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

Explanation:

Relation between electric potential and charge is as follows.

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

where,     V = electric potential

                q = charge

                r = distance

Putting the given values into the above formula as follows.

              $V = k \frac{q}{r}$

              $18 = 9 \times 10^{9} \times \frac{q}{1.20 }$

             q = $2.4 \times 10^{-9}$ C

Hence, charge on the sphere is $2.4 \times 10^{-9}$ C

Now, we will use the same formula  in order to calculate the electric potential as follows.

                 V = $k \frac{q}{R}$

where,       R = radius of the sphere

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

                 = 36 V

Thus, we can conclude that potential at the center of the given sphere is 36 V .