Question

A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemispherical surface? A diagram is required as part of your answer

Answer 1

We know that by gauss law the electric flux on a surface is equal to the dot product of the electric field with the surface element.

So the flux over whole hemisphere will be the integration of the flux over a single area element , throughput the hemisphere !

We take area element to be dS. ...refer figure 0

I'll be using int to represent integration.

# Flux = int( E .dS )

Doing dot product of E and dS

# Flux= int( E* cos(theta) * dS ).

dS * Cos(theta) is dA. ____ refer figure D

dA is basically the projection of dS over the base of the hemisphere.

# Flux= int( E*dA )

As E is constant so we can take ut out of integration.

# Flux= E* int(dA)

int(dA) is basically the integration of the element dA which is a part of the circular base of the hemisphere.

So, int(dA) = area of base of hemisphere of radius R

int(dA) = πR²

## Flux= E* πR²

So the flux is E* πR².

Hope it helps.

Your feedback is greatly appreciated !