Question

A charge of 2 μC is uniformly distributed on a spherical shell of radius 5 m. Electric potential at the centre of sphere is

Answer 1

Given:

A charge of 2 μC is uniformly distributed on a spherical shell of radius 5 m.

To find:

Electric potential at the centre of the sphere?

Calculation:

• First of all, since no charge is enclosed inside shell, the field intensity inside the shell will be zero.

• And potential on the surface of the shell will be kq/r (r - radius of sphere)

Now, we can say:

$\rm \displaystyle \: \int dV = - \int E dr$

Putting the limits:

$\rm \implies \displaystyle \: \int_{V_{s}}^{V_{c}} dV = - \int_{r}^{0} E dr$

• V_(c) is potential at centre, V_(s) is potential at surface.

$\rm \implies \displaystyle \: \int_{V_{s}}^{V_{c}} dV = \int_{0}^{r} E dr$

$\rm \implies \displaystyle V_{c} - V_{s} = \int_{0}^{r} E dr$

Now, value of E inside shell is 0.

$\rm \implies \displaystyle V_{c} - V_{s} = \int_{0}^{r} (0)dr$

$\rm \implies \displaystyle V_{c} - V_{s} = 0$

$\rm \implies \displaystyle V_{c} = V_{s}$

$\rm \implies \displaystyle V_{c} = \dfrac{kq}{r}$

$\rm \implies \displaystyle V_{c} = \dfrac{(9 \times {10}^{9} ) \times (2 \times {10}^{ - 6}) }{5}$

$\rm \implies \displaystyle V_{c} = 3.6 \times {10}^{3} \: volt$

So, potential at centre is 3.6 × 10³ V.