Equipotential Surfaces and Potential Gradient Problems and Solutions 2

Q#23.48

A metal sphere with radius $r_a$ = 1.20 cm is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_b$ = 9.60 cm. Charge +q is put on the inner sphere and charge $-q$ on the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 500 V with the inner sphere at higher potential. 

(a) Use the result of Exercise 23.47 (b) to calculate q.

(b) With the help of the result of Exercise 23.47(a), sketch the equipotential surfaces that correspond to 500, 400, 300, 200, 100, and 0 V.

(c) In your sketch, show the electric field lines. Are the electric field lines and equipotential surfaces mutually perpendicular? Are the equipotential surfaces closer together when the magnitude of $\vec{E}$ is largest? 

Answer:

shows that $V=\frac{1}{4\pi \epsilon_0}q\left(\frac{1}{r_a}-\frac{1}{r_b}\right)$ for $r <r_a$, 

$V=\frac{1}{4\pi \epsilon_0}q\left(\frac{1}{r}-\frac{1}{r_b}\right)$ for $r_a<r<r_b$ and 

$V_{ab}=\frac{1}{4\pi \epsilon_0}q\left(\frac{1}{r_a}-\frac{1}{r_b}\right)$

$E=\frac{kq}{r}$, radially outward, for $r_a\leqslant r \leqslant r_b$

(a) $V=\frac{1}{4\pi \epsilon_0}q\left(\frac{1}{r_a}-\frac{1}{r_b}\right)$ = 500 V gives 

$q=\frac{500 \ V}{k\left(\frac{1}{0.012 \ m}-\frac{1}{0.096}\right)}$

$q=7.62 \times 10^{-10} \ C$

(b) $V_b$ = 0 so $V_a$ = 500 V. The inner metal sphere is an equipotential with V = 500 V.

$\frac{1}{r}=\frac{1}{r_a}+\frac{V}{kq}$

V = 400 V at r = 1.45 cm, V = 300 V at r =1.85 cm, V = 200 V at r = 2.53 cm, V =100 V at r = 4.00 cm, V = 0 at r = 9.60 cm.  

The equipotential surfaces are sketched in Figure 23.48. 

(c) The equipotential surfaces are concentric spheres and the electric field lines are radial, so the field lines and equipotential surfaces are mutually perpendicular. The equipotentials are closest at smaller r, where the electric field is largest.

Q#23.49

A very long cylinder of radius carries a uniform charge density of 1.50 nC/m.

(a) Describe the shape of the equipotential surfaces for this cylinder. 

(b) Taking the reference level for the zero of potential to be the surface of the cylinder, find the radius of equipotential surfaces having potentials of 10.0 V, 20.0 V, and 30.0 V.

(c) Are the equipotential surfaces equally spaced? If not, do they get closer together or farther apart as r increases?

Answer:

Outside the cylinder it is equivalent to a line of charge at its center. 

The difference in potential between the surface of the cylinder (a distance R from the central axis) and a general point a distance r from the central axis is given by 

$\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(r/R)$

for r, gives $r=Re^{2\pi \epsilon_0 \Delta V/\lambda}$

For 10 V, the exponent is

$\frac{2\pi \epsilon_0 \Delta V}{\lambda}=\frac{2\pi(8.85 \times 10^{-12} \ C^2/N.m^2)(10 \ V)}{1.50 \times 10^{-9} \ C/m}$

which gives $r=(2.00 \ cm)e^{0.370}$ = 2.90 cm.

Likewise, the other radius are 4.20 cm (for 20 V) and 6.08 cm (for 30 V)

(c) $\Delta r_1=2.90 \ cm - 2.00 \ cm$ = 0.90 cm

$\Delta r_2=4.20 \ cm - 2.90 \ cm$ = 1.30 cm

$\Delta r_3=6.08 \ cm - 4.20 \ cm$ = 1.88 cm

As we can see, Δr increases, so the surfaces get farther apart. This is very different from a sheet of charge, where the surfaces are equally spaced planes.

Share :