INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

September 23, 2017

Electric Field of a Charged Hemisphere

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}$$

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.