How Should The Integral In Gauss’S Law Be Evaluated?

How Should The Integral In Gauss
How should the integral in Gauss’s law be evaluated? Over a Closed surface In the integral for Gauss’s law, the vector dA⃗ represents an infinitesimal surface element. The magnitude of dA⃗ is the area of the surface element.

Where does Gauss law can be evaluated?

Electromagnetic Theory Questions and Answers – Applications of Gauss Law This set of Electromagnetic Theory Multiple Choice Questions & Answers (MCQs) focuses on “Applications of Gauss Law”.1. Gauss law can be used to compute which of the following? a) Permittivity b) Permeability c) Radius of Gaussian surface d) Electric potential View Answer Answer: c Explanation: Gauss law relates the electric flux density and the charge density.

Thus it can be used to compute radius of the Gaussian surface. Permittivity and permeability are constants for a particular material.2. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 1m. a) 0 b) 1 c) 2 d) 3 View Answer Answer: a Explanation: Since 1m does not enclose any cylinder (three Gaussian surfaces of radius 2m, 4m, 5m exists), the charge density and charge becomes zero according to Gauss law.

Thus flux density is also zero.3. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 3m. a) 3 b) 10/3 c) 11/3 d) 4 View Answer Answer: b Explanation: The radius is 3m, hence it will enclose one Gaussian cylinder of R = 2m.

  1. By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 3) = σ(2π X 2), Thus D = 10/3 units.4.
  2. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ =-3 at R = 5m.
  3. Find the flux density at R = 4.5m.
  4. A) 4/4.5 b) 3/4.5 c) 2/4.5 d) 1/4.5 View Answer Answer: c Explanation: The Gaussian cylinder of R = 4.5m encloses sum of charges of two cylinders (R = 2m and R = 4m).

By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 4.5) = Q1 + Q2 = σ1(2π X 2) + σ2(2π X 4), here σ1 = 5 and σ2 = -2. We get D = 2/4.5 units.5. Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m.

  • Find the flux density at R = 6m.
  • A) 17/6 b) -17/6 c) 13/6 d) -13/6 View Answer Answer: d Explanation: The radius R = 6m encloses all the three Gaussian cylinders.
  • By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 = σ1(2π X 2) + σ2(2π X 4) + σ3(2π X 5), here σ1 = 5, σ2 = -2 and σ3 = -3.
  • We get D = -13/6 units.

Check this: | 6. Gauss law can be evaluated in which coordinate system? a) Cartesian b) Cylinder c) Spherical d) Depends on the Gaussian surface View Answer Answer: d Explanation: The Gauss law exists for all materials. Depending on the Gaussian surface of the material, we take the coordinate systems accordingly.

  1. Suppose if the material is a coaxial cable, the Gaussian surface is in the form of cylinder.
  2. Thus we take Cylinder/Circular coordinate system.7.
  3. Gauss law cannot be expressed in which of the following forms? a) Differential b) Integral c) Point d) Stokes theorem View Answer Answer: d Explanation: Gauss law can be expressed in differential or point form as, Div (D)= ρv and in integral form as ∫∫ D.ds = Q = ψ,

It is not possible to express it using Stoke’s theorem.8. The tangential component of electric field intensity is always continuous at the interface. State True/False. a) True b) False View Answer Answer: a Explanation: Consider a dielectric-dielectric boundary, the electric field intensity in both the surfaces will be Et1 = Et2, which implies that the tangential component of electric field intensity is always continuous at the boundary.9.

The normal component of the electric flux density is always discontinuous at the interface. State True/False. a) True b) False View Answer Answer: a Explanation: In a dielectric-dielectric boundary, if a free surface charge density exists at the interface, then the normal components of the electric flux density are discontinuous at the boundary, which means Dn1 = Dn2.10.

With Gauss law as reference which of the following law can be derived? a) Ampere law b) Faraday’s law c) Coulomb’s law d) Ohm’s law View Answer Answer: c Explanation: From Gauss law, we can compute the electric flux density. This in turn can be used to find electric field intensity., a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry, He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at, Subscribe to his free Masterclasses at & technical discussions at, : Electromagnetic Theory Questions and Answers – Applications of Gauss Law

What is a surface integral Gauss law?

Summary. A surface integral is used to calculate the quantity of a vector field that is penetrating through. a given surface. It is calculated by integrating the dot product of the vector field with a vector that is perpendicular to the surface at every point.

What is the main criteria to apply Gauss’s law?

Symmetry Avoids Integrals – The great irony of Gauss’s law is that the surface integral looks incredibly daunting, but this law is only really useful because no integration actually needs to be performed. As we will see, we will be able to use this law to compute electric fields of distributions of charge in cases where some degree of symmetry is present.

The basic approach is this: Construct an imaginary closed surface (called a gaussian surface ) around some collection of charge, then apply Gauss’s law for that surface to determine the electric field at that surface. This is a rather vague description, and glosses over a lot of important details, which we will learn through several examples.

There are two ingredients to the symmetry that need to be present to make using Gauss’s law so powerful:

A gaussian surface must exist where the electric field is either parallel or perpendicular to the surface vector. This makes the cosines in all the dot products equal to simply zero or one. The electric field that passes through the parts of the gaussian surface where the flux is non-zero has a constant magnitude.

These two conditions allow us to avoid an integral entirely, because the \(cos\theta\) in the integral goes away, and the electric field magnitude can be taken out of the integral, leaving only an integral of \(dA\), which is just the area of the surface. Then applying Gauss’s law is simple.

How do you use Gauss’s approach?

Carl Friedrich Gauss (1777-1855) is recognised as being one of the greatest mathematicians of all time. During his lifetime he made significant contributions to almost every area of mathematics, as well as physics, astronomy and statistics. Like many of the great mathematicians, Gauss showed amazing mathematical skill from an early age, and there are many stories which show how clever he could be.

  • The most well-known story is a tale from when Gauss was still at primary school.
  • One day Gauss’ teacher asked his class to add together all the numbers from $1$ to $100$, assuming that this task would occupy them for quite a while.
  • He was shocked when young Gauss, after a few seconds thought, wrote down the answer $5050$.

The teacher couldn’t understand how his pupil had calculated the sum so quickly in his head, but the eight year old Gauss pointed out that the problem was actually quite simple. He had added the numbers in pairs – the first and the last, the second and the second to last and so on, observing that $1+100=101$, $2+99=101$, $3+98=101$,,so the total would be $50$ lots of $101$, which is $5050$.

It is remarkable that a child still in elementary school had discovered this method for summing sequences of numbers, but of course Gauss was a remarkable child. Fortunately his talents were discovered, and he was given the chance to study at university. By his early twenties, Gauss had made discoveries that would shape the future of mathematics.

While the story may not be entirely true, it is a popular tale for maths teachers to tell because it shows that Gauss had a natural insight into mathematics. Rather than performing a great feat of mental arithmetic, Gauss had seen the structure of the problem and used it to find a short cut to a solution.

Gauss could have used his method to add all the numbers from $1$ to any number – by pairing off the first number with the last, the second number with the second to last, and so on, he only had to multiply this total by half the last number, just one swift calculation. Can you see how Gauss’s method works? Try using it to work out the total of all the numbers from $1$ to $10$.

What about $1$ to $50$? The answers are at the bottom of this page. Or why not challenge a friend to add up the numbers from $1$ to a nice large number, and then amaze them by getting the answer in seconds! The rest of the article explains how you could use algebra to write Gauss’s method – if you haven’t yet learned any algebra you may wish to skip this part. If we add both rows we get the sum of $1$ to $n$, but twice. Gauss added the rows pairwise – each pair adds up to $n+1$ and there are $n$ pairs, so the sum of the rows is also $n\times (n+1)$. It follows that $2\times (1+2+\ldots +n) = n\times (n+1)$, from which we obtain the formula. Answers: total from 1 to 10 = 55, total from 1 to 50 = 1275

For which type of field Gauss law is valid?

Gauss’s law is defined and valid only for a closed surface.

What does Gauss’s law measure?

Integral Equation  – Gauss’s law in integral form is given below: (34)  \ where:

\(\mathbf \) is the electric field \(Q\) is the enclosed electric charge \(\varepsilon_0\) is the electric permittivity of free space \(\hat }\) is the outward pointing unit-normal

Flux is a measure of the strength of a field passing through a surface. Electric flux is defined in general as (35)  \ We can think of electric field as flux density. Gauss’s law tells us that the net electric flux through any closed surface is zero unless the volume bounded by that surface contains a net charge.

How do you know which Gaussian surface to use?

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  • Gauss’ Law is a relation between the net flux through a closed surface and the amount of charge, \(Q^ \), in the volume enclosed by that surface: \ In particular, note that Gauss’ Law holds true for any closed surface, and the shape of that surface is not specified in Gauss’ Law. That is, we can always choose the surface to use when calculating the flux. For obvious reasons, we often call the surface that we choose a “gaussian surface”. But again, this surface is simply a mathematical tool, there is no actual property that makes a surface “gaussian”; it simply means that we chose that surface in order to apply Gauss’ Law. In Example 17.1.3 above, we confirmed that Gauss’ Law is compatible with Coulomb’s Law for the case of a point charge and a spherical gaussian surface. Physically, Gauss’ Law is a statement that field lines must begin or end on a charge (electric field lines originate on positive charges and terminate on negative charges). Recall, flux is a measure of the net number of lines coming out of a surface. If there is a net number of lines coming out of a closed surface (a positive flux), that surface must enclose a positive charge from where those field lines originate. Similarly, if there are the same number of field lines entering a closed surface as there are lines exiting that surface (a flux of zero), then the surface encloses no charge. Gauss’ Law simply states that the number of field lines exiting a closed surface is proportional to the amount of charge enclosed by that surface. Primarily, Gauss’ Law is a useful tool to determine the magnitude of the electric field from a given charge, or charge distribution. We usually have to use symmetry to determine the direction of the electric field vector. In general, the integral for the flux is difficult to evaluate, and Gauss’ Law can only be used analytically in cases with a high degree of symmetry. Specifically, the integral for the flux is easiest to evaluate if:

    1. The electric field makes a constant angle with the surface, When this is the case, the scalar product can be written in terms of the cosine of the angle between \(\vec E\) and \(d\vec A\), which can be taken out of the integral if it is constant: \ Ideally, one has chosen a surface such that this angle is \(0\) or \(180^ \),
    2. The electric field is constant in magnitude along the surface, When this is the case, the integral can be simplified further by factoring out, \(E\), and simply becomes an integral over \(dA\) (which corresponds to the total area of the surface, \(A\) ): \
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    Ultimately, the points above should dictate the choice of gaussian surface so that the integral for the flux is easy to evaluate. The choice of surface will depend on the symmetry of the problem. For a point (or spherical) charge, a spherical gaussian surface allows the flux to easily be calculated ( Example 17.1.3 ).

    1. Make a diagram showing the charge distribution.
    2. Use symmetry arguments to determine in which way the electric field vector points.
    3. Choose a gaussian surface that goes through the point for which you want to know the electric field. Ideally, the surface is such that the electric field is constant in magnitude and always makes the same angle with the surface, so that the flux integral is straightforward to evaluate.
    4. Calculate the flux, \(\oint \vec E\cdot d\vec A\),
    5. Calculate the amount of charge located within the volume enclosed by the surface, \(Q^ \),
    6. Apply Gauss’ Law, \(\oint \vec E\cdot d\vec A=\frac } \),

    Example \(\PageIndex \) An insulating sphere of radius, \(R\), contains a total charge, \(Q\), which is uniformly distributed throughout its volume. Determine an expression for the electric field as a function of distance, \(r\), from the center of the sphere.

    • Solution : Note that this is identical, mathematically, to the derivation that was done in Section 9.2 for the case of gravity.
    • When applying Gauss’ Law, we first need to think about symmetry in order to determine the direction of the electric field vector.
    • We also need to think about all possible regions of space in which we need to determine the electric field.

    In particular, for this case, we need to determine the electric field both inside ( \(r\leq R\) ) and outside ( \(r\geq R\) ) of the charged sphere. Figure \(\PageIndex \) shows the charged sphere of radius \(R\), If we consider the direction of the electric field outside the sphere (where \(\vec E_ \) is drawn), we realize that it can only point in the radial direction (towards or away from the center of the sphere), as this is the only choice that preserves the symmetry of the sphere. Figure \(\PageIndex \): For a spherical charge distribution, the electric field inside and outside must point n the radical direction, by symmetry. We now need to choose a gaussian surface that will make the flux integral easy to evaluate. Ideally, we can find a surface over which the electric field makes the same angle with the surface and over which the electric field is constant in magnitude. Figure \(\PageIndex \): A spherical gaussian surface to determine the electric field outside of a sphere of radius, R, holding charge, \(+Q\). In order to apply Gauss’ Law, we need to calculate:

    • the net flux through the surface.
    • the charge in the volume enclosed by the surface.

    The net flux through the surface is found in the same way as in Example 17.1.3, and is given by: \ where our choice of spherical surface led to \(\vec E\cdot d\vec A=EdA\), since \(\vec E\) and \(d\vec A\) are always parallel. Furthermore, by symmetry, the electric field must be constant in magnitude along the whole surface, or the spherical symmetry would be broken.

    This allowed us to factor the \(E\) out of the integral, leaving us with, \(\oint dA\), which is simply the area of our gaussian spherical surface, \(4\pi r^2\), The gaussian surface with \(r\geq R\) encloses the whole charged sphere, so the charge enclosed is simply the charge of the sphere, \(Q^ =Q\),

    Applying Gauss’ Law allows us to determine the magnitude of the electric field: \ which is the same as the electric field a distance \(r\) from a point charge. Thus, from the outside, a spherical charge distribution leads to the same electric field as if the charge were concentrated at the center of the sphere. Figure \(\PageIndex \): A spherical gaussian surface to determine the electric field inside of a sphere of radius, \(R\), holding charge, \(+Q\). The flux integral is trivial again, since the electric field always makes the same angle with the gaussian surface, and the magnitude of the electric field is constant in magnitude along the surface: \ In this case, however, the charge in the volume enclosed by the gaussian surface is less than \(Q\), since the whole charge is not enclosed.

    1. We are told that the charge is distributed uniformly throughout the spherical volume of radius \(R\),
    2. We can thus define a volume charge density, \(\rho\), (charge per unit volume) for the sphere: \ The volume enclosed by the gaussian surface is \(\frac \pi r^3\), thus, the charge, \(Q^ \), contained in that volume is given by: \ Finally, we apply Gauss’ Law to find the magnitude of the electric field inside the sphere: \ Note that the electric field increases linearly with radius inside of the charged sphere, and then decreases with radius squared outside of the sphere.

    Also, note that at the center of the sphere, the electric field has a magnitude of zero, as expected from symmetry. Discussion: In this example, we showed how to use Gauss’ Law to determine the electric field inside and outside of a uniformly charged sphere.

    We recognized the spherical symmetry of the charge distribution and chose to use a spherical surface in order to apply Gauss’ Law. This, in turn, allowed the flux to be easily calculated. We found that outside the sphere, the electric field decreases in magnitude with radius squared, just as if the entire charge were concentrated at the center of the sphere.

    Inside the sphere, we found that the electric field is zero at the center, and increases linearly with radius. Exercise \(\PageIndex \) A thin charged spherical shell carries a uniformly distributed charge of \(+Q\), Figure \(\PageIndex \): A charged spherical shell with a cubic device inside of it. If we place a cube inside the shell, as shown in Figure \(\PageIndex \), what is the total flux out of the surface of the cube?

    1. \(\frac \text \),
    2. \(\frac \text \),
    3. \(\frac \text \).
    4. \(0\text \),

    Answer Example \(\PageIndex \) An infinitely long straight wire carries a uniform charge per unit length, \(\lambda\), What is the electric field at a distance, \(R\), from the wire? Solution : We start by making a diagram of the charge distribution, as in Figure \(\PageIndex \), so that we can use symmetry arguments to determine the direction of the electric field vector.

    1. in the radial direction (point to/from the center of the wire).
    2. such that electric field lines form concentric circles with the wire.
    3. co-linear with the wire.

    In all three possibilities above, you would not be able to infer that one particular direction in the plane perpendicular to the wire is preferred. All three possibilities preserver the rotational symmetry of the wire (the wire looks the same from all directions in the plane perpendicular to the wire). Figure \(\PageIndex \): An infinite line of charge carrying uniform charge per unit length, \(λ\). The left panel shows a side view and the right panel a view from above. The electric field must be in the radial direction or there would be a preferred direction.

    1. Next, we need to choose a gaussian surface in order to apply Gauss’ Law.
    2. A convenient choice is a cylinder (a “pill box”) of radius, \(R\), and length, \(L\), as shown in Figure \(\PageIndex \), as this goes through a point that is a distance, \(R\), from the wire (where we are asked for the electric field).

    At all points on the cylindrical surface, the electric field vector is either perpendicular or parallel to the surface. Figure \(\PageIndex \): A cylindrical gaussian surface is used to calculate the flux from an infinite line of charge. We can think of the cylindrical surface as being composed of three surfaces: 2 disks on either end (the lids of the pill box), and the curved surface that makes up the side of the cylinder.

    The flux through the entire cylindrical surface will be the sum of the fluxes through the two lids plus the flux through the side: \ where you should note that the closed integral ( \(\oint\) ) was separated into three normal integrals ( \(\int\) ) corresponding to the three “open” surfaces that make up the closed surface.

    Again, remember that the flux is proportional to the net number of field lines exiting/entering the closed surface, so it make sense to count those lines over the three open surfaces and add them together to get the total number for the closed surface.

    1. The flux through the lids is identically zero, since the electric field is perpendicular to \(d\vec A\) everywhere on the lids.
    2. The total flux is thus equal to the flux through the curved side surface, for which the electric field vector is always parallel to \(d\vec A\), and for which the electric field vector is constant in magnitude: \ where we recognized that the side surface can be unfolded into a rectangle of height, \(L\), and width, \(2\pi R\), corresponding to the circumference of the cylinder, so that the area of the side of the cylinder is given by \(A=2\pi R L\),

    Next, we determine the charge inside the volume enclosed by the surface. Since the cylinder encloses a length, \(L\), of wire, the enclosed charge is given by: \ where \(\lambda\) is the charge per unit length on the wire. Putting this altogether into Gauss’ Law gives us the electric field at a distance, \(R\), from the wire: \ Note that this is the same result that we obtained in Example 16.3.3 when we took the limit of the finite line of charge having infinite length.

    1. Discussion: In this example, we applied Gauss’ Law to determine the electric field at a distance from an infinitely long charged wire.
    2. We used symmetry to argue that the field should be radial and in the plane perpendicular to the wire, and recognized that a cylindrical gaussian surface would exploit the symmetry so that the flux can easily be calculated.

    We obtained the same result as we did from integrating Coulomb’s Law in Example 16.3.3, However, using Gauss’ Law was much less work than integrating Coulomb’s Law. Exercise \(\PageIndex \) Why is it difficult to apply Gauss’ Law to a finite wire?

    1. It is easy to apply Gauss’ Law to a finite wire.
    2. Because the flux of a finite wire is undefined.
    3. Because we do not know the charge density of a finite wire.
    4. Because the symmetry argument does not hold.

    Answer josh’s thoughts Gauss’ Law requires us to choose a “gaussian” surface, but which surface should we choose? Generally, it is useful to choose a surface such that the flux can easily be determined, ideally without having to actually do an integral.

    If symmetry can be exploited such that \(\vec E\) has a constant magnitude and direction relative to \(d\vec A\) at every location of the gaussian surface, then \(\int \vec E \cdot d\vec A\) will be equal to \(E A\), This is why gaussian surfaces are often of the same shape as the charged object they are enclosing.

    For example, if I need to enclose a cylindrical charge, it would be reasonable to enclose the charge with a cylindrical gaussian surface, as shown in Figure \(\PageIndex \) Figure \(\PageIndex \): A cylindrical gaussian surface to enclose a cylindrical charge. When dealing with point charges which have no shape and are thus spherically symmetric, it makes sense to choose a spherical gaussian surface, as shown in Figure \(\PageIndex \), since the electric field is in the radial direction for a point charge. Figure \(\PageIndex \): A spherical gaussian surface to enclose a point charge. Finally, there are some cases of less than ideal choices for the gaussian surfaces. While never wrong, they may require rather complicated integrals to determine the flux. These cases will still provide a correct answer if the situation is modeled correctly.

    • Suppose that I enclose a spherical charge with a cylindrical gaussian surface, as shown in Figure \(\PageIndex \).
    • The electric field will be stronger near the middle of the cylinder’s length than at the center of its endcaps, which means that \(\vec E\) is not constant in \(\int \vec E \cdot d \vec A\), so the integral cannot be simplified to \(EA\),
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    A better choice for a gaussian surface in this case would be a sphere, which exploits the symmetry of the charge distribution and provides results in a \(\vec E\) of constant magnitude everywhere along the surface. Figures 17.2.7 and 17.2.8 give other examples of when we cannot assume \(\Phi\) to be equal to \(EA\), Figure \(\PageIndex \): A cylindrical surface is not a good choice to enclose a spherical charge. Example \(\PageIndex \) Determine the electric field above an infinitely large plane of charge with uniform surface charge per unit area, \(\sigma\), Solution : Figure \(\PageIndex \) shows a portion of the infinite plane. Figure \(\PageIndex \): The electric field above an infinite plane with uniform charge per unit area, \(σ\), must be perpendicular to the plane. A cylindrical or box-shaped gaussian surface would both lead to the flux integral being easy to calculate, as illustrated in Figure \(\PageIndex \). Figure \(\PageIndex \): A cylindrical surface or a box are both good choices for a gaussian surface above a plane, since only the parts of the surface parallel to the plane will have net flux through them. Let us choose a box (right panel of Figure \(\PageIndex \)) of length, \(L\), with a square cross-section of side, \(a\),

    • We place the box such that the plane intersects the center of the box (although this is not required, since we already know that the electric field will not depend on distance from the plane).
    • The flux through the box is simply the flux through the two horizontal planes (of area \(a^2\) ): \ The box encloses a section of the plane with area \(a^2\), so that the net charge enclosed by the surface is: \ Applying Gauss’ Law allows us to determine the magnitude of the electric field: \ which is the same result that we found in Example 16.3.4,

    Discussion: In this example, we used Gauss’ Law to determine the electric field above an infinite plane. We found that we had a choice of gaussian surfaces (cylinder, box) that allowed us to apply Gauss’ Law. We found the same result that we had found in Example 16.3.4 where we had integrated Coulomb’s Law (twice, once for a ring of charge, then for a disk, then took the limit of the disk radius going to infinity).

    What does ∫ mean in physics?

    Terminology and notation – In general, the integral of a real-valued function f ( x ) with respect to a real variable x on an interval is written as The integral sign ∫ represents integration. The symbol dx, called the differential of the variable x, indicates that the variable of integration is x, The function f ( x ) is called the integrand, the points a and b are called the limits (or bounds) of integration, and the integral is said to be over the interval, called the interval of integration. the integral is called an indefinite integral, which represents a class of functions (the antiderivative ) whose derivative is the integrand. The fundamental theorem of calculus relates the evaluation of definite integrals to indefinite integrals. There are several extensions of the notation for integrals to encompass integration on unbounded domains and/or in multiple dimensions (see later sections of this article). to express the linearity of the integral, a property shared by the Riemann integral and all generalizations thereof.

    What does the surface integral tell you?

    The line integral of a vector field $\dlvf$ could be interpreted as the work done by the force field $\dlvf$ on a particle moving along the path. The surface integral of a vector field $\dlvf$ actually has a simpler explanation. If the vector field $\dlvf$ represents the flow of a fluid, then the surface integral of $\dlvf$ will represent the amount of fluid flowing through the surface (per unit time).

    1. The amount of the fluid flowing through the surface per unit time is also called the flux of fluid through the surface.
    2. For this reason, we often call the surface integral of a vector field a flux integral,
    3. If water is flowing perpendicular to the surface, a lot of water will flow through the surface and the flux will be large.

    On the other hand, if water is flowing parallel to the surface, water will not flow through the surface, and the flux will be zero. To calculate the total amount of water flowing through the surface, we want to add up the component of the vector $\dlvf$ that is perpendicular to the surface.

    1. Let $\vc $ be a unit normal vector to the surface.
    2. The choice of normal vector orients the surface and determines the sign of the fluid flux.
    3. The flux of fluid through the surface is determined by the component of $\dlvf$ that is in the direction of $\vc $, i.e.
    4. By $\dlvf \cdot \vc $.
    5. Note that $\dlvf \cdot \vc $ will be zero if $\dlvf$ and $\vc $ are perpendicular, positive if $\dlvf$ and $\vc $ are pointing the same direction, and negative if $\dlvf$ and $\vc $ are pointing in opposite directions.

    Let’s illustrate this with the function \begin \dlsp(\spfv,\spsv) = (\spfv\cos \spsv, \spfv\sin \spsv, \spsv). \end that parametrizes a helicoid for $(\spfv,\spsv) \in \dlr = \times $. As shown in the following figure, we chose the upward point normal vector.

    What are limitations of Gauss’s law?

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  • Flux is a general and broadly applicable concept in physics. However, in this chapter, we concentrate on the flux of the electric field. This allows us to introduce Gauss’s law, which is particularly useful for finding the electric fields of charge distributions exhibiting spatial symmetry. The main topics discussed here are

    1. Electric flux. We define electric flux for both open and closed surfaces.
    2. Gauss’s law. We derive Gauss’s law for an arbitrary charge distribution and examine the role of electric flux in Gauss’s law.
    3. Calculating electric fields with Gauss’s law. The main focus of this chapter is to explain how to use Gauss’s law to find the electric fields of spatially symmetrical charge distributions. We discuss the importance of choosing a Gaussian surface and provide examples involving the applications of Gauss’s law.
    4. Electric fields in conductors. Gauss’s law provides useful insight into the absence of electric fields in conducting materials.

    Gauss’s law gives us an elegantly simple way of finding the electric field, and, as you will see, it can be much easier to use than the integration method described in the previous chapter. However, there is a catch—Gauss’s law has a limitation in that, while always true, it can be readily applied only for charge distributions with certain symmetries.

    • 6.1: Prelude to Gauss’s Law So far, we have found that the electrostatic field begins and ends at point charges and that the field of a point charge varies inversely with the square of the distance from that charge. These characteristics of the electrostatic field lead to an important mathematical relationship known as Gauss’s law. This law is named in honor of the extraordinary German mathematician and scientist Karl Friedrich Gauss.
    • 6.2: Electric Flux The electric flux through a surface is proportional to the number of field lines crossing that surface. Note that this means the magnitude is proportional to the portion of the field perpendicular to the area. The electric flux is obtained by evaluating the surface integral \ where the notation used here is for a closed surface S.
    • 6.3: Explaining Gauss’s Law if a closed surface does not have any charges inside the enclosed volume, then the electric flux through the surface is zero. Now, what happens to the electric flux if there are some charges inside the enclosed volume? Gauss’s law gives a quantitative answer to this question. Gauss’s law relates the electric flux through a closed surface to the net charge within that surface.
    • 6.4: Applying Gauss’s Law For a charge distribution with certain spatial symmetries (spherical, cylindrical, and planar), we can find a Gaussian surface over which \(\vec \cdot \hat = E\), where E is constant over the surface. The electric field is then determined with Gauss’s law.
    • 6.5: Conductors in Electrostatic Equilibrium The electric field inside a conductor vanishes. Any excess charge placed on a conductor resides entirely on the surface of the conductor. The electric field is perpendicular to the surface of a conductor everywhere on that surface. The magnitude of the electric field just above the surface of a conductor is given by \(E = \frac \).
    • 6.A: Gauss’s Law (Answers)
    • 6.E: Gauss’s Law (Exercises)
    • 6.S: Gauss’s Law (Summary)

    Thumbnail: Karl Friedrich Gauss (1777–1855) was a legendary mathematician of the nineteenth century. Although his major contributions were to the field of mathematics, he also did important work in physics and astronomy. (Public Domain; Christian Albrecht Jensen).

    Why is symmetry important in Gauss law?

    2.3 Applying Gauss’s Law By the end of this section, you will be able to:

    • Explain what spherical, cylindrical, and planar symmetry are
    • Recognize whether or not a given system possesses one of these symmetries
    • Apply Gauss’s law to determine the electric field of a system with one of these symmetries

    Gauss’s law is very helpful in determining expressions for the electric field, even though the law is not directly about the electric field; it is about the electric flux. It turns out that in situations that have certain symmetries (spherical, cylindrical, or planar) in the charge distribution, we can deduce the electric field based on knowledge of the electric flux. where is the area of the surface. Note that these symmetries lead to the transformation of the flux integral into a product of the magnitude of the electric field and an appropriate area. When you use this flux in the expression for Gauss’s law, you obtain an algebraic equation that you can solve for the magnitude of the electric field, which looks like The direction of the electric field at the field point is obtained from the symmetry of the charge distribution and the type of charge in the distribution.

    1. Identify the spatial symmetry of the charge distribution, This is an important first step that allows us to choose the appropriate Gaussian surface. As examples, an isolated point charge has spherical symmetry, and an infinite line of charge has cylindrical symmetry.
    2. Choose a Gaussian surface with the same symmetry as the charge distribution and identify its consequences, With this choice, is easily determined over the Gaussian surface.
    3. Evaluate the integral over the Gaussian surface, that is, calculate the flux through the surface, The symmetry of the Gaussian surface allows us to factor outside the integral.
    4. Determine the amount of charge enclosed by the Gaussian surface, This is an evaluation of the right-hand side of the equation representing Gauss’s law. It is often necessary to perform an integration to obtain the net enclosed charge.
    5. Evaluate the electric field of the charge distribution, The field may now be found using the results of steps 3 and 4.

    Basically, there are only three types of symmetry that allow Gauss’s law to be used to deduce the electric field. They are

    • A charge distribution with spherical symmetry
    • A charge distribution with cylindrical symmetry
    • A charge distribution with planar symmetry

    To exploit the symmetry, we perform the calculations in appropriate coordinate systems and use the right kind of Gaussian surface for that symmetry, applying the remaining four steps. A charge distribution has spherical symmetry if the density of charge depends only on the distance from a point in space and not on the direction.

    In other words, if you rotate the system, it doesn’t look different. For instance, if a sphere of radius is uniformly charged with charge density then the distribution has spherical symmetry ((a)). On the other hand, if a sphere of radius is charged so that the top half of the sphere has uniform charge density and the bottom half has a uniform charge density, then the sphere does not have spherical symmetry because the charge density depends on the direction ((b)).

    Thus, it is not the shape of the object but rather the shape of the charge distribution that determines whether or not a system has spherical symmetry. (c) shows a sphere with four different shells, each with its own uniform charge density. Although this is a situation where charge density in the full sphere is not uniform, the charge density function depends only on the distance from the centre and not on the direction. Figure 2.3.1 Illustrations of spherically symmetrical and nonsymmetrical systems. Different shadings indicate different charge densities. Charges on spherically shaped objects do not necessarily mean the charges are distributed with spherical symmetry.

    The spherical symmetry occurs only when the charge density does not depend on the direction. In (a), charges are distributed uniformly in a sphere. In (b), the upper half of the sphere has a different charge density from the lower half; therefore, (b) does not have spherical symmetry. In (c), the charges are in spherical shells of different charge densities, which means that charge density is only a function of the radial distance from the centre; therefore, the system has spherical symmetry.

    One good way to determine whether or not your problem has spherical symmetry is to look at the charge density function in spherical coordinates,, If the charge density is only a function of, that is, then you have spherical symmetry. If the density depends on or you could change it by rotation; hence, you would not have spherical symmetry. where is the unit vector pointed in the direction from the origin to the field point, The radial component of the electric field can be positive or negative. When, the electric field at points away from the origin, and when, the electric field at points toward the origin.

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    We can now use this form of the electric field to obtain the flux of the electric field through the Gaussian surface. For spherical symmetry, the Gaussian surface is a closed spherical surface that has the same centre as the centre of the charge distribution. Thus, the direction of the area vector of an area element on the Gaussian surface at any point is parallel to the direction of the electric field at that point, since they are both radially directed outward ().

    (Figure 2.3.2) Figure 2.3.2 The electric field at any point of the spherical Gaussian surface for a spherically symmetrical charge distribution is parallel to the area element vector at that point, giving flux as the product of the magnitude of electric field and the value of the area. According to Gauss’s law, the flux through a closed surface is equal to the total charge enclosed within the closed surface divided by the permittivity of vacuum, Let be the total charge enclosed inside the distance from the origin, which is the space inside the Gaussian spherical surface of radius, This gives the following relation for Gauss’s law: Hence, the electric field at point that is a distance from the centre of a spherically symmetrical charge distribution has the following magnitude and direction: (2.3.3) Direction: radial from to or from to, The direction of the field at point depends on whether the charge in the sphere is positive or negative. For a net positive charge enclosed within the Gaussian surface, the direction is from to, and for a net negative charge, the direction is from to,

    1. This is all we need for a point charge, and you will notice that the result above is identical to that for a point charge.
    2. However, Gauss’s law becomes truly useful in cases where the charge occupies a finite volume.
    3. The more interesting case is when a spherical charge distribution occupies a volume, and asking what the electric field inside the charge distribution is thus becomes relevant.

    In this case, the charge enclosed depends on the distance of the field point relative to the radius of the charge distribution, such as that shown in, (Figure 2.3.3) Figure 2.3.3 A spherically symmetrical charge distribution and the Gaussian surface used for finding the field (a) inside and (b) outside the distribution. If point is located outside the charge distribution—that is, if —then the Gaussian surface containing encloses all charges in the sphere. The field at a point outside the charge distribution is also called, and the field at a point inside the charge distribution is called. Focusing on the two types of field points, either inside or outside the charge distribution, we can now write the magnitude of the electric field as (2.3.4) (2.3.5) Note that the electric field outside a spherically symmetrical charge distribution is identical to that of a point charge at the centre that has a charge equal to the total charge of the spherical charge distribution. This is remarkable since the charges are not located at the centre only.

    Which statement is true for Gauss’s theorem?

    All the charges whether inside or outside the gaussian surface contribute to the electric flux. Text Solution All the charges whether inside or outside the gaussian surface contribute to the electric flux. Electric flux depends upon the geometry of the gaussian surface.Gauss theorem can be applied to non uniform electric field.

    The electric field over the gaussian surface remains continuous and uniform at every point. Answer : D Solution : From Gauss.s law of electrostatics, we know that only charges contain within the Gaussian surface contribute to the flux. Also, we know the electric flux depends only on total charge and nature of medium and has nothing to do with geometry.

    Furthermore Gauss theorem is applicable to the case of uniform electric field as that will allow the formation of a symmetrical Gauss surface. Hence, options (a), (b), (c) are also false. Turns out that electric field remains continuous at every point over a Gauss surface.

    What is the conclusion of Gauss law?

    According to Gauss Law, the charge enclosed divided by the permittivity equals the total electric flux out of a closed surface. The electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane parallel to the field.

    Why do we need to understand the concept of Gauss’s law?

    The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity, The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field. Gauss’s Law is a general law applying to any closed surface. It is an important tool since it permits the assessment of the amount of enclosed charge by mapping the field on a surface outside the charge distribution.

    For geometries of sufficient symmetry, it simplifies the calculation of the electric field. Another way of visualizing this is to consider a probe of area A which can measure the electric field perpendicular to that area. If it picks any closed surface and steps over that surface, measuring the perpendicular field times its area, it will obtain a measure of the net electric charge within the surface, no matter how that internal charge is configured.

    Index Electric field concepts

    How many Gauss methods are there?

    2.4.1 Gauss method – A linear equation in variables x 1, x 2,, x n has the form a 1 x 1 + a 2 x 2 + a 3 x 3 + ⋯ + a n x n = d, where the numbers a 1,, a n ∈ R are the equation’s coefficients and d ∈ R is a constant. An n -tuple ( s 1, s 2,, s n ) ∈ R n is a solution of that equation if substituting the numbers s 1, s 2,, s n for the variables gives a true statement: a 1 s 1 + a 2 s 2 + a 3 s 3 + ⋯ + a n s n = d,

    A system of linear equations a 1, 1 x 1 + a 1, 2 x 2 + ⋯ + a 1, n x n = d 1, a 2, 1 x 1 + a 2, 2 x 2 + ⋯ + a 2, n x n = d 2, ⋯ a n, 1 x 1 + a n, 2 x 2 + ⋯ + a n, n x n = d n has the solution ( s 1, s 2,, s n ) if that n -tuple is a solution of all of the equations in the system. Theorem 2.4.1 Gauss method If a linear system is changed to another by one of the following operations: • an equation is swapped with another, • an equation has both sides to be multiplied by a nonzero constant, • an equation is replaced by the sum of itself and a multiple of another, then the two systems have the same set of solutions.

    Note that each of the above three operations has a restriction. Multiplying a row by 0 is not allowed because obviously that can change the solution set of the system. Similarly, adding a multiple of a row to itself is not allowed because adding −1 times the row to itself has the effect of multiplying the row by 0.

    What are the two applications of Gauss Law?

    Major Gauss law applications are the following: Electric field due to a uniformly charged infinite straight wire. Electric field due to a uniformly charged infinite plate sheet. Electric field due to a uniformly charged thin spherical shell.

    How does electric flux connect with Gauss’s law?

    Applications of Gauss’s Law –

    • Gauss’s Law can be used to find the electric field of point charge, infinite line of charge, infinite sheet of charge or sphere of charge. All these charge distributions are symmetrical in nature.
    • We can also use Gauss’s Law to find the electric flux passing through a closed surface.
    • The following is the value of electric field calculated using Gauss’s Law for three different distributions of charges:
      • Electric Field due to infinite line of charge = λ / 2πε 0 r
      • Electric Field due to infinite sheet of charge = σ / 2ε 0
      • Electric Field due to spherical shell = kq / r 2

    : Electric Flux and Gauss’s Law Basics | Definition, Equation

    What is Gauss’s formula?

    Integral form – Electric flux through an arbitrary surface is proportional to the total charge enclosed by the surface. No charge is enclosed by the sphere. Electric flux through its surface is zero. Gauss’s law may be expressed as: Φ E = Q ε 0 = }}} where Φ E is the through a closed surface S enclosing any volume V, Q is the total enclosed within V, and ε 0 is the, The electric flux Φ E is defined as a of the : Φ E = =} S } E ⋅ d A \cdot \mathrm \mathbf } where E is the electric field, d A is a vector representing an element of of the surface, and · represents the of two vectors. In a curved spacetime, the flux of an electromagnetic field through a closed surface is expressed as Φ E = c =c} S } F κ 0 − g d S κ }\,\mathrm S_ } where c is the ; F κ 0 } denotes the time components of the ; g is the determinant of ; d S κ = d S i j = d x i d x j S_ =\mathrm S^ =\mathrm x^ \mathrm x^ } is an orthonormal element of the two-dimensional surface surrounding the charge Q ; indices i, j, κ = 1, 2, 3 and do not match each other.

    • Since the flux is defined as an integral of the electric field, this expression of Gauss’s law is called the integral form,
    • A tiny Gauss’s box whose sides are perpendicular to a conductor’s surface is used to find the local surface charge once the electric potential and the electric field are calculated by solving Laplace’s equation.

    The electric field is perpendicular, locally, to the equipotential surface of the conductor, and zero inside; its flux πa 2 · E, by Gauss’s law equals πa 2 · σ / ε 0, Thus, σ = ε 0 E, In problems involving conductors set at known potentials, the potential away from them is obtained by solving, either analytically or numerically.

    The electric field is then calculated as the potential’s negative gradient. Gauss’s law makes it possible to find the distribution of electric charge: The charge in any given region of the conductor can be deduced by integrating the electric field to find the flux through a small box whose sides are perpendicular to the conductor’s surface and by noting that the electric field is perpendicular to the surface, and zero inside the conductor.

    The reverse problem, when the electric charge distribution is known and the electric field must be computed, is much more difficult. The total flux through a given surface gives little information about the electric field, and can go in and out of the surface in arbitrarily complicated patterns.

    1. An exception is if there is some in the problem, which mandates that the electric field passes through the surface in a uniform way.
    2. Then, if the total flux is known, the field itself can be deduced at every point.
    3. Common examples of symmetries which lend themselves to Gauss’s law include: cylindrical symmetry, planar symmetry, and spherical symmetry.

    See the article for examples where these symmetries are exploited to compute electric fields.

    How do you use Gauss’s law to find electric flux?

    Gauss law says the electric flux through a closed surface = total enclosed charge divided by electrical permittivity of vacuum.