Question

A sphere of radius $R$ has a uniform distribution of electric charge in its volume. At a distance $x(<R)$ from its centre, 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 $x$

Let the total charge in the sphere be $Q$.
Volume charge density of the sphere is, $\rho =\frac{Q}{\frac{4}{3}\pi R^3}$

Let us take a spherical Gaussian surface at a distance $x$ from the centre of the sphere $(x<R)$.
Applying Gauss's law,
$\oint \overrightarrow{E}.\overrightarrow{dA}=\frac{q_{in}}{{ϵ}_0}$
$\Rightarrow \oint \overrightarrow{E}.d\overrightarrow{A}=\frac{\left(\rho \right)\left(\frac{4}{3}\pi x^3\right)}{{ϵ}_0}$
$\Rightarrow EA=\frac{\left(\rho \right)\left(\frac{4}{3}\pi x^3\right)}{{ϵ}_0}$
$\Rightarrow E\left(4\pi x^2\right)=\frac{\left(\rho \right)\left(\frac{4}{3}\pi x^3\right)}{{ϵ}_0}$
$\Rightarrow E{x}^2=\frac{\left(\frac{Q}{\frac{4}{3}\pi R^3}\right)\frac{4}{3}\pi x^3}{4\pi {ϵ}_0}$
$\Rightarrow E=\frac{Qx}{4\pi {ϵ}_0{R}^3}$
$\therefore E=\frac{kQx}{R^3}$

Thus, $E\propto x$ for $x<R$


Hence, option (c) is correct.