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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.

Discussion:

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, $$E=\int dE_z=\frac{q}{4\pi\epsilon_0 a^2}\int_{0}^{\pi/2}sin \theta cos\theta d\theta =\frac{q}{8\pi \epsilon_0 a^2}$$