Question

The electric field at the centre 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 figure. If $\alpha =\beta =\frac{\pi }{3}$, then electric field intensity at centre due to remaining portion is $\frac{E_0}{n},$ then the value of $n$ is

Answer 1

As $\alpha =\beta =\frac{\pi }{3},$ we can consider that hemispherical shell is made of three identical subparts and magnetic field due to each part be $E$


From symmetry (Fig) we can say that resultant of above 3 vectors must be $E_0$
$\Rightarrow E\cos {60}^{\circ }+E\cos {60}^{\circ }+E=E_0$
$\Rightarrow E=\frac{E_0}{2}$