Question

A Gaussian surface in the form of a hemisphere of radius R = 5.51 cm lies in a uniform electric field of magnitude E = 1.08 N/C. The surface encloses no net charge. At the (flat) base of the surface, the field is perpendicular to the surface and directed into the surface. What is the flux through (a) the base and (b) the curved portion of the surface?

Answer 1

Answer:

$Flux_{base}=-0.0103Nm^2/C$

$Flux_{curvePortion}=-Flux_{base}=0.0103Nm^2/C$

Explanation:

a)At the base of the surface the Electric flux is easy to find, because the Electric Field is constant in magnitude and perpendicular to the flat surface:

$Flux_{base}=-E*S=-E*\pi*R^2=-1.08*\pi*0.0551^2=-0.0103Nm^2/C$

The Flux is negative because the Electric field goes into the surface

b) The Gaussian surface in form of a hemisphere encloses no net charge. The Gauss law says that the Flux of the electric field is proportional to the net charge enclosed. At this case, the charge is zero, then the total Flux is zero too.

$Flux_{total}=Flux_{base}+Flux_{curvePortion}=0$

Then:

$Flux_{curvePortion}=-Flux_{base}=0.0103Nm^2/C$

The Flux is positif because the Electric field goes out of the surface.