Question

The electric field at the center of a uniformly charged hemispherical shell is $E_0$. Now two portions of the hemisphere are cut from either side and remaining portion is shown in the figure. If $\alpha =\beta =\frac{\pi }{3}$ and electric field intensity at centre due to remaining portion is $\frac{E_0}{x}$, then $x$ is (answer upto two decimal places)

Answer 1

Consider the whole hemisphere as three portions
If electric field due to one portion is $E_1$, then $2E_1\sin {30}^0+E_1=E_0$
$2E_1=E_0$
$E_1=\frac{E_0}{2}$