Question

A hollow spherical conductor with a \(5.00-\mathrm{cm}\) radius has a surface charge of \(8.00 \mathrm{nC}\) a) What is the potential \(8.00 \mathrm{~cm}\) from the center of the sphere? b) What is the potential \(3.00 \mathrm{~cm}\) from the center of the sphere? c) What is the potential at the center of the sphere?

Short answer

Answer: The electric potentials are 899 V at 8.00 cm, 1438 V at 3.00 cm, and 1438 V at the center.

Step-by-step solution

Convert units

First, we need to convert the given distances and charge into their corresponding SI units: \(r_1 = 8.00\ \mathrm{cm} = 0.08\ \mathrm{m}\) \(Q = 8.00\ \mathrm{nC} = 8.00\times10^{-9}\ \mathrm{C}\)

Calculate potential

Next, we use the formula for the electric potential due to a point charge: \(V_1 = \frac{kQ}{r_1} = \frac{8.99\times10^9 \ \mathrm{N\ m^2/C^2} \cdot 8.00\times10^{-9}\ \mathrm{C}}{0.08\ \mathrm{m}} = 899\ \mathrm{V}\) The potential \(8.00\ \mathrm{cm}\) from the center of the sphere is \(899\ \mathrm{V}\). #b) Potential 3.00 cm from the center of the sphere#

Convert units

First, we need to convert the given distance into its corresponding SI unit: \(r_2 = 3.00\ \mathrm{cm} = 0.03\ \mathrm{m}\)

Determine potential inside the sphere

Since the sphere is a conductor and \(r_2\) is smaller than the sphere's radius, the charge will distribute uniformly on the surface. Hence, the potential inside the sphere is constant and equal to the potential at the sphere's surface. To calculate the potential at the surface, we can use the formula for the electric potential due to a point charge at a distance equal to the sphere's radius: \(r_s = 5.00\ \mathrm{cm} = 0.05\ \mathrm{m}\) \(V_s = \frac{kQ}{r_s} = \frac{8.99\times10^9 \ \mathrm{N\ m^2/C^2} \cdot 8.00\times10^{-9}\ \mathrm{C}}{0.05\ \mathrm{m}} = 1438\ \mathrm{V}\) Thus, the potential \(3.00\ \mathrm{cm}\) from the center of the sphere is also \(1438\ \mathrm{V}\). #c) Potential at the center of the sphere#

Determine potential at the center

Since the potential inside the conductor is constant and equal to the potential at the sphere's surface, the potential at the center of the sphere is also \(1438\ \mathrm{V}\).