Question

A disc of radius $R$ has surface charge density $\sigma ={\sigma }_{0}r$, where $r$ is the distance from the centre of the disc. Find the total charge on the disc.

• A. $2\pi {\sigma }_0{R}^3$
• B. $\frac{2\pi {\sigma }_0{R}^3}{3}$
• C. $\frac{4\pi {\sigma }_0{R}^3}{3}$
• D. $\frac{\pi {\sigma }_0{R}^3}{3}$

Answer 1

The correct option is B $\frac{2\pi {\sigma }_0{R}^3}{3}$


Let us choose a small area element $dA$ of radius $r$ and thickness $dr$.

$dA=2\pi rdr$

$dq=\sigma dA={\sigma }_{0}r\left(2\pi rdr\right)$

To obtain the total charge, we will integrate $dq$ as shown.

$\Rightarrow q=\underset{r=0}{\overset{r=R}{\int }}{\sigma }_{0}r\left(2\pi rdr\right)$

$\Rightarrow q=2\pi {\sigma }_0\underset{r=0}{\overset{r=R}{\int }}{r}^{2}dr$

$q=2\pi {\sigma }_0{\left[\frac{r^3}{3}\right]}_{r=0}^{r=R}$

$q=\frac{2\pi {\sigma }_0{R}^3}{3}$

Hence, option (b) is the correct answer.