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 \varepsilon_0 r}$
• B. $ - \frac{1}{5} \cdot \frac{q^2}{4 \pi \varepsilon_0 r}$
• C. $\frac{7}{5} \cdot \frac{q^2}{4 \pi \varepsilon_0 r}$
• D. $\frac{1}{5} \cdot \frac{q^2}{4 \pi \varepsilon_0 r}$

Answer 1

Answer: (B)

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

From the figure, $d=\sqrt{x^{2}+r^{2}} \quad \ldots (i)$
$\therefore $ Potential due to a uniform ring of positive charge $q$ at point $P$,
$V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d}$
$=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{\sqrt{x^{2}+r^{2}}}$ [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 \varepsilon_{0}} \cdot \frac{q}{C B}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{C A}\right] $
$=\frac{-1}{4 \pi \varepsilon_{0}} \cdot q^{2}\left[\frac{1}{\frac{5}{4} r}-\frac{1}{5} r\right]$
$=\frac{-1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{5 r}[4-3]$
$=-\frac{1}{5} \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{r}$