Question

An isolated solid metallic sphere of radius $R$ is given an electric charge. Which of the graphs below correctly shows the way in which the electric field $E$ varies with distance $x$ from the centre of the sphere?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let us consider a spherical Gaussian surface at a distance $x$ from the centre of the solid metallic sphere.


Due to symmetry of the Gaussian surface w.r.t. the sphere containing charge, the direction of $\overrightarrow{E}$ will be radially outwards and of the same magnitude everywhere on the Gaussian surface.
For the region outside the sphere i.e $x\ge R$, applying Gauss's law:
$\oint \overrightarrow{E}\cdot \overrightarrow{A}=EA=\frac{q_{in}}{{ϵ}_0}$
$(\because q_{in}=q\text{for}x\ge R)$
$\Rightarrow E\left(4\pi x^2\right)=\frac{q}{{ϵ}_0}$
$\Rightarrow E=\frac{q}{4\pi {ϵ}_0{x}^2}$
$\Rightarrow E=\frac{kq}{x^2}$
$\Rightarrow E\propto \frac{1}{x^2}$
Thus, graph of $Evsx$ will be hyperbolic for $x\ge R$.

Applying Gauss's law for region inside sphere $(x<r)$, the net charge enclosed by the Gaussian surface will be zero.
($\because$ charge resides only on the surface of a metallic conductor)

$\Rightarrow E\left(4\pi x^2\right)=\frac{q_{in}}{{ϵ}_0}$
$\Rightarrow E\left(4\pi x^2\right)=0$
$\therefore E=0$
i.e Inside the metallic sphere, electric field is zero.

Therefore,
$E=0\text{for}x<R$
$E=\frac{kq}{x^2}\text{for}x\ge R$
The graph of electric field $E$ vs radial distance $x$ will be as follows:


Hence, option (c) is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Caution: At the surface of the metallic sphere,}\\ \text{an electric field exists due to the presence}\\ \text{of charge. Thus,}E=\frac{kq}{R^2}\text{at the surface}(x=R)\\ \text{The electric field is zero inside the metallic sphere.}\end{array}$