Question

A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 18.0 V . Assume that V = 0 at an infinite distance from the sphere.

What is the electric potential at the center of the sphere?

Answer 1

Answer:

The electric potential at the center of the sphere is 36 V.

Explanation:

Given that,

Radius R= 0.600 m

Distance D = 1.20 m

Electric potential = 18.0 V

We need to calculate the electric potential

Using formula of electric potential

$V=\dfrac{kq}{r}$

Put the value into the formula

$18.0=\dfrac{9\times10^{9}\times q}{1.20}$

$q=\dfrac{18.0\times1.20}{9\times10^{9}}$

$q=2.4\times10^{-9}\ C$

We need to calculate the electric potential at the center of the sphere

Using formula of potential

$V=\dfrac{kq}{r}$

Put the value into the formula

$V=\dfrac{9\times10^{9}\times2.4\times10^{-9}}{0.600}$

$V=36\ V$

Hence, The electric potential at the center of the sphere is 36 V.

Answer 2

The electric potential at the center of the sphere is 36 V.

The given parameters;

• radius of the, R = 0.6 m
• distance of the charge, r = 1.2  m
• potential difference, V = 18 V

The magnitude of the charge is calculated as follows;

$V = \frac{kq}{r} \\\\q = \frac{Vr}{k} \\\\q = \frac{18 \times 1.2}{9\times 10^9} \\\\q = 2.4 \times 10^{-9} \ C$

The electric potential at the center of the sphere;

$V = \frac{kq}{r} \\\\V = \frac{9\times 10^9 \times 2.4 \times 10^{-9}}{0.6} \\\\V = 36 \ V$

Thus, the electric potential at the center of the sphere is 36 V.

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