PHYS 1102 Electrostatic Potential Solutions

Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Electrostatic Potential Solutions

(a) E = 5x²i. Calculate ΔV between x = 0 m and x = 4 m.

(b) E = (2/y)j. Calculate ΔV between y = 2 m and y = 5 m.

The potential difference between two points is defined as

ΔV = -ò_i^fE • dl ,

where dl is the parallel to the path of integration. The potential difference is independent of the path taken, so one may choose whichever path makes the integration easiest.

(i) We will integrate along the line joining the two points so that dl = idx.

ΔV = -ò_i^fE • dl

= -ò₀⁴ 5x²i • i dx

= -ò₀⁴ 5x²dx

= -(5/3) x³ |₀⁴

= -320/3 Volts

Therefore V(x=4) is at a lower potential than V(x=0).

(b) We will integrate along the line joining the two points so that dl = jdy.

ΔV	= -ò_i^fE • dl
	= -ò₂⁵ (2/y) j • j dy
	= -ò₂⁵ 2dy / y
	= -2 ln(y) \|₂⁵
	= -2 ln(5/2) Volts

Therefore V(y=5) is at a lower potential than V(y=2).

Find the electrostatic potential difference between points A and B which are distances r_A = 2.0 m and r_B = 1.0 m from an infinitely long thin wire with λ = 1.0 μC/m. The result E = λ/2p₀r is useful. If an electron (q = -e = -1.602 × 10^-19 C and mass m_e = 9.11 × 10^-31 kg) is released from rest at point A, what is it's speed at point B?

We are given that the electric field is

E = E(r)r = (λ/2pe₀r)r ,

where r is the unit radial vector pointing outward from the wire.

The potential difference between two points is defined as

ΔV = V(A) - V(B) = -ò_i^fE • dl ,

where dl is the parallel to the path of integration. Since the electric field is radial, the dot product picks out portion of dl which is parallel to r.

V	= -ò_i^fE • dl
	= -ò_A^B E(r) dr
	= -(λ/2pe₀) ò₁² dr/r
	= -(2kλ) ln(r) \|₁²
	= -(2kλ) ln(2)
	= -2(8.99 × 10⁹ N-m²/C²)(1 × 10^-6 C/m) ln(2)
	= -1.2463 × 10⁴ Volts

The minus sign indicates that point A is at a lower potential than point B. Therefore a negative particle released from rest at point A will accelerate towards point B. Using conservation of energy

½m(v_B)² - ½m(v_a)² = U_A - U_B = qΔV .

Since v_A = 0, solving for v_B yields

v_B	= [2qV/m]^½
	= [2(-1.602 × 10^-19 C)(-1.2463 × 10⁴ V)/(9.11 × 10^-31 kg)]^½
	= 6.6 × 10⁷ m/s

The charge will be moving at 6.6 × 10⁷ m/s when it reaches point B.

Find the electrostatic potential between points A and B which are distances x_A = 2.0 m and x_B = 1.0 m from an infinitely large thin plate with Σ = 1.0 μC/m². The result E = Σ/2ε₀ is useful. If an electron (q = -e = -1.602 × 10^-19 C and mass m_e = 9.11 × 10^-31 kg) is released from rest at point A, what is it's speed at point B?.

We are given that the electric field is

E = E(x)i = (Σ/2ε₀)i ,

where i is the unit vector pointing outward from the plane.

The potential difference between two points is defined as

ΔV = V(A) - V(B) = -ò_i^fE • dl ,

where dl is the parallel to the path of integration. Since the electric field only has an i component, the dot product picks out portion of dl which is parallel to i.

V	= -ò_i^fE • dl
	= -ò_A^B E(x) dx
	= -(Σ/2ε₀) ò₁² dx
	= -(2kΣ) x \|₁²
	= -2kΣ
	= -2(8.99 × 10⁹ N-m²/C²)(1 × 10^-6 C/m²)
	= -5.6486 × 10⁴ Volts

The minus sign indicates that point A is at a lower potential than point B. Therefore a negative particle released from rest at point A will accelerate towards point B. Using conservation of energy

½m(v_B)² - ½m(v_a)² = U_A - U_B = qΔV .

Since v_A = 0, solving for v_B yields

v_B	= [2qV/m]^½
	= [2(-1.602 × 10^-19 C)(-5.6486 × 10⁴ V)/(9.11 × 10^-31 kg)]^½
	= 1.4 × 10⁸ m/s

If started from rest, an electron will be moving at 1.4 × 10⁸ m/s when it reaches point B. This is almost half the speed of light!

In Gauss’ Law question #7, we found the electric field due to a uniformly charge long thick plate to be

where d was the thickness of the plate, ρ₀ was the volume charge density, and a is the distance from the centre of the plate. Determine V(x) relative to V(0) where x is the distance from the centre of the plate. Assume V(0) = 0. Plot V as a function of x.

The potential difference between the two points a = x and a = 0 is defined as

ΔV = V(x) - V(0) = -ò_i^fE • dl ,

where dl is the parallel to the path of integration. The potential difference is independent of the path taken, so one may choose whichever path makes the integration easiest. We will integrate along the line joining the two points so that dl = ida. Setting V(0) = 0, we have

V(x) = -ò_i^fE dl = -ò₀^x E(a) i • i da = -ò₀^x E(a)da .

Since the electric field is symmetric, E(-a) = E(a), for this problem, then V(-x) = V(x). Thus we only need to evaluate V(x) for two regions.

(i) For 0 £ x £ ½d

V(x)	= -ò₀^x E(a)da
	= -ò₀^x (ρ₀/ε₀) a da
	= -(ρ₀/ε₀) ò₀^x a da
	= -(ρ₀/ε₀) ½a² \|₀^x
	= -(ρ₀x² / 2ε₀)

(ii) For x ³ ½d

V(x)	= -ò₀^x E(a)da
	= -{ò₀^½d (ρ₀/ε₀) a da + ò_½d^x (ρ₀d/2ε₀) da }
	= -(ρ₀/ε₀)ò₀^½d a da - (ρ₀d/2ε₀) ò_½d^x da
	= -(ρ₀/ε₀) ½a² \|₀^½d - (ρ₀d/2ε₀) a \|_½d^x
	= -(ρ₀d²/8ε₀) - (ρ₀d/2ε₀)(x - ½d)
	= -(ρ₀d/2ε₀)(x - ¼d)

Thus the potential looks like

Note that V(x) = 0 at x = 0, which is consistent with our assumption that V(0) = 0.

In Gauss’ Law question #8, we found the electric field due to a long thin wire of uniform charge density λ to be E(a) = (λ/2pe₀a)r where a is the radial distance from the wire and r is the unit radial vector. Determine V(r) relative to V(r=1) where r is the distance from the centre of the cylinder. Assume V(1) = 0. Plot V as a function of r.
The potential difference between the two points a = r and a = 1 is defined as

ΔV = V(r) - V(1) = -ò_i^fE • dl ,

where dl is the parallel to the path of integration. The potential difference is independent of the path taken, so one may choose whichever path makes the integration easiest. We will integrate along the line joining the two points so that dl = rda, where r is the unit radial vector. Setting V(1) = 0, we have

V(r) = -ò_i^fE • dl = -ò₁^r E(a) r • r da = -ò₁^r E(a)da .

Evaluating V(r), we find

V(r) = -ò₁^r E(a)da

= -ò₁^r (λ/2ε₀a) da

= -(λ/2ε₀) ò₁^r da/a

= -(λ/2ε₀) ln(a) |₁^r

= -(λ/2ε₀) ln(r)

Note that V(r) is negative for r > 1 and positive for r < 1. The potential looks like

Also note that V(x=1) = 0 consistent with our initial assumption.
In Gauss’ Law question #9, we found the electric field due to a long thin cylindrical shell of radius R and negative surface charge density Σ to be

where the field is radial (i.e. in the r direction). Determine V(r) relative to V(r = 0) where r is the distance from the centre of the cylinder. Assume V(0) = 0. Plot V as a function of r.

The potential difference between the two points a = r and a = 0 is defined as

ΔV = V(r) - V(0) = -ò_i^fE • dl ,

V(r) = -ò_i^fE • dl = -ò₀^r E(a) r • r da = -ò₀^r E(a)da .

Since E(a) has two regions, we need to evaluate V(r) separately for each region.

(i) For 0 £ r £ R

V(r)	= -ò₀^r E(a)da
	= -ò₀^r (0) da
	= 0

(ii) For r ³ R

V(r)	= -ò₀^r E(a)da
	= -ò₀^r E(a) da - ò_R^r E(a) da
	= -ò₀^r (0) da - ò_R^r (R/ε₀) da/a
	= 0 - (R/ε₀) ln(a) \|_R^r
	= - (R/ε₀) ln(r/R)

The potential looks like

Note that V(r=1) = 0 consistent with our initial assumption.

In Gauss’ Law question #10, we found the electric field due to a solid sphere of radius R and total charge Q to be radial and have the form

where the field is radial (i.e. in the r direction). Determine V(r) relative to V(¥) where r is the distance from the centre of the sphere. Assume V(r=¥) = 0. Plot V as a function of r.

The potential difference between the two points a = r and a = ¥ is defined as

ΔV = V(r) - V(¥) = -ò_i^fE • dl ,

V(r) = -ò_i^fE • dl = -ò_¥^r E(a) r • r da = -ò_¥^r E(a)da .

Since E(a) has two regions, we need to evaluate V(r) separately for each region.

(i) For 0 £ r £ R

V(r)	= -ò_¥^r E(a)da
	= -ò_¥^r E(a)da - ò_R^r E(a)da
	= -ò_¥^r (Q/4pe₀) da/a² - ò_R^r (Q/4pe₀R²) ada
	= (Q/4pe₀) 1/a \|_¥^r - (Q/4pe₀R²) ½a² \|_R^r
	= Q/4pe₀R - (Q/8pe₀R²)(r² - R²)
	= 3Q/8pe₀R - Qr²/8pe₀R²

(ii) For r ³ R

V(r)	= -ò_¥^rE(a)da
	= -ò_¥^r (Q/4pe₀) da/a²
	= (Q/4pe₀) 1/a \|_¥^r
	= Q/4pe₀r

The potential looks like

Note that V(r=¥) = 0, consistent with our initial assumption.

In Gauss’ Law question #11, we found the electric field of a thick spherical shell of total charge Q and inner and outer radii of R and 2R to be

where the field is radial (i.e. in the r direction). Determine V(r) relative to V(¥)where r is the distance from the centre of the sphere. Assume V(¥) = 0. Plot V as a function of r.

The potential difference between the two points a = r and a = ¥ is defined as

ΔV = V(r) - V(¥) = -ò_i^fE • dl ,

V(r) = -ò_i^fE • dl = -ò_¥^r E(a) r • r da = -ò_¥^r E(a)da .

Since E(a) has three regions, we need to evaluate V(r) separately for each region.

(i) For 0 £ r £ R

V(r)	= -ò_¥^r E(a)da
	= -ò_¥^2R E(a)da -ò_2R^R E(a)da - ò_R^r E(a)da
	= -ò_¥^2R (Q/4pe₀) da/a² -ò_2R^R (Qa/28pe₀R³-Q/28pe₀a²) da -ò_R^r 0 da
	= (Q/4pe₀) 1/a \|_¥^2R - {(Q/28pe₀R³) ½a² + (Q/28pe₀) 1/a }\|_2R^R
	= Q/8pe₀R - {(Q/56₀R³)(R² - 4R²) + (Q/28pe₀)(1/R - 1/2R)}
	= (9/14)( Q/4pe₀R)

(ii) For R £ r £ 2R

V(r)	= -ò_¥^r E(a)da
	= -ò_¥^2R E(a)da -ò_2R^R E(a)da
	= -ò_¥^2R (Q/4pe₀) da/a² -ò_2R^R (Qa/28pe₀R³-Q/28pe₀a²) da
	= (Q/4pe₀) 1/a \|_¥^2R - {(Q/28pe₀R³) ½a² + (Q/28pe₀) 1/a }\|_2R^r
	= Q/8pe₀R - {(Q/56₀R³)(r² - 4R²) + (Q/28pe₀)(1/r - 1/2R)}
	= (12/14)( Q/4pe₀R) - (1/14)(Qr²/4pe₀R³) - (2/14)(Q/4pe₀r)

(iii) For r ³ 2R

V(r)	= -ò_¥^r E(a)da
	= -ò_¥^r E(a)da
	= -ò_¥^r (Q/4pe₀) da/a²
	= (Q/4pe₀) 1/a \|_¥^r
	= Q/4pe₀r

The potential looks like

Note that V(r=¥) = 0, consistent with our initial assumption.

In Gauss’ Law question #12, we found the electric field due to a thin spherical shell of radius R and surface charge -s to be

where the field is radial (i.e. in the r direction). Determine V(r) relative to V(r=¥) where r is the distance from the centre of the sphere. Assume V(r=¥) = 0. Plot V as a function of r.

The potential difference between the two points a = r and a = ¥ is defined as

ΔV = V(r) - V(¥) = -ò_i^fE • dl ,

V(r) = -ò_i^fE • dl = -ò_¥^r E(a) r • r da = -ò_¥^r E(a)da .

Since E(a) has two regions, we need to evaluate V(r) separately for each region.

(i) For 0 £ r £ R

V(r)	= -ò_¥^r E(a)da
	= -ò_¥^R E(a)da - ò_R^r E(a)da
	= -ò_¥^R (ΣR²/ε₀) da/a² - ò_R^r 0 da
	= (ΣR²/ε₀) 1/a \|_¥^R
	= ΣR/ε₀

(ii) For r ³ R

V(r)	= -ò_¥^r E(a)da
	= -ò_¥^r (ΣR²/ε₀) da/a²
	= (ΣR²/ε₀) 1/a \|_¥^r
	= ΣR² / ε₀r

The potential looks like

Note that V(r=¥) = 0, consistent with our initial assumption.

The electric field due to a large plate is E = Σ/2ε₀ as we have seen. A capacitor consists of two identical plates with equal but opposite charge distributions.

_net

₀

(a) Let's first sketch the field due to each plate. The electric field for the positive plate is shown by the dashed red lines leaving the plate. The electric field for the negative plate is shown by the solid green red lines entering the plate.

As can be seen, there are an equal number of solid and dashed lines. Outside the plates, the dashed and solid lines are anti-parallel and thus cancel. Between the plates, both sets of lines point in the same direction and thus add. To be more formal, we can make use of Gauss' Law,

òE • ndA = Q_enclosed/ε₀ ,

where n is the outward-looking normal on each side of the surface. The pillbox has two ends and one side, so that the integral has three parts

òE • ndA = ò_leftE • n_left dA + ò_sideE • n_sidedA + ò_rightE • n_rightdA .

By symmetry the field on either side is horizontal, with no lines crossing the sides of the "gaussian pillbox". Thus E • n_side = 0 and the integral for the side of the pillbox vanishes. Symmetry also demands that if the electric field is not zero it must be equal and opposite on either side of the capacitor. Thus E • n_left = E • n_right = E_outside. Hence the integrals reduce to

ò_leftE • n_left dA + ò_sideE • n_sidedA + ò_rightE • n_rightdA = 2E_outsideòdA .

The integral over the area is just the area of the end of the pillbox, A. So Gauss's law is

2E_outsideA = Q_enclosed/ε₀ .

Since the plate have equal and opposite charges, we know Q_enclosed = 0. Hence E_outside = 0 as we would expect from the diagram.

(b) The potential difference between the two points x_final = d and x_initial = 0 is defined as

ΔV = -ò_i^fE • dl,

E_between = Σ/2ε₀ + Σ/2ε₀ = Σ/ε₀ .

Hence we find the potential difference to be

V(r) = -ò_i^fE • dl = -ò₀^d E_between i • i da = -E_betweenò₀^d da = -E_betweend = -Σd/ε₀ .

So the potential difference between the plates is directly proportional to the separation d of the plates. The minus sign indicates that the negative plate is at a lower potential than the positive plate which is what we would expect.

Two point charges of q₁ = 3.0 μC and q₂ = 4.0 μC are situated at the opposite corners of a rectangle as shown below. The short side has length L = 0.25 m. Find the total potential at points A and B. If a free particle of charge q_f= 1.0 μC and mass M = 15 g has speed v_A = 2.50 m/s at point A and it follows the indicated path, what will be its speed at point B?

Points A and B are at different electrostatic potentials because of the arrangement of the charges. A positive charge will speed up if it moves from high potential to low or slow down if it moves from low potential to high. It does this because a high potential area has relatively more positive charge in the area than the low potential area and a positive charge is thus repelled.

So to find v_B, we find determine the potential at A and B. The potential at point A is given by the two fixed point charges and is equal to the sum of the potential of each charge

V_A	= kq₁/L + kq₂/2L
	= (8.99 × 10⁹ N-m²/C²)[(3 × 10^-6C)/(0.25 m) + (3 × 10^-6C)/(0.50 m)]
	= 1.798 × 10⁵ Volts

Similarly the potential at point B is

V_B	= kq₁/2L + kq₂/L
	= (8.99 × 10⁹ N-m²/C²)[(3 × 10^-6C)/(0.50 m) + (3 × 10^-6C)/(0.25 m)]
	= 1.978 × 10⁵ Volts

Using Conservation of Energy,

½m(v_B)² - ½m(v_A)² = q(V_A - V_B) .

Solving for v_B, we find

v_B	= [(v_A)² + 2q(V_A - V_B)/m]^½
	= [(2.50 m/s)² + 2(1.00 × 10^-6 C)(1.79810⁵ - 1.97810⁵ Volts)/(0.015 kg]^½
	= 2.25 m/s

Five charges are arranged as shown below. What is the electrostatic potential energy of each configuration? The separation between charges is L.

The electrostatic potential of a charge configuration is given by the formula

The charges in the diagrams below have been numbered and various value of r_ij have been shown.

(a)

U =

K(Q)(-Q)/L + K(Q)(Q)/2L + K(Q)(-Q)/3L + K(Q)(Q)/4L + K(Q)(-Q)/5L +

K(-Q)(Q)/L + K(-Q)(-Q)/2L + K(-Q)(Q)/3L + K(-Q)(-Q)/4L +

K(Q)(-Q)/L + K(Q)(Q)/2L + K(Q)(-Q)/3L +

K(-Q)(Q)/L + K(-Q)(-Q)/2L +

K(Q)(-Q)/L

-(35/12) KQ²/L

(b)

U =

K(Q)(-Q)/L + K(Q)(Q)/2L + K(Q)(Q)/3L + K(Q)(Q)/4L + K(Q)(-Q)/5L +

K(-Q)(Q)/L + K(-Q)(Q)/2L + K(-Q)(Q)/3L + K(-Q)(-Q)/4L +

K(Q)(Q)/L + K(Q)(Q)/2L + K(Q)(-Q)/3L +

K(Q)(Q)/L + K(Q)(-Q)/2L +

K(Q)(-Q)/L

+(3/4) KQ²/L

In case (a) net energy was gained in assembling the charges while work had to be done to assemble the charges in (b). If left free to move, the charges in (a) would collapse in on themselves and would explode away from each other in (b).

A rod is bent into a semi-circular arc of radius R. The rod has a uniform linear charge distribution λ. Find the potential at the centre of the arc, point P, by direct integration. The identity S= Rθ and dS = Rdθ may be of use.

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

dq = λdS = Rdθ .

The electrostatic potential is a scalar given by

dV = kdq/R = kλdθ .

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

ΔV	= ò₀^π dV
	= (λk) ò₀^π dθ
	= (plk)

Note that the potential at the centre of the arc is independent of the radius of the arc. As well, since the potential of a point charge is set to be zero at r = #163;, and we built the arc out of point charge infinitesimals, this potential is also relative to infinity.

A total charge Q is distributed uniformly along a straight rod of length L. Find the potential at a point P at a distance h from the midpoint of the rod. (Hint: ò[x²+k²]^-½dx = ln[x+[x²+k²]^½] + C). Use the gradient with respect to y to find the electric field at that point.

First, the wire has a uniform linear charge density = 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 = dx = (Q/L)dx .

The electrostatic potential is a scalar given by

dV = kdq/r = k(Q/L)dx/r .

Before we can integrate, we need to find r in terms of x and y. Using trigonometry,

r = [y² + x²]^½ .

Hence

dV = (kQ/L)dx/[y² + x²]^½ .

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

ΔV	= ò_-½L^½L dV
	= (kQ/L) ò_-½L^½L dx / [a² + x²]^½
	= (kQ/L) ln\|x + [y²+x²]^½\| \|_-½L^½L }
	= (kQ/L) ln\|(½L + [y²+¼L²]^½)/(-½L + [y²+¼L²]^½)\|

The electric field along the y-axis is given by

E_y = -j¶V/¶y = +j(KQ/y)[y²+¼L²]^-½ .

This is the same result as question 2 in the Gauss' Law handout.

Three thin rods of glass of length L carry charges uniformly distributed along their lengths. The charges on the three rods are +Q, +Q, and -Q, respectively. The rods are arranged along the sides of an equilateral triangle. What is the electrostatic potential at the midpoint of this triangle? (Hint: Use the result of the question 14.)

First we use trigonometry to find y in terms of L, y = L / 2tan(30°) = L / 2Ö3.

The potential of the triangle is the sum of the potentials due to its sides

V	= V₁ + V₂ + V₃
	= (K/L)(Q + Q + -Q) ln\|(½L + [y²+¼L²]^½)/(-½L + [y²+¼L²]^½)\|
	= (KQ/L) ln\|(L/2 + [L²/12+L²/4]^½)/(-L/2 + [L²/12+L^2+/4]^½)\|
	= (kQ/L) ln\| (2 + Ö3) / (2 - Ö3)\|

The electric potential over a certain region is given by V = 3x²y-4xz-5y² volts. Determine the components of the electric field and evaluate at the point (+1,0,+2).

The electric field is given by the negative gradient of the electrostatic potential

E	= -ÑV
	= -i¶V/¶x + -j¶V/¶y + -k¶V/¶z
	= -i(6xy - 4z) + -j(3x² - 10y) + -k(-4x)

At (1,0,2) the electric field is

E = 8i - 3 j + 4k.

Over a certain region of space, the electric potential is V = 5x-3x²y+2yz². Determine the components of the electric field and evaluate electric field at the point (1,0,-2).

The electric field is given by the negative gradient of the electrostatic potential

E	= -ÑV
	= -i¶V/¶x + -j¶V/¶y + -k¶V/¶z
	= -i(5 - 6xy) + - j(-3x² + 2z²) + -k(4yz)

At (1,0,-2) the electric field is

E = -5i - 5j .

Questions?mike.coombes@kwantlen.ca

V	= V₁ + V₂ + V₃
	= (K/L)(Q + Q + -Q) ln\|(½L + [y²+¼L²]^½)/(-½L + [y²+¼L²]^½)\|
	= (KQ/L) ln\|(L/2 + [L²/12+L²/4]^½)/(-L/2 + [L²/12+L^2+/4]^½)\|
	= (kQ/L) ln\| (2 + Ö3) / (2 - Ö3)\|

ΔV	= -ò_i^fE • dl
	= -ò₀⁴ 5x²i • i dx
	= -ò₀⁴ 5x²dx
	= -(5/3) x³ \|₀⁴
	= -320/3 Volts

V(r)	= -ò₁^r E(a)da
	= -ò₁^r (λ/2ε₀a) da
	= -(λ/2ε₀) ò₁^r da/a
	= -(λ/2ε₀) ln(a) \|₁^r
	= -(λ/2ε₀) ln(r)