Question

A sphere of radius $R$ has electric charge uniformly distributed in its entire volume. At a distance $x$ from its centre for $x<R$, the electric field is directly proportional to

• A. $\frac{1}{x^2}$
• B. $\frac{1}{x}$
• C. $x$
• D. $x^2$

Answer 1

The correct option is C $x$

Volume charge density of sphere is,
$\rho =\frac{Q}{V}$
where,
$Q$ is the total charge distributed on the entire volume of sphere,
$V$ is the volume of the sphere.

$\Rightarrow \rho =\frac{Q}{\frac{4}{3}\pi R^3}$

Let us construct a spherical surface at a distance $x$ from centre of sphere $(x<R)$.


Applying Gauss's law,
$$\varphi =\frac{q_{in}}{{ϵ}_o}$$
here, charge enclosed by the sphere of radius $x$,
$q_{in}=\left(\frac{4}{3}\pi x^3\right)$ and electric flux through the sphere,

$\Rightarrow \varphi =\overrightarrow{E}.d\overrightarrow{A}$ = $\frac{\rho \left(\frac{4}{3}\pi x^3\right)}{{ϵ}_o}$

$\Rightarrow EA=\frac{\rho \left(\frac{4}{3}\pi x^3\right)}{{ϵ}_o}$

$\Rightarrow E\times 4\pi \times x^2=\frac{\rho \left(\frac{4}{3}\pi x^3\right)}{{ϵ}_o}$

Substituting the value of $\rho$,
$\Rightarrow E\times 4\pi \times x^2=\frac{\left(\frac{3Q}{4\pi R^3}\right)\left(\frac{4}{3}\pi x^3\right)}{{ϵ}_o}$

$\Rightarrow E=\frac{Qx}{4\pi {ϵ}_o{R}^3}$

Thus, for given radius $\left(R\right)$ and charge distribution $\left(Q\right)$,
$$E\propto x$$Hence, option (c) is correct answer.