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 are respectively:
(A) Zero and $\dfrac{Q}{{4\pi {\varepsilon _0}{R^2}}}$
(B) $\dfrac{Q}{{4\pi {\varepsilon _0}R}}$ and zero
(C) $\dfrac{Q}{{4\pi {\varepsilon _0}R}}$ and $\dfrac{Q}{{4\pi {\varepsilon _0}{R^2}}}$
(D) Both are zero

Answer 1

A conducting sphere has all its charges residing at the surface of the sphere. The Electric field can be considered as the gradient of the potential.

Formula used: In this solution we will be using the following formula;
$\Rightarrow E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{{{r^2}}}$ where $E$ is the electric field, $Q$ is the charge whose electric field at a point is of interest, $r$ is the distance considered, and ${\varepsilon _0}$ is the permittivity of free space.
$\Rightarrow EA = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$ where $A$ is the surface area of a Gaussian surface, ${Q_{enc}}$ is the charge enclosed by the surface.
$\Rightarrow V = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{r}$ where $V$ is the electric potential.
$\Rightarrow E = \dfrac{{dV}}{{dr}}$ where $E$ is the electric field, and $\dfrac{{dV}}{{dr}}$ shows the differential rate of electric potential with radial distance.

Complete step by step answer
In the question, a conducting sphere is to possess a charge $Q$, the electric field and electric potential at the surface are to be found at the centre.
A conducting sphere always has its charges concentrated at the surface of the sphere, hence at the centre there are no charges. Thus, using gauss’s law given in its constant form as
$EA = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}}$ where $E$ is the electric field, $A$ is the surface area of a Gaussian surface, and ${\varepsilon _0}$ is the permittivity of free space. Hence, since there are no charges at the centre ${Q_{enc}} = 0$ (picking a small spherical Gaussian surface around centre) then
$\Rightarrow EA = \dfrac{{{Q_{enc}}}}{{{\varepsilon _0}}} = 0$
$\Rightarrow E = 0$
The electric field can be given as
$\Rightarrow E = \dfrac{{dV}}{{dr}}$ where $\dfrac{{dV}}{{dr}}$ shows the differential rate of electric potential with radial distance.
Since, $E = 0$ then $V = {\text{constant}}$ (integrating zero gives a constant), and it is given by
$\Rightarrow V = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{Q}{R}$
Hence, the correct answer is B.

Note
For clarity the value for $V$ can be gotten by integrating the Electric field from infinity to zero because electric potential at any point can be defined as the work done per unit charge in moving the charge from infinity to that point. Since, the electric field outside the sphere and inside are different it is given as
$V = \left( {\int_0^R {{E_{r < R}}dr} + \int_R^\infty {{E_{r > R}}} dr} \right)$
$\Rightarrow V = \left[ K \right]_0^R + \left[ { - \dfrac{Q}{{4\pi {\varepsilon _0}r}}} \right]_R^\infty$ since electric field at any point $r > R$ is $\dfrac{Q}{{4\pi {\varepsilon _0}r}}$, and at $r < R$ is zero
Hence,
$V = \left( {K - K} \right) + \left( { - \dfrac{Q}{{4\pi {\varepsilon _0}\infty }} - \left( { - \dfrac{Q}{{4\pi {\varepsilon _0}R}}} \right)} \right)$
$\Rightarrow V = \dfrac{Q}{{4\pi {\varepsilon _0}R}}$.