Question

For the hemisphere of radius 'R' shown in the figure the surface charge density ${}^{\prime }{\sigma }^{\prime }$ varies with '$\theta$ ' (shown in figure) as $\sigma ={\sigma }_0\cos \theta .$ The total charge on the sphere is

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

Answer 1

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

The hemisphere of radius $=R$

charge density $=a$

angle $=\theta$

$\sigma ={\sigma }_0\cos \theta$

The total charge on the sphere

$q={\sigma }_{0}2\pi R^2$

$=$ charge density $\times$ surface area.