Question

A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are:
A. zero and $\dfrac{Q}{{4\pi { \in _0}{R^2}}}$
B. $\dfrac{Q}{{4\pi { \in _0}R}}$ and zero
C. $\dfrac{Q}{{4\pi { \in _0}R}}$ and $\dfrac{Q}{{4\pi { \in _0}{R^2}}}$
D. both are zero

Answer 1

Hint: In a conducting sphere, the charge is distributed over the whole surface of the sphere. The electric field can be found using the Gauss law and electric potential is related to electric as electric field is space derivative of the electric potential.

Detailed step by step solution:
We are given a conducting sphere of radius R which is given a charge Q. This charge gets evenly distributed on the surface of this sphere.
First, let's find the electric field due to this charged sphere. We know that the Gauss’ law is given as
$\oint {\overrightarrow E .\overrightarrow {dS} } = \dfrac{{{Q_{net}}}}{{{ \in _0}}}$
Here E signifies the electric field passing through a certain area dS. ${Q_{net}}$ is the total amount of charge and ${ \in _0}$ signifies the permittivity of the vacuum.
Now as we know that net charge inside the sphere is zero because charge is only on the surface of the sphere. Therefore, for a Gaussian sphere drawn inside the sphere we have
${Q_{net}} = 0 \\\ \Rightarrow E = 0 \\\$
Hence the electric field at the centre of the sphere is equal to zero.
Secondly, the electric potential is given in terms of the electric field as follows:
$E = \dfrac{{dV}}{{dr}}$
Now if E = 0 at centre then V = constant at the centre of the sphere. Therefore the electric potential due to charge Q on the surface on sphere at distance equal to the radius of sphere is given as
$V = \dfrac{Q}{{4\pi { \in _0}R}}$
Hence, the correct answer is option B.

Note: The student must note that if the sphere is solid then the electric field inside the sphere is not equal to zero. According to Gauss law, the electric field is non-zero if the charge enclosed within a Gaussian surface is also non-zero.