Question

A metal sphere of radius 0.16m carries an electric charge of +0.5mewC, which is spread uniformly over the spheres surface. This sphere is fixed, by means of insulating supports, at the center of a larger metal shell, with inner radius 0.68m and outer radius 5.49m. Now a charge of -0.54mewC is placed on the metal shell and distribute itself uniformly over the shells inner surface.

a) what is the potential on the outer surface of the spherical shell, in megavolt? you can assume the potential is zero at infinity.
b) what is the potential on the surface of the sphere, in megavolts?

Answer 1

a. The potential on the outer surface of the spherical shell is approximately
`\( 818.5 \, \text{kV} \)`.

b. The potential on the surface of the sphere is approximately
`\( 28.11 \, \text{MV} \)`.

To solve this problem, you can use the formula for the electric potential
`\( V \)` due to a point charge:

`\[ V = (k \cdot |Q|)/(r) \]`

where:

• 
  `\( V \)` is the electric potential,
• 
  `\( k \)` is Coulomb's constant
  `(\( k \approx 8.9875 * 10^9 \, \text{N m}^2/\text{C}^2 \))`,
• 
  `\( Q \)` is the charge, and
• 
  `\( r \)` is the distance from the charge.

part a):

a) Potential on the outer surface of the spherical shell:

The charge on the inner surface of the shell induces an equal and opposite charge on the outer surface due to the conductive property of the metal. Therefore, the potential on the outer surface of the shell is due to the charge on the sphere alone.

Given:

• Charge on the sphere,
  `\( Q_{\text{sphere}} = +0.5 \, \mu C = 5 * 10^(-7) \, C \)`
• Radius of the spherical shell,
  `\( r_{\text{outer}} = 5.49 \, \text{m} \)`

Using the formula for the potential:

`\[ V_{\text{outer}} = \frac{k \cdot |Q_{\text{sphere}}|}{r_{\text{outer}}} \]`

`\[ V_{\text{outer}} = \frac{8.9875 * 10^9 \, \text{N m}^2/\text{C}^2 * 5 * 10^(-7) \, \text{C}}{5.49 \, \text{m}} \]`

`\[ V_{\text{outer}} \approx 8.185 * 10^5 \, \text{V} = 818.5 \, \text{kV} \]`

Therefore, the answer is approximately
`\( 818.5 \, \text{kV} \)`.

part b):

b) Potential on the surface of the sphere:

The potential on the surface of the sphere is due to its own charge.

Given:

• Charge on the sphere,
  `\( Q_{\text{sphere}} = +0.5 \, \mu C = 5 * 10^(-7) \, C \)`
• Radius of the sphere,
  `\( r_{\text{sphere}} = 0.16 \, \text{m} \)`

Using the formula for the potential:

`\[ V_{\text{sphere}} = \frac{k \cdot |Q_{\text{sphere}}|}{r_{\text{sphere}}} \]`

`\[ V_{\text{sphere}} = \frac{8.9875 * 10^9 \, \text{N m}^2/\text{C}^2 * 5 * 10^(-7) \, \text{C}}{0.16 \, \text{m}} \]`

`\[ V_{\text{sphere}} \approx 2.811 * 10^7 \, \text{V} = 28.11 \, \text{MV} \]`

Therefore, the answer is approximately
`\( 28.11 \, \text{MV} \)`.