Problem

A hollow sphere is charged to $14 \, \mu \mathrm{C}$. Find the potential:

(a) at its surface,
(b) inside the sphere, and
(c) at a distance of $0.2 \, \mathrm{m}$ from the surface.

The radius of the sphere is $0.3 \, \mathrm{m}$.

Data

• Charge: $q = 14 \, \mu \mathrm{C} = 14 \times 10^{-6} \, \mathrm{C}$
• Radius: $r = 0.3 \, \mathrm{m}$
• Distance from surface: $d = 0.2 \, \mathrm{m}$
• Coulomb’s constant: $k = 9 \times 10^9 \, \mathrm{Nm^2C^{-2}}$

Prerequisite Concepts

• Potential at Surface:

$$V_{\text{surface}} = k \frac{q}{r}$$

• Potential Inside: $V_{\text{inside}} = V_{\text{surface}}$
• Potential at Distance:

$$V = k \frac{q}{r + d}$$

Answer

Part (a): Potential at Surface

$$V_{\text{surface}} = k \frac{q}{r}$$

Substitute values:

$$V_{\text{surface}} = \frac{(9 \times 10^9)(14 \times 10^{-6})}{0.3}$$ $$V_{\text{surface}} = 42 \times 10^4 \, \mathrm{V}$$ $$V_{\text{surface}} = 4.2 \times 10^5 \, \mathrm{V}$$

Part (b): Potential Inside

The potential inside a hollow sphere is the same as that on the surface:

$$V_{\text{inside}} = V_{\text{surface}} = 4.2 \times 10^5 \, \mathrm{V}$$

Part (c): Potential at Distance $0.2 \, \mathrm{m}$ from Surface

The total distance from the center is $r + d = 0.3 + 0.2 = 0.5 \, \mathrm{m}$.

$$V = k \frac{q}{r + d}$$

Substitute values:

$$V = \frac{(9 \times 10^9)(14 \times 10^{-6})}{0.5}$$ $$V = 25.2 \times 10^4 \, \mathrm{V}$$ $$V = 2.52 \times 10^5 \, \mathrm{V}$$

Final Answer

• (a) Potential at surface: $4.2 \times 10^5 \, \mathrm{V}$.
• (b) Potential inside: $4.2 \times 10^5 \, \mathrm{V}$.
• (c) Potential at $0.2 \, \mathrm{m}$ from surface: $2.52 \times 10^5 \, \mathrm{V}$.

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