Question

A point charge $-q$ is carried from a point $A$ to another point $B$ on the axis of a charged ring of radius ‘$r$’ carrying a charge $+q$. If the point $A$ is at a distance $\frac{4}{3}r$ from the centre of the ring and the point $B$ is $\frac{3}{4}r$ from the centre but on the opposite side, what is the net work that need to be done for this ?

• A. $-\frac{7}{5}\cdot \frac{q^2}{4\pi {\epsilon }_{0}r}$
• B. $-\frac{1}{5}\cdot \frac{q^2}{4\pi {\epsilon }_{0}r}$
• C. $\frac{7}{5}\cdot \frac{q^2}{4\pi {\epsilon }_{0}r}$
• D. $\frac{1}{5}\cdot \frac{q^2}{4\pi {\epsilon }_{0}r}$

Answer 1

Answer: (B)

A charged circular ring of radius $R$ is shown in figure.

From the figure, $d=\sqrt{x^2+r^2}\dots (i)$
$\therefore$ Potential due to a uniform ring of positive charge $q$ at point $P$,
$V=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{d}$
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{\sqrt{x^2+r^2}}$ [from Eq. (i)]
Now,

$\therefore CB=\sqrt{O{C}^2+O{B}^2}=\sqrt{r^2+{\left(\frac{3}{4}\right)}^2{r}^2}=\frac{5}{4}r$
Similarly, $AC=\frac{5}{3}r$
$\therefore$ Work done in bringing a point charge $-q$ from $A$ to $B$,
$W=-q\left(V_B-V_A\right)$
$=-q\left[\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{CB}-\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{CA}\right]$
$=\frac{-1}{4\pi {\epsilon }_0}\cdot q^2\left[\frac{1}{\frac{5}{4}r}-\frac{1}{5}r\right]$
$=\frac{-1}{4\pi {\epsilon }_0}\cdot \frac{q^2}{5r}[4-3]$
$=-\frac{1}{5}\cdot \frac{1}{4\pi {\epsilon }_0}\cdot \frac{q^2}{r}$