Question

For an infinite line of charge having charge density $\lambda$ lying along x-axis, the work required in moving charge q from C to A along arc CA is:

• A. $\frac{q\lambda}{\pi\epsilon_0} log_e \sqrt{2}$
• B. $\frac{q\lambda}{4\pi\epsilon_0} log_e \sqrt{2}$
• C. $\frac{q\lambda}{4\pi\epsilon_0} log_e 2$
• D. $\frac{q\lambda}{2\pi\epsilon_0} log_e \frac{1}{2}$

Answer 1

Answer: (C)

$\frac{q\lambda}{4\pi\epsilon_0} log_e 2$

Answer 2

The electric potential at a distance $r$ from an infinite line of charge with linear charge density $\lambda$ is given by:

$V(r) = -\frac{\lambda}{2\pi\epsilon_0} \log_e r + K$

where $K$ is an arbitrary constant.

The work done in moving a charge $q$ from point C to point A is:

$W_{C \to A} = q(V_A - V_C)$

The potential difference between points A and C is:

$V_A - V_C = \left(-\frac{\lambda}{2\pi\epsilon_0} \log_e r_A + K\right) - \left(-\frac{\lambda}{2\pi\epsilon_0} \log_e r_C + K\right) = -\frac{\lambda}{2\pi\epsilon_0} (\log_e r_A - \log_e r_C) = \frac{\lambda}{2\pi\epsilon_0} (\log_e r_C - \log_e r_A) = \frac{\lambda}{2\pi\epsilon_0} \log_e \left(\frac{r_C}{r_A}\right)$

From the figure, assume that point C is located such that its distance from the x-axis is $r_C = \sqrt{2}a$, and point A is at a distance $r_A = a$ from the x-axis. This would make the ratio $r_C/r_A = \sqrt{2}$.

Then the work done is:

$W_{C \to A} = q(V_A - V_C) = q \frac{\lambda}{2\pi\epsilon_0} \log_e \left(\frac{r_C}{r_A}\right) = q \frac{\lambda}{2\pi\epsilon_0} \log_e \left(\frac{\sqrt{2}a}{a}\right) = q \frac{\lambda}{2\pi\epsilon_0} \log_e \sqrt{2}$

Since $\log_e 2 = 2 \log_e \sqrt{2}$, then $\log_e \sqrt{2} = \frac{1}{2} \log_e 2$.

Therefore, $W_{C \to A} = \frac{q\lambda}{2\pi\epsilon_0} \log_e \sqrt{2} = \frac{q\lambda}{2\pi\epsilon_0} \frac{1}{2} \log_e 2 = \frac{q\lambda}{4\pi\epsilon_0} \log_e 2$