Question

Consider a long uniformly charged cylinder having constant volume charge density '$\lambda$' and radius '$R$'. A Gaussian surface is in the form of a cylinder of radius '$r$' such that vertical axis of both the cylinders coincide. For a point inside the cylinder ($r < R$), electric field is directly proportional to

• A. $r^{-1}$
• B. r
• C. $r^2$

Answer 1

Answer: (B)

r

Answer 2

For a uniformly charged cylinder with volume charge density $\rho$, consider a Gaussian cylindrical surface of radius r (< R) and length L. The enclosed charge is

$Q_{\text{enc}} = \rho \cdot \pi r^2 L$.

By Gauss’s law,

$E \cdot (2\pi r L) = \frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{\rho \pi r^2 L}{\varepsilon_0}$.

Solving for E,

$E = \frac{\rho r}{2\varepsilon_0}$.

Thus, for points inside the cylinder, the electric field is directly proportional to r.