Question

The surface charge density of a hollow hemisphere varies with $\theta$ as $\sigma ={\sigma }_0\cos \theta$. The situation is shown in the figure. Find the total charge on the hemisphere.

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

Answer 1

The correct option is C ${\sigma }_0\pi R^2$

Consider the area element $dA$ having the radius $r$ which is at an angle $\theta$ as we can see in the figure. Let $R$ be the radius of the hemisphere.

The area element will form a ring of radius $r$ and width $dr$ where $dr=Rd\theta$.

Thus $dA=2\pi rdr=2\pi rRd\theta .....\left(1\right)$



Now, $\sin \theta =\frac{r}{R}\Rightarrow r=R\sin \theta$

Using this in $\left(1\right)$

$dA=2\pi R^2\sin \theta d\theta$

Thus, the value of charge on this area element will be,

$dq=\sigma dA$

As it is given that $\sigma ={\sigma }_0\cos \theta$

$dq=\left({\sigma }_0\cos \theta \right)\left(2\pi R^2\sin \theta d\theta \right)$

$dq={\sigma }_0\pi R^2\sin 2\theta d\theta$ $(\because \sin 2\theta =2\sin \theta \cos \theta )$

To determine the total charge on the hemisphere, we will integrate $dq$ from $0$ to $\pi /2$

$\int dq=\underset{\theta =0}{\overset{\theta =\frac{\pi }{2}}{\int }}{\sigma }_0\pi R^2\sin 2\theta d\theta$

$\int dq={\sigma }_0\pi R^2\underset{\theta =0}{\overset{\theta =\frac{\pi }{2}}{\int }}\sin 2\theta d\theta$

$q=-\frac{\pi {\sigma }_0{R}^2}{2}(\cos 2\theta {)}_{\theta =0}^{\theta =\frac{\pi }{2}}$

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

Hence, option (c) is the correct answer.