A3.5-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:

The flux through the flat base is:

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

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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As per Gauss Law

Net flux through enclosed surface is

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

given that

now we have

luster is not a good characteristic to use by itself because luster is the property of minerals that indicates how much the surface of a mineral reflects light. the luster of a mineral is affected by the brilliance of the light used to observe the mineral surface.

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