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

To solve this problem, we first determine the expression for the electric field at the center of a uniformly charged semicircular arc.

Let the total charge on the half ring be $Q$ and its radius be $R = 35 \text{ cm} = 0.35 \text{ m}$. The linear charge density $\lambda$ is defined as the charge per unit length. For a semicircular arc of radius $R$, the length is $\pi R$, so:
$$\lambda = \frac{Q}{\pi R}$$
Consider an infinitesimal element of the ring subtending an angle $d\theta$ at the center. The charge on this element is $dq = \lambda dl = \lambda (R d\theta)$. The electric field $dE$ at the center due to this element is:
$$dE = \frac{1}{4\pi\epsilon_o} \frac{dq}{R^2} = \frac{1}{4\pi\epsilon_o} \frac{\lambda R d\theta}{R^2} = \frac{\lambda}{4\pi\epsilon_o R} d\theta$$
Due to symmetry, the horizontal components of the electric field from opposite sides of the ring cancel out. We only need to integrate the vertical components (along the axis of symmetry):
$$E = \int dE \sin\theta = \int_{0}^{\pi} \frac{\lambda}{4\pi\epsilon_o R} \sin\theta d\theta$$
$$E = \frac{\lambda}{4\pi\epsilon_o R} [-\cos\theta]_{0}^{\pi} = \frac{\lambda}{4\pi\epsilon_o R} (1 - (-1)) = \frac{2\lambda}{4\pi\epsilon_o R} = \frac{\lambda}{2\pi\epsilon_o R}$$
Substituting $\lambda = \frac{Q}{\pi R}$ into the formula:
$$E = \frac{Q}{2\pi^2 \epsilon_o R^2}$$
Given $E = 100 \text{ V/m}$, $R = 0.35 \text{ m}$, $\pi = 3.14$, and $\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$. Rearranging for $Q$:
$$Q = E \cdot 2\pi^2 \epsilon_o R^2$$
$$Q = 100 \times 2 \times (3.14)^2 \times 8.85 \times 10^{-12} \times (0.35)^2$$
$$Q = 200 \times 9.8596 \times 8.85 \times 0.1225 \times 10^{-12}$$
$$Q \approx 2137.8 \times 10^{-12} \text{ C} = 2.1378 \times 10^{-9} \text{ C}$$
Since $1 \text{ nC} = 10^{-9} \text{ C}$, the value is approximately $2.14 \text{ nC}$.