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

Answer:

$3.037037037\times 10^{-16}\ C$

Explanation:

dr= Width = 30 μm

R = Radius = 1.8 cm

Q = $2.05\times 10^6\times 1.6\times 10^{-19}$

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

$\sigma=\dfrac{Q}{A}\\\Rightarrow \sigma=\dfrac{Q}{\pi r^2}$

Charge is given by

$q=\sigma dA\\\Rightarrow q=\dfrac{Q}{\pi r^2} 2\pi rdr\\\Rightarrow q=\dfrac{2Qrdr}{R^2}\\\Rightarrow q=\dfrac{2\times 2.05\times 10^6\times 1.6\times 10^{-19}\times 0.005\times 30\times 10^{-6}}{0.018^2}\\\Rightarrow q=3.037037037\times 10^{-16}\ C$

The charge contained in the ring is $3.037037037\times 10^{-16}\ C$