Question

One fourth of a sphere of radius $R$ is removed as shown. An electric field $E$ exists parallel to x – y plane. Find the flux through remaining curved part.

• A. $\pi R^{2}E$
• B. $\sqrt{2}\pi R^{2}E$
• C. $\pi R^{2}E/\sqrt{2}$
• D. None of these

Answer 1

The correct option is C

$\pi R^{2}E/\sqrt{2}$

In the given setup the cut sphere is placed as shown. The shaded region is open. Any field line entering from anywhere else will contribute 0 to the net flux. Because it will enter from one surface and exit from another.

so the only field lines contributing to the net flux of the hollow sphere should be entering through the open space.

So let’s find the flux through the open surface imagining it closed.

so as you can see there are two surfaces normal to each other through which field lines are passing.

Let’s calculate through each surface one by one.

Area of this section will be $\frac{\pi R^2}{2}$ angle between Area and $E$ vector $={45}^0$

so ${\varphi }_1=E\frac{\pi R^2}{2}cos{45}^0$

$=E\frac{\pi R^2}{2}\frac{1}{\sqrt{2}}$

Similarly for the other surface

${\varphi }_2=E\frac{\pi R^2}{2}\frac{1}{\sqrt{2}}$

${\varphi }_{net}={\varphi }_1+{\varphi }_2=\frac{E\pi R^2}{\sqrt{2}}$