Question

A Gaussian surface in the form of a hemisphere of radius R = 3.04 cm lies in a uniform electric field of magnitude E = 1.64 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:

Part a)

$\phi = -4.76 \times 10^{-3} Nm^2/C$

Part b)

$\phi_{curved} = 4.76 \times 10^{-3} Nm^2/C$

Explanation:

Part a)

Electric flux entering into the base

so it is given as

$\phi = E.A$

$\phi = - EA$

$\phi = -(1.64)(\pi r^2)$

$\phi = -(1.64)(\pi\times 0.0304^2)$

$\phi = -4.76 \times 10^{-3} Nm^2/C$

Part b)

Now since we know that there is no enclosed charge in the hemisphere

so net flux must be zero

$\phi_{curved} + \phi_{flat} = 0$

$\phi_{curved} - 4.76 \times 10^{-3} = 0$

$\phi_{curved} = 4.76 \times 10^{-3} Nm^2/C$