’t put the capacitor too close to the edge of the table!
2. Place the power supply behind the variable capacitor. Plug the power supply in, but do not turn it on.
3. Connect the red and black jumper leads to the red and black terminals of the power supply. Simply clip the alligator clip to the hole and leave the other end of the leads free.
4. Place the electrometer to the left of the capacitor.
5. Attach the spade terminals of the electrometer leads to the binding posts on the back side of each capacitor plate. The red lead goes to the right plate, the black lead goes to the left plate.
6. Plug the BNC connector into the electrometer.
7. Put the plates at the minimum1 mm separation. The white bumpers prevent the plates from being set closer together. If the plates are not parallel to each other, use the adjustment knobs in the middle of the right support to align the plates. The left edge of the plastic tab that extends toward the scale should be aligned with the 1mm mark.

Collecting Experimental Data

1. Make sure that the equipment setup is complete and correct.
2. Turn all four controls on the power supply fully counterclockwise.
3. Turn the leftmost knob (Fine Current) to the 12 o’clock position (straight up!)
4. Turn the power supply on. The displays should light.
5. Use the Fine and Coarse Voltage knobs (the two rightmost knobs) to set the voltage to 10.0V.
6. Put the plates at their minimum setting
7. Set the Electrometer to the 30V scale.
8. Push the power button on the Electrometer. The 30V LED should light.
9. Push the zero button on the electrometer. This zeros the meter and makes sure that the plates are at zero volts relative to each other.
10. Momentarily touch the leads from the power supply to the plates, black to the left plate and red to the right plate.
11. The electrometer should read 12 volts at this point (12V is the first small mark above “1” on the lower scale. If it doesn’t check your setup and try again. Sometimes you have to touch the leads to the plates several times to get the proper 12V reading.
12. From this point on, you must be careful not to touch the plates. If you touch them you will alter the charge in the plates and spoil the data!
13. Watch the electrometer to make sure that the charge is being held. If you see a drop of more than a volt in 30 seconds stop and figure out what is wrong before proceeding.
14. Switch the electrometer to the 100V setting. The meter should still read 12V, but on the 100V scale.
15. Carefully slide the plates apart to the 5mm setting.
16. Take a reading from the electrometer and record it in the table under that Measured Voltage column.
17. Repeat the previous two steps for the other plate spacings and record the appropriate data.
 Plate separation ( mm ) Calculated Capacitance ( pF ) Calculated Charge (pC) Calculated Voltage (V) Measured Voltage (V) 1 5 10 15 20 25 30 35 40

Data Analysis:

1. On graph paper, plot the Calculated Capacitance on the x (horizontal) axis versus the Voltage on the y (vertical) axis. Plot both the calculated value and the measured value for voltage, using either different colors or line styles to distinguish the two curves. Make sure that you choose appropriate scales and label the axes and scales clearly. It is best to orient the paper with its long axis in the horizontal direction (“landscape mode”).