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Gauss' Law Solutions

1. Using the integral form of Coulomb's Law for the following.

(a) Find the electric field at a point a distance a from one end of a long thin wire of length L and total charge Q. Examine the limit a >> L and show that your result is identical to that of a point charge.
(b) If the charge distribution is λ(x) = 2Qx/L², find the electric field. Examine the limit a >> L and show that your result is identical to that of a point charge.



(a) First, the wire has a uniform linear charge density λ = Q/L. Next, we consider a small portion of the wire, dx, located at a distance x from the left end of the wire having charge dq. These infinitesimals are related by

dq = λdx = (Q/L)dx .

Assuming that the charge on the wire is positive, the electric field at x = L + a due to dq, dE, is directed to the right. Using Coulomb's Law, dE is given by

dE = kdq r/r³ .

Here r = i(L+a-x) and r = L+a-x.  Thus

dE = ikdq/r² = i(kQ/L)dx/(L+a-x)² .

Integrating over the entire length of the wire, we get the total electric field to be
 

E	= ò0L dE
	= i(kQ/L) ò0L dx/(L+a-x)2
	= i(kQ/L){1/(L+a-x) |0L}
	= i(kQ/L){1/a - 1/(L+a)}
	= i(kQ/a2)(a / L+a)

In the limit a >> L, a/(L+a) ® 1, and

E = i(kQ/a²) .

At a great distance from the wire, the wire produces a field like that of a point charge.

(b) Here we are given the linear charge density λ(x). Next, we consider a small portion of the wire, dx, located at a distance x from the left end of the wire having charge dq. These infinitesimals are related by

dq = λ(x)dx = (2Q/L²) xdx .

Assuming that the charge on the wire is positive, the electric field at x = L + a due to dq, dE, is directed to the right. Using Coulomb's Law, dE is given by

dE = ikdq/r² = i(2Q/L²) dx {x/(L+a-x)²} .

Integrating over the entire length of the wire, we get the total electric field to be
 

E	= ò0L dE
	= i(2Q/L2) ò0L dx{x/(L+a-x)2}
	= i(2Q/L2){ (L+a)/(L+a-x) + ln(L + a - x) |0L}
	= i(2Q/L2){L/a - ln(l + L/a)}

In the limit a >> L, {L/a - ln(1 + L/a)} ® ½(L/a)², and

E = i(kQ/a²) .

At a great distance from the wire, the wire produces a field like that of a point charge.



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2. Using the integral form of Coulomb's Law, find the electric field at a point a distance a from the centre of a long thin wire of length L and total charge Q. The identity òdu[u²+v²]^3/2 = u/{v²[u²+v²]^½} + C may be of use. Show that your result reduces to that of a point charge in the limit a >> L. Also show that your answer reduces to E = 2kλ/a in the limit L >> a, the well-known and very useful result for a long thin wire. Note λ = Q/L.






First, the wire has a uniform linear charge density λ = Q/L. Next, we consider a small portion of the wire, dx, located at a distance x from the left end of the wire having charge dq. These infinitesimals are related by

dq = λdx = (Q/L)dx .

Assuming that the charge on the wire is positive, the electric field at r points away from dq. Using Coulomb's Law, dE is given by

dE = kdq r/r³ .

Here r = −ix + ja and r = (x² + a²)^½.  Thus

 dE = [−ix + ja] (kQ/L)dx/[x² + a²]^3/2 .

Integrating over the entire length of the wire, we get the total electric field to be
 

E	= ò-½L½L dE
	= (kQ/L) {-iò-½L½L dx x/[a2 + x2]3/2 + jò-½L½L dx a/[a2 + x2]3/2}
	= (kQ/L){0 + j x / a[a2+x2]½ |-½L½L }
	= j (kQ/L){½L / a[a2+¼L2]½ - (-½L) / a[a2+¼L2]½}
	= j (kQ/L){2L / a[4a2+L2]½ }
	= j (2kQ / La){1 / [1+4(a/L)2]½ }

In the limit L >> a, 1 / [1+4(a/L)²]^½® 1, and

E = i(2kQ/La) = i(2kλ/a) .

Note that there is no i component to the electric field as can be seen from the symmetry of the problem.



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3. Using the integral form of Coulomb's Law, find the electric field at a point midway between two long thin wires of length L and total charge Q and off to one side a distance h. The distance between the wires is a. The identity òdu[u²+v²]^-3/2 = u/{v²[u²+v²]^½} + C may be of use. Examine the limit h >> L and h >> a.






First, each wire has a uniform linear charge density λ = Q/L. Next, we consider a small portion of one wire, dx, located at a distance x from the midpoint of the two wires having charge dq. These infinitesimals are related by

dq = λdx = (Q/L)dx .

Assuming that the charge on the wire is positive, the electric field at r points away from dq. Using Coulomb's Law, dE is given by

dE = kdq r/r³ .

Here r = −ix + jh and r = (x² + h²)^½.  Thus

dE = [−ix + jh](kQ/L)dx/[h² + x²]^3/2 .

The symmetry of the problem can help reduce the amount of work we have to do. First the i components from each wire cancel, so we need only consider the j contributions. Second the j contributions must be the same. As a result we need only integrate over the length of would wire and double the result. We find that the total electric field is
 

E	= 2ò½aL+½a dEy
	= j (2kQ/L) { ò½aL+½a h/[h2 + x2]3/2 }
	= j (2kQ/L){ x / h[h2+x2]½ |½aL+½a }
	= j (2kQ/L){(L+½a) / h[h2+(L+½a)2]½ - (½a) / h[h2+¼a2]½}

In the limit h >> L and h >> a, (L+½a) / h[h²+(L+½a)²]^½− (½a) / h[h²+¼a²]^½® L/h², and

E = i(2kQ/h²) .

This is same result as from a single point of charge 2Q.



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4. A wire has been bent into a semicircle of radius R. It has a linear charge density λ. Determine the electric field at point P, at the centre of the circle. The identity S = Rθ and dS = Rdθ may help.






We consider a small portion of the wire, dS, located at an angle θ, having charge dq. These infinitesimals are related by

dq = λdS = λRdθ .

Assuming that the charge on the wire is positive, the electric field at P points away from dq. Using Coulomb's Law, dE is given by

dE = kdq r/r³ .

Here r = −iRcos(θ) − jRsin(θ) and r = R.  Thus

dE = [−iRcos(θ) − jRsin(θ)]kλRdθ/R³ = [−icos(θ) − jsin(θ)] kλdθ/R .

Integrating over the entire angular length of the wire, we get the total electric field to be
 

E	= ò0π dE
	= -(kλ/R) { iò0πcos(θ)dθ + jò0πsin(θ)dθ }
	= -(kλ/R){ isin(θ)|0p- jcos(θ)|0π }
	= -(kλ/R){0 − 2j}
	= -j(2kλ/R)

Of course, we could have used the symmetry of the problem to predict that the i component had to be zero.



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5. Three long thin wires of length L and charges Q, Q, and −Q are arranged to form an equilateral triangle. Use the result of question # 2 to find the electric field at the centre of the triangle.






The magnitude of the electric field from each wire is the same and is directed as shown in the diagram. The net electric field will clearly be downwards, the i contributions will cancel. Using trigonometry, the centre point is distance x = ½Ltan(30°) = L / 2Ö3 from each wire.

Using the result of Question 2,

E_wire = (2kQ / Lx){1 / [1+4(x/L)²]^½ } = 6kQ/L² .

The total electric field is

E_net = -j [E_wire + 2E_wirecos(60°)] = - j 2E_wire = - j 12kQ/L² .



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6. The "Gaussian Surface" for an infinitely large charged plate is a pillbox of surface area A and length 2a centred about the origin. The plate has a positive surface charge Σ. (a) How much charge is contained in the pillbox?

(b) Symmetry demands that the electric field point in the +x direction on the right side of the plate and -x on the other side. Explain why this is so.
(c) How much electric field passes through the sides of the pillbox?
(d) The pillbox has a thickness a on either side of the plate. How do the magnitudes of the electric field compare at x = -a and x = +a?
(e) Use Gauss's Law to determine the electric field at a.
(f) Sketch the electric field E(x) as a function of x.

(a) The charge contained is Q = A.

(b) A diagram let's us see why the net field is only in the i directions. 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.

(c) From part (b) we have seen that the E is in the ±i direction. Therefore no fields pass through the sides of the pillbox as shown in the diagram below.



(d) By symmetry, the electric fields must have the same magnitude on either side of the plate, E(-a) = E(a).

(e)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 ε₀, or

òE • ndA = Q_enclosed/ε₀,           (1)

where n is the outward-looking normal on each side of the surface. From part (a) we know Q_enclosed = A. The pillbox has two ends and one side, so that the integral has three parts

òE • ndA = ò_leftE • n_leftdA + ò_sideE • n_sidedA + ò_rightE • n_rightdA .

Examining the diagram below, we see that E • n_side = 0 since the vectors are at right angles. Similarly, E • n_left = E • n_right = E since the vectors are parallel.



Thus equation (1) reduces to

2E ò_end dA = ΣA/ε₀ .            (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 = ΣA/ε₀ .

Hence the field due to an infinitely large plate is E = Σ / 2ε₀. This result is good for real plates if one is much closer to the plate than the diameter of the plate.



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7. A very long thick plate has a uniform positive volume charge density give by ρ(x) = ρ₀ for -½d £ x £ ½d, where ρ₀ is a positive constant, d is the thickness of the plate and x = 0 is the centre of the plate.

(a) What is the symmetry of the electric field for this object?
(b)What is the electric field at x = 0? Explain why.
(c) How much charge is contained in the a "gaussian pillbox" of surface area A and thickness 2a where a < ½d. Use Gauss' Law to find E(a) .
(d) How much charge is contained in the a "gaussian pillbox" of surface area A and thickness 2a where a > ½d. Use Gauss' Law to find E(a) .
(e) Sketch the electric field for all a.

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

òE • ndA = Q_enclosed/ε₀ ,

reduces to

2 E(a) A = Q_enclosed/ε₀ .

(b) 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.

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



The charge inside is given by

Q_enclosed = ρ₀V = ρ₀(2a)A = 2aAρ₀ .

Thus Gauss' Law yields,

2 E(a) A = 2aAρ₀/ε₀ .

Thus the electric field is

E(a) = ρ₀a/ε₀ .

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



The charge inside is given by

Q_enclosed = ρ₀V = ρ₀Ad ,

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

Thus Gauss' Law yields,

2 E(a) A = Adρ₀/ρ₀.

Thus the electric field is E(a) = ρ₀d / 2ε₀ .

(e) The electric field as a function of a looks like





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8. The "Gaussian Surface" for cylinders is also a cylinder. It has length L and radius a. Consider an infinitely long wire of uniform charge per unit length λ.

(a) How much electric field passes through the ends of the "gaussian cylinder". Explain.
(b) How much charge is contains in the "gaussian cylinder"?
(c) Use Gauss's Law to determine the electric field at a distance a.



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

òE • ndA = ò_leftE • n_leftdA + ò_sideE • n_sidedA + ò_rightE • n_rightdA .

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

(b) Only a length L of the wire is inside the "gaussian cylinder", so Q_enclosed = λL .

(c) 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 ε₀, or

òE • ndA = Q_enclosed/ε₀ ,

where n is the outward-looking normal on each side of the surface. From parts (a) and (b), we know that this reduces to

ò_side E(a)dA = λL/ε₀ .

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

E(a) ò_side dA = λL/ε₀ ,

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

E(a) 2πaL = λL/ε₀ .

Solving for the electric field yields

E(a) = λ / 2pe₀a .



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9. A very long thin cylindrical shell of radius R carries a negative surface charge density Σ.

(a) If the radius of the "gaussian cylinder" is smaller than R, i.e. a < R, how much charge is contained by the gaussian surface?
(b) What, therefore, is the electric field everywhere inside the cylindrical shell?
(c) How much charge is contained in the "gaussian cylinder" when a > R?
(d) Use Gauss's Law to determine the electric field at a.
(e) Sketch the electric field as a function of a.

For cases of cylindrical symmetry, Gauss' Law

òE • ndA = Q_enclosed/ε₀ ,

reduces to the much simpler

E(a) 2πaL = Q_enclosed/ε₀ ,

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

(a) We draw an end view of the shell and "gaussian cylinder".



We see that there is no charge inside the "gaussian cylinder", so Q_enclosed = 0.

(b) Thus we find that the electric field for a < R is 0. There is no electric inside the shell.

(c) Considering the case of a > R, we have the following diagram.



The "gaussian cylinder" encircles a length L of the shell, so Q_enclosed = Σ2πRL.

(d) Thus Gauss' Law reduces to

E(a) 2πaL = Σ2πRL / ε₀ ,

and hence the electric field is

E(a) = ΣR / aε₀ ,

for a > R.

(e) The electric field as a function of a looks like



Note the jump in the electric field at the shell.



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10. A solid sphere of radius R carries a total positive charge Q uniformly distributed throughout the sphere. The "Gaussian Surface" for a sphere is a sphere of radius a concentric with the sphere.

(a) What is the electric field at the centre of the sphere? Explain?
(b) If a < R, how much charge is contained inside the "gaussian sphere"?
(c) Use Gauss's Law to determine the electric field at the surface of the "gaussian sphere".
(d) If a > R, how much charge is contained inside the "gaussian sphere"?
(e) Use Gauss's Law to determine the electric field at the surface of the "gaussian sphere".
(f) Sketch the electric field.

(a) The charge is arranged symmetrically about the centre, so the electric field from a piece of charge on one side is cancelled by the electric field from the diametrically opposed piece of charge on the other side. Thus the electric field is zero at the centre.

(b) We sketch the problem



The "gaussian sphere" contains only a portion of the total charge given by

Q_enclosed = Q V_gauss/V_sphere = Q (4πa³/3 / 4πR³/3) = Q(a/R)³ .

(c) Gauss' Law is stated

òE • ndA = Q_enclosed/ε₀ ,

where n is the outward looking normal. For cases of spherical symmetry, as we have here, E and n are parallel and E • n = E(a). The electric field at the surface, E(a), is a constant because of the spherical charge distribution and may be taken outside the integral. We determined Q_enclosed in part (b), thus we have

E(a) ò dA = (Q/ε₀)(a/R)³ .

The integral is simply the surface area of the "gaussian sphere", hence we have

E(a) 4πa² = (Q/ε₀)(a/R)³ .

Thus the electric field is

E(a) = Qa / 4pe₀R³ ,

for 0 £ a £ R.

(d) We sketch the "gaussian sphere" and see that it contains the entire sphere therefore Q_enclosed = Q.



(e) Gauss' Law reduces to

E(a) 4πa² = Q/ε₀ ,

and the electric field is

E(a) = Q / 4pe₀a² ,

for a ³ R.

(f) The electric field looks like





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11. A spherical shell of inner radius R and outer radius 2R, has a uniform charge distribution and total charge Q.

(a) Determine the charge inside the "gaussian sphere" for the three regions
  (i)   0 < a < R,
  (ii)  R < a < 2R,
  (iii) 2R < a.
(b) Use Gauss's Law to determine the electric field at the surface of the "gaussian surface" when
  (i)   0 < a < R,
  (ii)  R < a < 2R,
  (iii) 2R < a.
(c) Sketch the electric field for all a.

(a) First we sketch the "gaussian sphere" in each case.



(i) Here we see that there is no charge inside the "gaussian sphere", so Q_enclosed = 0.

(ii) Here there is only a portion of the total charge inside the "gaussian sphere" given by the ratio of volumes



(iii) Here the entire shell is enclosed, so Q_enclosed = Q.

(b) In cases of spherical symmetry, Gauss' Law

òE • ndA = Q_enclosed/ε₀ ,

reduces to

E(a) 4πa² = Q_enclosed / ε₀ .

For our three cases, we have

(i) For a £ R,

E(a) 4πa² = 0 .

Thus E(a) = 0.

(ii) For R £ a £ 2R,

E(a) 4πa² = Q(a³-R³)/(7ε₀R³) .

Thus E(a) = (Q/4pe₀R³)(a³-R³)/(7a²).

(iii) For a £ 2R,

E(a) 4πa² = Q/ε₀ .

Thus E(a) = Q/4pe₀a².

(c) The electric field looks like





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12. Use Gauss's Law to determine the electric field as a function of distance a from the centre of a thin spherical shell of radius R and negative surface charge -Σ. Sketch the electric field as a function of a. By looking at the electric field, can we distinguish between a sphere with all its charge on the surface and a sphere with the charge smeared throughout the sphere?






For cases of spherical symmetry, Gauss' Law

òE • ndA = Q_enclosed/ε₀ ,

reduces to

E(a) 4πa² = Q_enclosed/ε₀ .

Here we have two cases as shown in the diagram below: (i) 0 £ a £ R and (ii) a ³ R.



(i) For 0 £ a £ R, Q_enclosed = 0. Hence the electric field is zero.

(ii) For a ³ R, the total shell is enclosed so Q_enclosed = Q. We are not given Q but rather the charge density so

Q_enclosed = -Σ 4πR² .

Thus Gauss' Law is

E(a) 4πa² = -Σ 4πR² / ε₀ .

Hence the electric field is

E(a) = -(Σ/ε₀)(R/a)² .

Thus the electric field looks like



As long as the charge distribution is symmetrical, we can find the electric field from Gauss' Law. Gauss' Law says that E(a) depends only on the total charge enclosed and thus is independent of how that charge is distributed throughout the material. Hence we cannot distinguish between a sphere with all its charge on the surface and a sphere with the charge smeared throughout the sphere.

Questions?mike.coombes@kwantlen.ca