Answer:
[tex]3.037037037\times 10^{-16}\ C[/tex]
Explanation:
dr= Width = 30 μm
R = Radius = 1.8 cm
Q = [tex]2.05\times 10^6\times 1.6\times 10^{-19}[/tex]
r = 0.5 cm
Area is given by
[tex]A=\pi r^2[/tex]
Differentiating with respect to r
[tex]dA=2\pi rdr[/tex]
Surface charge density is given by
[tex]\sigma=\dfrac{Q}{A}\\\Rightarrow \sigma=\dfrac{Q}{\pi r^2}[/tex]
Charge is given by
[tex]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[/tex]
The charge contained in the ring is [tex]3.037037037\times 10^{-16}\ C[/tex]
