Question

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

• A. 2.14
• B. 2.44
• C. 3.25
• D. 0.7

Hint:

Use the formula for the electric field at the center of a charged semicircular arc: $E = \frac{2k\lambda}{R}$, where $\lambda = \frac{Q}{\pi R}$.

Answer 1

The Correct Option is A

The magnitude of the electric field at the center of a circular arc of radius $R$ subtending an angle $2\alpha$ at the center is given by the general formula:
$$E = \frac{2k\lambda \sin\alpha}{R}$$
where $k = \frac{1}{4\pi\epsilon_o}$ and $\lambda$ is the linear charge density. For a half ring, the total angle subtended is $\pi$ radians, which means $2\alpha = \pi$ or $\alpha = \pi/2$.
Substituting $\alpha = \pi/2$ into the formula:
$$E = \frac{2k\lambda \sin(\pi/2)}{R} = \frac{2k\lambda}{R}$$
Given the total charge $Q$ is distributed over the length $\pi R$, the charge density is:
$$\lambda = \frac{Q}{\pi R}$$
Substituting this into the electric field formula:
$$E = \frac{2}{4\pi\epsilon_o R} \left( \frac{Q}{\pi R} \right) = \frac{Q}{2\pi^2\epsilon_o R^2}$$
We are given:
$E = 100 \text{ V/m}$
$R = 35 \text{ cm} = 0.35 \text{ m}$
$\epsilon_o = 8.85 \times 10^{-12} \text{ F/m}$
$\pi = 3.14$
Calculate $Q$:
$$Q = E \times 2 \times \pi^2 \times \epsilon_o \times R^2$$
$$Q = 100 \times 2 \times 3.14 \times 3.14 \times 8.85 \times 10^{-12} \times 0.35 \times 0.35$$
Performing the multiplication:
$$Q = 200 \times 9.8596 \times 8.85 \times 0.1225 \times 10^{-12}$$
$$Q \approx 2.138 \times 10^{-9} \text{ C}$$
Converting to nanocoulombs ($10^{-9} \text{ C}$):
$$Q \approx 2.14 \text{ nC}$$
The closest option is 2.14.