Question

The diagram shows an open surface consisting of a half-sphere of radius $R$. The base of the half-sphere is on the xy-plane. There is an electric field in this region which has uniform strength $E$ and points in the $+z$-direction.
What is the net flux of electric field through the open hemispherical surface?

• A. 0
• B. $E.\frac{1}{2}\pi R^2$
• C. $E.\pi R^2$
• D. $E.2\pi R^2$
• E. $E.4\pi R^2$

Answer 1

The correct option is C $E.\pi R^2$

Area of the base of hemispherical surface of radius R ,

$A=\pi R^2$ ,

as this shape (with base) in not enclosing any charge , therefore net flux through this shape is zero i.e.

flux through circular base -flux through hemisphere=0 ,

or flux through circular base =flux through hemisphere ,

or flux through hemisphere$=EA\cos 0=E.\pi R^2$

therefore flux linked with , hemispherical surface will be ,

$\varphi =E.\pi R^2$