Question

A uniformly charged semicircular arc of radius $r$ has linear charge density $\lambda$. What is the electric field at its centre? ($\epsilon_0 =$ permittivity of free space)

• A. $\frac{\lambda}{2\pi\epsilon_0 r}$
• B. $\frac{\lambda}{4\pi\epsilon_0 r}$
• C. $\frac{\lambda}{4\epsilon_0 r}$
• D. $\frac{2\epsilon_0}{\lambda}$

Hint:

For an arc subtending any general angle $\alpha$ at the center, the electric field is $E = \frac{\lambda}{2\pi\epsilon_0 r} \sin\left(\frac{\alpha}{2}\right)$. For a semicircle, $\alpha = \pi$, so $\sin(\pi/2) = 1$, giving the simplified result immediately.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the net electric field at the center of curvature of a uniformly charged semicircular arc.

Step 2: Key Formula or Approach:
Consider a small elemental angle $d\theta$ on the arc. The charge on this element is $dq = \lambda(r d\theta)$. The electric field due to this element at the center is $dE = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2}$.
Due to symmetry, the components of the electric field parallel to the diameter cancel out. Only the perpendicular components ($dE \cos\theta$, assuming angle $\theta$ is measured from the axis of symmetry) add up.

Step 3: Detailed Explanation:
Let the axis of symmetry be the y-axis. The angle $\theta$ ranges from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.
The net electric field $E$ is the integral of the y-components:
$$E = \int dE \cos\theta$$
$$E = \int_{-\pi/2}^{\pi/2} \left( \frac{1}{4\pi\epsilon_0} \frac{\lambda r d\theta}{r^2} \right) \cos\theta$$
$$E = \frac{\lambda}{4\pi\epsilon_0 r} \int_{-\pi/2}^{\pi/2} \cos\theta d\theta$$
Integrate $\cos\theta$:
$$E = \frac{\lambda}{4\pi\epsilon_0 r} [\sin\theta]_{-\pi/2}^{\pi/2}$$
$$E = \frac{\lambda}{4\pi\epsilon_0 r} \left[ \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) \right]$$
$$E = \frac{\lambda}{4\pi\epsilon_0 r} [1 - (-1)]$$
$$E = \frac{\lambda}{4\pi\epsilon_0 r} (2)$$
$$E = \frac{\lambda}{2\pi\epsilon_0 r}$$

Step 4: Final Answer:
The electric field at the centre is $\frac{\lambda}{2\pi\epsilon_0 r}$, matching option (A).