Questions: 1 2 3 4 5 6 7 8 9 10 11

## PHYS 2420 Gauss' Law Solutions

1. Use Gauss' Law to determine the electric field at a distance r from the centre of an infinitely large plate of surface charge Σ.

A diagram will let us determine the symmetry of the electric field. Consider identical small pieces of charge which are diametrically opposed and the electric field each produces at x = ± a.

If the pieces have the same charge, the electric fields they produce will have the same magnitude. Looking at the above diagram, we see that the j components will cancel. Thus the electric field can only be in the ± i direction.

In the diagram below, we sketch a side view of the field and the Guassian pillbox. Not e that no field lines pass through the sides of the pillbox.

Symmetry also requires that the electric fields must have the same magnitude on either side of the plate, E(–a) = E(a).

Gauss’ Law states that the total flux of the electric field through all sides of the pillbox is equal to the enclosed charge divided by e0, or

ò EndA = Qenclosed/e0 ,         (1)

where n is the outward–looking normal on each side of the surface. For a uniformly charged surface, we know Qenclosed = s A. The pillbox has two ends and one side, so that the integral has three parts

ò EndA = ò left E nleft dA + ò side E nside dA + ò right E nright dA .

Examining the diagram below, we see that E nside = 0 since the vectors are at right angles. Similarly, E nleft = E nright = E since the vectors are parallel.

Thus equation (1) reduces to

2E ò end dA = s A/e0 .          (2)

Since E is constant at the ends, it was taken out side the integral. The integral itself then reduces to the surface area of the end of the pillbox,

2EA = s A/e0 .

Hence the field due to an infinitely large plate is E = s / 2e0. This result is good for real plates if one is much closer to the plate than the diameter of the plate.

2. A large plate has a charge density give by r (x) = r0x2 for –d/2 £ x £ d/2, where r0 is a positive constant, d is the thickness of the plate and x = 0 is the centre of the plate. The plate is much longer than it is wide. Use Gauss' Law to find E(x). Sketch the result.

Symmetry indicates that the electric field only has an i component and no j component. This means that Gauss’ Law,

ò E ndA = Qenclosed/e0 ,

reduces to

2 E(a) A = Qenclosed/e0 .

The electric field at x = 0 is zero since the charge is distributed symmetrically about the y–axis. The electric field from a piece of charge on one side will exactly cancel with the field from a piece of charge on the other side.

There is one boundary and two distinct regions here. Thus we need to use apply Gauss' Law twice. Once for a pillbox totally inside the boundary and once for a pillbox which crosses the boundary.

Case i. a < d/2.

We sketch the pillbox below. We need to find the charge inside this pillbox.

The charge inside is given by

Qenclosed = ò pillbox ρ dV = Aò–a+a ρ(x) dx = Aρ0 ò –a+a x2 dx = (2/3)Aρ0a3 .

Thus Gauss’ Law yields,

2 E(a) A = (2/3)Aρ0a3/e0.

Thus the electric field is E(a) = (1/3)ρ0a3/e0.

Case ii. a > d/2.

We need to find the charge inside the pillbox shown in the diagram below.

Note that there is charge inside the pillbox only up to a = d/2. Thus the charge inside is given by

Qenclosed = ò pillbox ρ dV = Aò –d/2+d/2 ρ(x) dx = Aρ0 ò–d/2+d/2 x2 dx = (1/12)Aρ0d3,

since the charge is confined to –½d £ x £ ½d.

Thus Gauss’ Law yields,

2 E(a) A = (1/12)Aρ0d3/e0 .

Thus the electric field is E(a) = (1/24) ρ0d3/e0 .

The electric field as a function of a looks like

3. Use Gauss’ Law to determine the electric field at a distance R to the side of an infinitely long wire of charge per length λ.

The electric field is radially symmetric about the axis of the wire as shown in the diagram below. The Guassian surface must have the same symmetry, i.e. it must be a cylinder.

The amount of flux that passes through the cylinder is given by

ò E ndA = ò left E nleft dA + ò side E nside dA + ò right E nright dA .

Examining the diagram above, we see that E nside = E(a) since the vectors are parallel. Similarly, E nleft = E nright = 0 since the vectors are at right angles. So there is no flux through the ends of the "gaussian cylinder".

Only a length L of the wire is inside the "gaussian cylinder", so Qenclosed = l L .

Gauss’ Law states that the total flux of the electric field through all sides of the pillbox is equal to the enclosed charge divided by e0, or

ò E ndA = Qenclosed/e0 ,

where n is the outward–looking normal on each side of the surface. For this case this reduces to

ò side E(a) dA = l L/e0 .

Since E(a) is a constant for a given radius a, it may be taken outside the integral,

E(a) ò side dA = l L/e0 ,

and we are left with an integral for the surface area of the side of the "gaussian cylinder" which is A = 2p aL. Thus we have

E(a) 2p aL = l L/e0 .

Solving for the electric field yields E(a) = l / 2p e0a .

4. A very long thin cylindrical shell of radius R carries a surface charge density Σ. Use Gauss’ Law to determine the electric field as a function of distance r from the centre of the cylinder at a point very far from either end. Sketch the electric field as a function of r.

For cases of cylindrical symmetry, Gauss’ Law

ò E ndA = Qenclosed/e0 ,

reduces to the much simpler

E(a) 2p aL = Qenclosed/e0 ,

where a is the radius of the "gaussian cylinder" and L is its length.

The surface of the cylindrical shell forms a boundary. We will need to do two cases; one where the Gaussian surface is smaller than the boundary and one where it is larger.

Case i. a < R

We sketch a side view of the shell and the Gaussian surface below.

We see that there is no charge inside the "gaussian cylinder", so Qenclosed = 0. Hence E = 0 for a < R is 0. There is no electric field inside the shell.

Case ii. a > R

We have the following diagram.

The "gaussian cylinder" encircles a length L of the shell, so Qenclosed = s A = s 2p RL. The area of a cylindrical shell is the circumference, 2p R, times the length L.

Thus Gauss’ Law reduces to

E(a) 2p aL = s 2p RL /e0 ,

and hence the electric field is

E(a) = s R/ae0 ,

for a > R.

The electric field as a function of a looks like

Note the jump in the electric field at the shell. This is the signature of a surface charge.

5. Use Gauss’ Law to determine the electric field as a function of distance a from the centre of a long cylinder of radius R and charge density ρ(r) = Br2. Sketch the electric field as a function of a.

For cases of cylindrical symmetry, Gauss’ Law

ò E ndA = Qenclosed/e0 ,

reduces to the much simpler

E(a) 2p aL = Qenclosed/e0 ,

where a is the radius of the "gaussian cylinder" and L is its length.

The surface of the cylinder forms a boundary. We will need to do two cases; one where the Gaussian surface is smaller than the boundary and one where it is larger.

Case i. a < R

We sketch a side view of the shell and the Gaussian surface below.

The charge inside is given by

Qenclosed = ò pillbox ρ dV = 2p Lò 0a ρ(r) rdr = 2p Lò 0a Br3 dr = ½p LBa4.

Thus Gauss’ Law yields,

E(a) 2p aL = ½p LBa4/e0 .

Thus the electric field is E(a) = ¼Ba3/e0.

Case ii. a > R

We have the following diagram.

The "gaussian cylinder" encircles a length L of the cylinder, but the charge only extends to R. The charge inside is given by

Qenclosed = ò pillbox ρ dV = 2p Lò 0R ρ(r) rdr = 2p Lò 0R Br3 dr = ½p LBR4.

Thus Gauss’ Law yields,

E(a) 2p aL = ½p LBR4/e0.

Thus the electric field is E(a) = ¼BR4/ae0.

The electric field as a function of a looks like

6. A sphere of radius R carries a total charge Q. What is the charge density of the sphere? Use Gauss’ Law to determine the electric field as a function of distance a from the centre of the sphere. Sketch the electric field as as a function of a.

The charge density will be ρ = Q/V = 3Q / 4p R3.

For cases of spherical symmetry, Gauss' Law

ò E ndA = Qenclosed/e0 ,

becomes

E(a) 4p a2 = Qenclosed/e0 .

First consider the case a > R.

For this particular problem we are asked to apply Gauss' Law at a particular distance. We show the Gaussian surface in the diagram below.

We see that the Gaussian surface contains the entire sphere so that Qenclosed = Q. Therefore for this case Gauss's Law yields

E(a) = Q/4p e0a2 .

Next we consider the case when a < R.

The charge inside is given by

Qenclosed = ò pillbox ρ dV = 4p ò 0a ρ(r) r2dr = 4p Q/(4p R3/3) ò 0a r2 dr = Q(a/R)3.

Thus Gauss’ Law yields,

E(a) 4p a2 = Q(a/R)3/e0 .

Thus the electric field is E(a) = (Q/4p e0)(a/R3) .

The electric field looks like

7. A spherical shell of inner radius R and outer radius 2R, has a uniform charge distribution and total charge Q. Determine the charge density ρ(r) for the three regions. Find the electric field everywhere. Sketch the electric field.

Obviously, ρ = 0 for 0 £ a < R and for a > 2R. The density in the region R £ a £ 2R is uniform and can be calculated using the formula ρ = Q/V, where V is the volume of space that has charge in it. In this case,

V = Vouter sphere – Vinner sphere = (4/3)p (2R)3 – (4/3)p(R)3 = (28/3)p R3 .

Hence the density is

ρ = 3Q/28p R3

Gauss' Law for spheres becomes

E(a) 4p a2 = Qenclosed/e0 .

Since we have two boundaries, one at R and the other at 2R, we have three cases to examine.

Case i. a < R.

We sketch the Gaussian surface as shown in the diagram below.

There is no charge enclosed so E = 0.

Case ii. R < a < 2R.

We sketch the Gaussian surface as shown in the diagram below.

The charge inside is given by

Qenclosed = ò pillbox ρ dV = 4p ò Ra ρ(r) r2dr = 4p 3Q/(28p R3) ò Ra r2 dr = (Q/7)[(a/R)3–1].

Note that the lower limit of integration is R because ρ = 0 for 0 < r < R. Thus Gauss’ Law yields,

E(a) 4p a2 = (Q/7)[(a/R)3–1]/e0 .

Thus the electric field is E(a) = (Q/28p e0)[(a/R3)–1/a2].

Case iii. a > 2R.

We sketch the Gaussian surface as shown in the diagram below.

The Gaussian surface contains the entire sphere, so Qenclosed = Q. Thus Gauss’ Law yields,

E(a) 4p a2 = Q/e0 .

Thus the electric field is E(a) = Q/4p e0a2.

The electric field will look like

8. Use Gauss’ Law to determine the electric field as a function of distance a from the centre of a sphere of radius R and charge density ρ(r) = Ar2. Find and sketch the electric field as a function of r.

For cases of spherical symmetry, Gauss' Law

ò E ndA = Qenclosed/e0 ,

becomes

E(a) 4p a2 = Qenclosed/e0 .

The surface of the sphere forms a boundary. We will need to do two cases; one where the Gaussian surface is smaller than the boundary and one where it is larger.

Case i. a < R

We sketch a side view of the sphere and the Gaussian surface below.

The charge inside is given by

Qenclosed = ò pillbox ρ dV = 4p ò 0a ρ(r) r2dr = 4p ò 0a Ar4 dr = (4/5)p Aa5.

Thus Gauss’ Law yields,

E(a) 4p a2 = (4/5)p Aa5/e0 .

Thus the electric field is E(a) = Aa3/5e0.

Case ii. a > R

We have the following diagram.

The "Gaussian sphere" encircles the entire sphere, but the charge in the sphere only extends to R. The charge inside is given by

Qenclosed = ò pillbox ρ dV = 4p ò 0R ρ(r) r2dr = 4p ò 0R Ar4 dr = (4/5)p AR5.

Thus Gauss’ Law yields,

E(a) 4p a2 = 4p AR5/5e0 .

Thus the electric field is E(a) = AR5/5a2e0.

The electric field as a function of a looks like

9. Consider the following electrostatic fields. Determine the charge density ρ .
(a) E = –iky – jkx.
(b) E = –ikx – jky – kkz.
(c) E = rr + zz.

Now ρ(x,y,z) = –e0Ñ •E, where

Ñ •E = ∂Ex/∂x + j ∂Ey/∂ y + k ∂Ez/∂z

in Cartesian coordinates. As well

Ñ •E = (1/r)∂(rEr)/∂r + (1/r)∂Eq/∂q + (1/r)∂Ez/∂z

in cylindrical coordinates. Hence

(a) ρ(x,y,z) = 0 .

(b) ρ(x,y,z) = –ke0(x + y + z) .

(c) ρ(x,y,z) = –e0(2 + 1/r) .

10. (a) E =5x2yi. Integrate along between the path given by (3,2) to (6,2) to (6,8) to (3,8) to (3,2).

(b) E = xi+(2/y)j. Integrate along between the path given by (2,2) to (2,6) to (5,6) to (5,2) to (2,2).

(a) First we sketch the path of integration

The line integral breaks into four parts

–ò C E·dL = –ò1 E·dL – ò2 E·dL – ò3 E·dL– ò4 E·dL.

In paths 1 and 3, dL = i dx ; in path 2 and 4, dL = j dy. Thus the integral reduces to

–ò C E·dL = –ò(3,2)(6,2) E·i dx – ò(6,2)(6,8) E·j dy – ò(6,8)(3,8) E·i dx – ò(3,8)(3,2) E·j dy .

Since we know that the field point only in the i direction, the second and fourth integrals are zero and we are left with

 –òC E·dL = –ò(3,2)(6,2) Edx – ò(6,8)(3,8) Edx = –ò(3,2)(6,2) 5x2y dx – ò(6,8)(3,8) 5x2y dx = –5x3y/3 |(3,2)(6,2) - 5x3y/3 |(6,8)(3,8) = –5[63 – 33]2/3 – 5[63 – 33]8/3 = –1050 Volts.

(b) The path of integration is

The line integral breaks into four parts

–ò C E·dL = –ò1 E·dL – ò2 E·dL – ò3 E·dL– ò4 E·dL.

In paths 1 and 3, dL = j dy ; in path 2 and 4, dL = i dx. Thus the integral reduces to

–ò C E·dL = –ò(2,2)(2,6) E·j dy – ò(2,6)(5,6) E·i dx – ò(5,6)(5,2) E·j dy – ò(5,2)(2,2) E·i dx .

Doing the dot products, we are left with

–ò C E·dL = –ò26 (2/y) dy – ò25 x dx – ò62 (2/y) dy – ò52 x dx .

Evaluating each integral on the right hand side yields

 –ò C E·dL = –2 ln(y)| 26 – ½x2|25 - 2 ln(y)| 62 – ½x2|52 = –2 ln(3) – ½(21) + 2 ln(3) + ½(21) = 0

11. Determine which of the following electric fields are conservative. If conservative, determine a potential for the field. Determine the charge density ρ (x,y,z).
(a) Ex = x+y, Ey = –x+y, and Ez = –2z,
(b) Ex = 2y, Ey = 2x+3z, and Ez = 3y,
(c) Ex = x2–z2, Ey = 2, and Ez = –2xz,

An electric field E is conservative if Ñ × E = 0.

(a)We evaluate the curl of the electric field by

So this field is not conservative.

(b) Treating this field the same way, we find

This field is conservative.

(c) Again

This field is conservative.

If an electric field is conservative, we can define an electrostatic potential V such that E = –ÑV. In terms of the x, y, and z components of the electric field Ex = –∂V/∂x, Ey = –∂V/∂y, and Ez = –∂V/∂z. Since we are given Ex, Ey, and Ez we can integrate to find V.

For field (b), We have

 ∂V/∂x = –2y \ V = –2xy + C1(y,z) ∂V/∂y = –(2x + 3z) \ V = –2xy – 3yz + C2(x,z) ∂V/∂z = –3y \ V = –3yz + C3(x,y)

where C1, C2, and C3 are functions of integration. The potential which is consistent with all three results is

V = –2xy – 3yz + C ,

where C is a constant.

For field (c) we have

 ∂V/∂x = –(x2 – z2) \ V = –x3/3 + xz2 + C1(y,z) ∂V/∂y = –2 \ V = –2y + C2(x,z) ∂V/∂z = 2xz \ V = xz2 + C3(x,y)

where C1, C2, and C3 are functions of integration. The potential which is consistent with all three results is

V = –x3/3 – 2y + xz2 + C ,

where C is a constant.

The charge density can be found from Poisson’s Equation, Ñ2V = –ρ /e0. In Cartesian coordinates, the Poisson’s Equation is defined

2V/∂x2 + ∂ 2V/∂y2 + ∂2V/∂z2 = –ρ /e0 .

Taking the derivatives of the potential that we found for field (b), yields ρ = 0. This field has no free charges. For field (c), we find ρ = 4e0x.

Questions? mike.coombes@kwantlen.ca