Gauss’s Law

When we have to find an electric field due to extended charge distribution, it involves the integration of charge elements and is sometimes very tedious to do. But it could be done easily in case of certain symmetry.

  • Spherical symmetry- Uniform volume charge spread in the volume of a sphere.
  • Cylindrical symmetry - Uniform volume charge in volume of a cylinder.
  • Infinite line charge - Uniform line charge spread in an infinitely long wire.
  • Infinite surface charge - Uniform charge distributed on an infinite large plane sheet.

In case of any one of the above symmetry, the calculation of electric field can be done using Gauss’s law.

Gauss’s law statement :

Gauss’s law states that electric flux through any surface enclosing charges is equal to 1/ϵo times the total charge enclosed by the surface. Mathematically,

Gaussian surface: A hypothetical closed surface enclosing a charge is called the gaussian surface of that charge.

By clever choice of Gaussian surface, we can easily find the electric field produced by certain symmetry charge configurations which are otherwise difficult to evaluate using Coulomb force and principle of superposition.

Two things that must be taken care while using gauss’s law is that

  • The charge must be enclosed.
  • We take only those charges which are inside the Gaussian surface

Great Significance of Gauss Law:

We already have discussed some of them but let's just gather them all under a single heading.

  •  Gauss law is true for any closed surface no matter what the shape or size.
  •  The term q on the right side of gauss’s law includes the sum of all the charges enclosed by the surface. The charges may be located anywhere inside the surface.
  • In any situation, there may be some charge inside and outside the chosen surface. On the left side of Gauss’s law,  the electric field is due to all the charges both inside and outside the surface. The term q  on the right side however represents only the total charge inside the surface.
  • Gauss’s Law Is based on the Inverse-square dependence on distance contained in  Coulomb's law.
  • Gauss’s Law is often useful to word a much easier calculation of electrostatic field when the system has some symmetries as discussed above. This is facilitated by the choice of Gaussian surface

Note:

Gauss’s law is always valid, but it is useful only when there is any one of the above-mentioned symmetry and we can take out E outside the integral due to symmetry and so only surface integration is left over the chosen gaussian surface.

Application of gauss law

As we have discussed the importance of gauss law and also discussed in which symmetry situation this will be useful.

Note for the students

I will focus less on derivation in this part as anyone can get it in every book.

My focus in this section would be to give my readers a good understanding of the concept and its applications. I will directly give the Electric field of some symmetry cases without derivation but will focus on giving you a good conceptual understanding of the results and how we will use it in problem-solving.

Electric flux due to charge in a cube

Case -1  Charge is placed inside the cube.

Suppose we have a cube and a charge is placed inside the cube. Since the cube is a closed surface, we can directly use Gauss's law. electric flux coming out of the cube will be q/ϵ0

from the six faces of the cube. If we wish to calculate electric flux through one face of this cube, we will divide the whole flux by the number of faces of the cube ( here 6 ).

Electric flux through one the face when the charge is placed inside the cube is= q/6ϵo

Case -2  Charge placed on one of the faces of the cube.

Now we can have another case where the charge is kept on one of the faces of the cube.  This will not be a closed surface as the charge is not closed completely. But we can close it by putting one more cube over the charge so now the charge will be fully enclosed inside two cubes.

So the total electric flux through the close surface ( two cubes here) =q/ϵ0

The flux through one cube will be total flux divided by no of cubes. So in this case =  q/2ϵ0

Now If you imagine 2 cubes joined together, there will be 10 faces of the cubes outside.

So if we wish to calculate electric flux through each face then we will divide the total flux by no of faces on the outer side = q/10ϵ0

Case -3 when the charge is at the edge of the cube.

In this case, we would require 4 cubes to enclose the charge.

Total flux through the closed surface (4 cubes in this case) will be = q/ϵo

Electric flux through one cube is total flux divided by no of cubes = q/2ϵ0

there will be a total of 12 faces of the cube through which flux will be coming out in this case.

Electric flux passing through one face will be Total flux / no of faces toward the outer region = q/12ϵo

Case -4  Charge placed on one of the vertices.

No of cubes required to enclose this charge = 8

Electric flux coming out of 8 cubes ( enclosed surface) = q/ϵ0

Electric flux coming out of 1 cube = q/8ϵ0

No of faces on the outer side of these 8 cubes joined together = 24

Electric flux coming through one of the flux is=q/24ϵ0

Field due to an infinitely long straight uniformly charged wire.

Consider an infinitely long straight wire having charges uniformly distributed over it. The linear charge density of this wire is λ.

  • Suppose this is a positive charge distribution then from previous knowledge of  Electric  field lines we can say that the direction of electric field will be radially outward from the wire as shown in figure below
  • As Electric field due to this wire has cylindrical symmetry; the right choice of the gaussian surface would be a cylinder.
  • Draw a cylinder of saying length L and radius ‘r’ round the wire. You will notice that this cylinder would have three surfaces, Two circles at the top and bottom and one curved surface around the wire.
  • The next step would be to draw the area vector for all three surfaces. From our previous discussion, we must know that the area vector must be perpendicular to the surface. Reader may refer to the direction of Area vector for each surface  ( denoted by n )and also take a look at direction of electric field lines coming out of three surfaces

  • Now you will notice that for the circular area, the area vector and Electric field are perpendicular (makes 90 degree) and for the curve surface, area vector and electric field is parallel ( angle 0).
  • Now In the formula for Gauss law in the integral we have  Eds, Which means Edscosθ and we also know that cos 0=1 and cos 90=0. So if we break this integral for 3 surfaces ( two circular surfaces and one curved surface), the integral of the circular surface would vanish and we will be left with only one integral with curve surface integration.
  • Now that we have simplified the left side, let's talk about the right side now.  The right side of gauss’s law is actually q/ϵ0. Here q is the charge enclosed by the gaussian surface. Here since the length of the gaussian cylinder is “L” and we have linear charge density λ. Then the total charge enclosed by the cylinder  q=λL

Eds around curved surface = λL/ ϵo

  • Since the curve surface of the gaussian area is at “r” distance from the wire and so E at all the points on the curve surface is constant . We can now take constant E outside the integral and can only integrate over the area of the curve surface.
  • Note that curve surface area of cylinder of length L and radius “r” is  =2ΠrL

 Ecurve surfaceds = E * 2ΠrL= λL/ϵ0

E=λL/2Πϵ0*rL

 E= λ/2Πϵ0*r…. Electric field due to the infinitely long wire of uniform linear charge density

The above result is for the electric field for an infinitely long wire having uniform linear charge density. Let us analyze this result.

  1.  The electric field is proportional to linear charge density.
  2. The field is inversely proportional to the distance from the wire. The Further you go from the wire, the less will be the value of the electric field.

Electric field due to an infinitely long plane sheet having uniform surface charge density.

Consider an infinitely long plane sheet having uniform surface charge density σ

Suppose this is a positive charge distribution than from previous knowledge of Electric field lines we can say that the direction of 

  • Electric field will be radially outward from the plane sheet in both directions as shown in figure below.
  • As Electric field due to this plane sheet has cylindrical symmetry; the right choice of gaussian surface would be a cylinder. Consider a gaussian surface of length L and cross -section area “A” as shown in figure.
  • Just like in the previous case, we have discussed that this cylinder has three surfaces and draw the area vector for each surface.
  • Then we will see the direction of E and the area vector through each surface. In this case you will notice that for the curve surface E and area vector is perpendicular  (angle=90) and for two circular cross-sections, E and area vector are parallel ( angle 0).

  • So in this case when we break the closed surface integral on LHS, we will get three integrals; two for cross-section area ( will contribute as angle =0) and one for curved surface area ( this will vanish as angle=90).
  • Now for the RHS of Gauss’s law where q represents the charge enclosed inside the gaussian surface. As you can see from the figure, the area of the plane sheet which is equal to the cross-section area of the cylinder is enclosed inside the gaussian surface.
  • So Left-hand side of Gauss’s equation will be 

Eds = Eclosed surfaceds= 2Ecross-sectionds= 2EA

  • Since we have uniform surface charge density σ and area enclosed is A, then the charge enclosed q=σA. So right side of equation q/ϵ0  = σA/ϵ0.

When we will equate the LHS and RHS we will get 2EA=σA/ϵ0

E= σ/2ϵ0……………( Electric field due to infinitely long plane sheet)

The above relation is the expression of Electric field due to an infinitely long plane sheet having uniform surface charge density.

Now Let's analyze it.

  1.  Like the previous case, Electric field is proportional to surface charge density.
  2. But the thing to be noticed here is actually that Electric field due to the infinitely plane sheet is actually independent of distance.
  3.  We can conclude from the above discussion that Electric field due to the infinite plane sheet is constant in magnitude.
  4. The direction of Electric field will be pointing away from the sheet from both sides of the plane in case of positive surface charge density.
  5. The direction of Electric field will be pointing towards the sheet from both sides of the plane in case of negative surface charge density.

Application of Electric field due to Infinitely long plane sheet.

Electric field due to two parallel infinite plane sheet having equal and opposite surface charge density ±σ

When we have two parallel infinite plane sheets. The whole space is divided into three regions

  • First region at the left of both the plates  ( say P)
  • Second region is in between the plates. ( say Q)
  • Third region on the right of both the plates. ( say R)

  1. Since from our analysis we know that Electric field due to the infinite plane sheet is independent of the distance from the sheet, then we can take away very necessary information from this fact.The magnitude of electric field at any of these regions due two plane sheets will be equal.
  2.  Now Lets talk about direction of E due to Plate A and Plate B , in these three regions. Referring to points 4 and 5 of our analysis above. We can conclude that direction of EA will be away from the plates and direction of EB will be toward the plates in all three regions. Refer above figure.
  3. From the figure above you can say that at regions P and R, the direction of EA and EB is opposite and their magnitudes are equal; they would cancel each other out.
  4. Above discussion concludes that Electric fields in region P and R are zero.
  5. In region between the plate, EA and EB are in same directions and hence they will add to give E= σ/ϵo  between the plates.

The above result is very important, always remember that the electric field between two parallel plates of opposite surface charge density is E=σ/ϵ0

And E=0 everywhere except in between the plates.

Now you might be thinking about what would be the case if both the plates will have charges of the same sign and have the same surface charge density like each plate has   +σ  charge density.

  • The magnitude of electric field in every region due to any of the two plates will be the same as it doesn’t depend on distance from the plates.
  • The direction of the field between the plates due to two plates is pointing in opposite directions and hence cancel each other. So the net field in between the plates in region 2 is zero if the plates have the same charge densities.
  • If we look at regions 1 and 3, we will see that the direction of the electric field due to plates in these regions are in the same direction and hence they will add. The value of electric field in region 1 and 3 will be E=σ/ϵ0.

There can be one more possibility, what if the charge densities on the plates are not the same? The answer is very simple in that case they will neither completely cancel in the region where they are in the opposite direction, nor they will double in the region where they are in the same direction.

Just simply add and subtract them with their magnitude.