Question

A 3.5-cm radius hemisphere contains a total charge of 6.6 × 10–7

c. the flux through the rounded portion of the surface is 9.8 × 104 nm2/c. the flux through the flat base is:

Answer 1

As per Gauss Law

Net flux through enclosed surface is

$\phi = \frac{Q}{\epsilon_0}$

here through this hemisphere total flux will pass through two portions

1). from the curved surface

2). from flat circular base

so now we have

$\phi_{base} + \phi_{surface} = \frac{Q}{\epsilon_0}$

given that

$Q = 6.6 * 10^{-7} C$

$\phi_{surface} = 9.8 * 10^4$

now we have

$\phi_{base} + 9.8*10^4 = \frac{6.6*10^7}{8.85 * 10^{-12}}$

$\phi_{base} = - 9.8 *10^4 + 7.46 * 10^4$

$\phi_{base} = - 2.34 * 10^4 N*m^2/C$

Answer 2

The flux through the flat base is:

• -2.34 * 10⁴N * m^2/C

What is Gauss' Law?

This states that the net flux of an electric field in a closed surface is directly proportional to the electric charge.

To calculate the flux through the flat base, we would make use of Gauss's Law

φ$base + surface = Q/eo$

We input the value of Q

Q= 6.6 x 10^-7C

Next, we make the addition

-9.8 x 10⁴ + 7.46 * 10⁴

=-2.34 * 10⁴N * m^2/C

Therefore, the flux through the flat base is given as -2.34 * 10⁴N * m^2/C

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