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Question \#7: You construct a spherical conducting shell with an excess charge ...

Jun 14, 2024

Solution by Steps

step 1

To solve the problem, we first need to draw a picture of the spherical conducting shell with the given radii and charge

step 2

The inner radius

R_1

10 \, \text{cm}

and the outer radius

R_2

12 \, \text{cm}

. The excess charge on the shell is

4.5 \, \text{C}

step 3

For part (b), we need to determine the voltage in each region. We will use the reference voltage at infinity,

V(\infty) = 0

step 4

Inside the shell (r < R_1), the electric field is zero because it is a conductor. Therefore, the voltage is constant and equal to the voltage at

R_1

step 5

For

R_1 \leq r \leq R_2

, the voltage

V(r)

can be found using Gauss's law. The electric field

E

in this region is given by

E = \frac{Q}{4 \pi \epsilon_0 r^2}

step 6

Integrating the electric field to find the voltage, we get

V(r) = \frac{Q}{4 \pi \epsilon_0} \left( \frac{1}{R_1} - \frac{1}{r} \right)

step 7

Outside the shell (r > R_2), the shell can be treated as a point charge. The voltage

V(r)

is given by

V(r) = \frac{Q}{4 \pi \epsilon_0 r}

step 8

For part (c), we need to draw a plot of the voltage as a function of distance. The plot will show a constant voltage inside the shell, a decreasing voltage between

R_1

and

R_2

, and a further decreasing voltage outside the shell

step 9

For part (d), we need to determine the voltage at each boundary and the origin. At

r = 0

V(0) = V(R_1)

. At

r = R_1

V(R_1) = \frac{Q}{4 \pi \epsilon_0 R_1}

. At

r = R_2

V(R_2) = \frac{Q}{4 \pi \epsilon_0} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

Answer

The voltage at

r = 0

is the same as at

R_1

. The voltage at

R_1

\frac{Q}{4 \pi \epsilon_0 R_1}

. The voltage at

R_2

\frac{Q}{4 \pi \epsilon_0} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

Key Concept

Voltage distribution in a spherical conducting shell

Explanation

The voltage inside a spherical conducting shell is constant, decreases between the inner and outer radii, and further decreases outside the shell.