Question

A non conducting semicircular disc (as shown in figure) has a uniform surface charge density $\sigma$. The electric field intensity at the centre of the disc

• A. $\frac{\sigma(b-a)}{4\pi\epsilon_0}$
• B. $\frac{\sigma(b-a)}{2\pi\epsilon_0}$
• C. $\frac{\sigma \ln(b-a)}{2\pi\epsilon_0(b-a)}$
• D. $\frac{\sigma}{2\pi\epsilon_0} \ln(\frac{b}{a})$

Answer 1

Answer: (D)

$\frac{\sigma}{2\pi\epsilon_0} \ln(\frac{b}{a})$

Answer 2

The semicircular disc has uniform surface charge density $\sigma$, inner radius $a$ and outer radius $b$. We want to find the electric field at the center of the disc.

Consider a semicircular ring of radius $r$ and infinitesimal thickness $dr$. The area of this ring is $dA = \frac{1}{2} \pi (r+dr)^2 - \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (r^2 + 2r dr + (dr)^2 - r^2) = \frac{1}{2} \pi (2r dr) = \pi r dr$, neglecting the $(dr)^2$ term.

The charge on this semicircular ring is $dq = \sigma dA = \sigma \pi r dr$. This charge is distributed along a semicircular arc of radius $r$. The linear charge density of this arc is $\lambda = \frac{dq}{\text{length of arc}} = \frac{\sigma \pi r dr}{\pi r} = \sigma dr$.

The electric field at the center of a uniformly charged semicircular arc of radius $r$ and linear charge density $\lambda$ is given by $E_{arc} = \frac{2k\lambda}{r}$, where $k = \frac{1}{4\pi\epsilon_0}$. The direction of the electric field is perpendicular to the diameter, pointing towards the arc if the charge is positive. Assuming the semicircle is in the upper half plane and the center is at the origin, the electric field is directed along the positive y-axis.

Substituting $\lambda = \sigma dr$ and $k = \frac{1}{4\pi\epsilon_0}$, the electric field at the center due to the semicircular ring of radius $r$ and thickness $dr$ is $dE = \frac{2 (\frac{1}{4\pi\epsilon_0}) (\sigma dr)}{r} = \frac{\sigma dr}{2\pi\epsilon_0 r}$.

To find the total electric field at the center, we integrate this expression from the inner radius $a$ to the outer radius $b$: $E = \int_{a}^{b} dE = \int_{a}^{b} \frac{\sigma dr}{2\pi\epsilon_0 r}$. Since $\sigma$ and $\epsilon_0$ are constants, we can take them out of the integral: $E = \frac{\sigma}{2\pi\epsilon_0} \int_{a}^{b} \frac{1}{r} dr$. The integral of $\frac{1}{r}$ with respect to $r$ is $\ln|r|$. Since $r$ is a radius, $r > 0$, so $|r| = r$. $\int_{a}^{b} \frac{1}{r} dr = [\ln r]_{a}^{b} = \ln b - \ln a = \ln\left(\frac{b}{a}\right)$.

So, the total electric field at the center is $E = \frac{\sigma}{2\pi\epsilon_0} \ln\left(\frac{b}{a}\right)$. The direction of the electric field is perpendicular to the diameter and points towards the semicircular disc.