Question

The diagram shows a cylinder of radius $R$ that carries a charge that is uniformly distributed throughout its volume. The cylinder has a charge per unit length of $\lambda$.
Which of the following expressions gives the electric field strength at point $P$, a point that lies inside the cylinder and is a distance $r$ from the axis of the cylinder?

• A. $E=\frac{\lambda R^2}{\pi {ϵ}_0{r}^4}$
• B. $E=\frac{\lambda R^2}{2\pi {ϵ}_{0}r}$
• C. $E=\frac{\lambda r}{2\pi {ϵ}_0{R}^2}$
• D. $E=\frac{\lambda R^2}{2\pi {ϵ}_0{R}^3}$

Answer 1

Answer: (C)

$E=\frac{\lambda r}{2\pi {ϵ}_0{R}^2}$

Let the charge per unit volume of the cylinder be $\rho$ and length of the cylinder be $l$.

Thus charge density $\rho =\frac{Q}{\pi R^{2}l}=\frac{\lambda }{\pi R^2}$ where $\lambda =\frac{Q}{l}$

$\therefore$ Charge enclosed by Gaussian cylinder of radius $r$, $q_{enc}=\pi r^{2}l\times \rho =\frac{\lambda l{r}^2}{R^2}$

Using Gauss's law : $\oint \overrightarrow{E}.\overrightarrow{da}=\frac{q_{enc}}{{ϵ}_o}$

$\therefore$ $E_p\times 2\pi rl=\frac{\lambda l{r}^2}{R^2{ϵ}_o}$ $⟹E_p=\frac{\lambda r}{2\pi {ϵ}_o{R}^2}$