Question

A charge +q is distributed over a thin ring of radius r with line charge density

$\lambda = \frac{q \sin^2 \theta}{\pi r}$. Note that the ring is in the xy-plane and $\theta$ is the angle made by r with the x-axis. The work done by the electric force in displacing a point charge +Q from the center of the ring to infinity is

• A. equal to $\frac{qQ}{8\pi \epsilon_0 r}$.
• B. equal to zero only if the path is a straight line perpendicular to the plane of the ring.
• C. always zero because electrostatic field is conservative.
• D. equal to $\frac{qQ}{4\pi \epsilon_0 r}$.

Answer 1

Answer: (D)

equal to $\frac{qQ}{4\pi \epsilon_0 r}$.

Answer 2

The work done by the electric force is $W = Q(V_{initial} - V_{final})$. Here, the initial point is the center of the ring, and the final point is infinity. So, $W = Q(V_{center} - V_{\infty})$. Since $V_{\infty} = 0$, $W = Q V_{center}$.

For any charge distribution, the potential at a point is given by $V = \int \frac{dq}{4\pi\epsilon_0 R}$, where $R$ is the distance from $dq$ to the point.

At the center of the ring, every charge element $dq$ is at a distance $r$ from the center. Thus, the potential at the center is $V_{center} = \int \frac{dq}{4\pi\epsilon_0 r} = \frac{1}{4\pi\epsilon_0 r} \int dq$.

The integral $\int dq$ represents the total charge on the ring, which is given as $q$.

Therefore, $V_{center} = \frac{q}{4\pi\epsilon_0 r}$.

The work done is $W = Q V_{center} = Q \left(\frac{q}{4\pi\epsilon_0 r}\right) = \frac{qQ}{4\pi\epsilon_0 r}$.