A total charge \(q\) is spread uniformly over the inner surface of a non-conducting hemispherical cup of inner radius \(a\). Calculate the electric field.


Consider a circular strip symmetric about \(z-axis\) of radius \(r\) and width \(ad\theta\)
The charge on the strip is $$ dq=q\frac{2\pi r ad\theta}{2\pi a^2}=\frac{qr d\theta}{a}=q sin\theta d\theta$$
(a) At the centre of the hemisphere, the \(x-component\) of the field will be cancelled for reasons of symmetry. The entire field will be contributed by the \(z-component\) alone.
$$ dE=dE_z= \frac{q sin\theta d\theta cos\theta}{4\pi\epsilon_0 a^2}
Therefore, $$ E=\int dE_z=\frac{q}{4\pi\epsilon_0 a^2}\int_{0}^{\pi/2}sin \theta cos\theta d\theta =\frac{q}{8\pi \epsilon_0 a^2}$$