Question

A circular plastic disk with radius R = 1.80 cm has a uniformly distributed charge of Q = +(2.05 ✕ 106)e on one face. A circular ring of width 30 μm is centered on that face, with the center of the ring at radius r = 0.50 cm. What charge is contained within the width of the ring?

Answer 1

`3.037037037* 10^(-16)\ C`

Step-by-step explanation:

dr= Width = 30 μm

R = Radius = 1.8 cm

Q =
`2.05* 10^6* 1.6* 10^(-19)`

r = 0.5 cm

Area is given by

`A=\pi r^2`

Differentiating with respect to r

`dA=2\pi rdr`

Surface charge density is given by

`\sigma=(Q)/(A)\\\Rightarrow \sigma=(Q)/(\pi r^2)`

Charge is given by

`q=\sigma dA\\\Rightarrow q=(Q)/(\pi r^2) 2\pi rdr\\\Rightarrow q=(2Qrdr)/(R^2)\\\Rightarrow q=(2* 2.05* 10^6* 1.6* 10^(-19)* 0.005* 30* 10^(-6))/(0.018^2)\\\Rightarrow q=3.037037037* 10^(-16)\ C`

The charge contained in the ring is
`3.037037037* 10^(-16)\ C`