Question 139: Electrostatics & Capacitance

Question

A thin half ring of radius 35 cm is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100 \mathrm{~V} / \mathrm{m}$, then the value of $Q$ is $\_\_\_\_$ nC .
$$ \left(\epsilon_0=8.85 	imes 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2 	ext { and } \pi=3.14ight) $$

Accepted Answer

For a uniformly charged semicircular ring with linear charge density $\lambda$ and radius R , the electric field E at the center is:
$$ E=\frac{2 \mathrm{k} \lambda}{\mathrm{R}}=\frac{2 \lambda}{4 \pi \epsilon_0 \mathrm{R}}=\frac{\lambda}{2 \pi \epsilon_0 \mathrm{R}} $$
Where $\lambda=\frac{	ext { Total Charge }}{	ext { Length }}=\frac{\mathrm{Q}}{\pi \mathrm{R}}$.
Substituting $\lambda$ into the field formula:
$$ E=\frac{Q / \pi R}{2 \pi \epsilon_0 R}=\frac{Q}{2 \pi^2 \epsilon_0 R^2} $$
$$\Rightarrow $$ $$ \mathrm{Q}=\mathrm{E} \cdot 2 \pi^2 \epsilon_0 \mathrm{R}^2 $$
The magnitude of electric field at the centre is $\mathrm{E}=100 \mathrm{~V} / \mathrm{m}$
The radius of the semi-circle is $\mathrm{R}=35 \mathrm{~cm}=0.35 \mathrm{~m}$
The value of $\epsilon_0=8.85 	imes 10^{-12}$ and $\pi=3.14$
$$ Q=100 	imes 2 	imes(3.14)^2 	imes 8.85 	imes 10^{-12} 	imes(0.35)^2 $$
$$\Rightarrow $$ $Q=100 	imes 2 	imes 9.8596 	imes 8.85 	imes 10^{-12} 	imes 0.1225$
$$\Rightarrow $$ $Q \approx 2.14 	imes 10^{-9} C$
$$\Rightarrow $$ $$ \mathrm{Q}=2.14 \mathrm{nC} $$
Therefore, the value of Q is 2.14 nC . Thus, the correct option is (A).

Answer

2.14

Answer

2.44

Answer

3.25

Answer

0.7

Electrostatics & Capacitance

Choose the correct answer: