Question

A positive charge Q2 is uniformly distributed over a nonconducting disc of radius a which has a concentric circular hole of radius b. At the center of the hole there is another nonconducting disc of radius d where a charge Q1 is uniformly distributed.

a) Find the surface charge density of the disc with the hole σ2.
b) Find the surface charge density 01 of the disc of radius d.
c) Find the total charge enclosed by the circle of radius

Answer 1

a) σ =
`(Q_1)/( a^2 - b^2)` , b) σ =
`(Q_2)/(d^2)` , c) Q_ {total} = Q₁ + Q₂, σ_ {net} =
`(Q_1 + Q_2)/(\pi \ a^2)`

Step-by-step explanation:

a) The very useful concept of charge density is defined by

σ = Q / A

In this case we have a circular disk

The are of a circle is

A = π r²

in this case we have a hole in the center of radius r = b, so

A_net = π r² - π r_ {hollow} ²

A_ {net} = π (a² - b²)

whereby the density is

σ =
`(Q_1)/( a^2 - b^2)`

b) The density of the other disk is

σ = Q₂ / A₂

σ =
`(Q_2)/(d^2)`

c) The total waxed load is requested by the larger circle

Q_ {total} = Q₁ + Q₂

the net charge density, in the whole system is

σ =
`(Q_(total) )/( A_(total) )`

the area is

A_{total} = π a²

since the other circle is inside, we are ignoring the space between the two circles

σ_ {net} =
`(Q_1 + Q_2)/(\pi \ a^2)`