Question

The work done by electric field during the displacement of a negatively charged particle towards a fixed positively charged particle is $9\, J$. As a result the distance between the charges has been decreased by half. What work is done by the electric field over the first half of this distance?

• A. 3 J
• B. 6 J
• C. 1.5 J
• D. 9 J

Answer 1

Answer: (A)

Here, $U_{1} = \frac{Q\left(-q\right)}{4 \pi \varepsilon_{0} r} ; U_{2} = \frac{Q\left(-q\right)}{4 \pi \varepsilon_{0} (r/2)} $
$U_1 - U_2 = \frac{Q\left(-q\right)}{4 \pi \varepsilon_{0}} \left[ \frac{1}{r} - \frac{2}{r} \right]$
= $ \frac{Qq}{4 \pi \varepsilon_{0}r} = 9$ ..(i)
When negative charge travels first half of distance, $i.e., r/4$, potential energy of the system
$ U_3 = \frac{Q\left(-q\right)}{4 \pi \varepsilon_{0}(3r /4)} = \frac{-Qq}{4 \pi \varepsilon_{0}r} \times \frac{4}{3}$
$\therefore $ Work done = $U_1 - U_3$
$ = \frac{Q\left(-q\right)}{4 \pi \varepsilon_{0}r} + \frac{Q\,r}{4 \pi \varepsilon_{0}r} \times \frac{4}{3} $
$ = \frac{Qq}{4 \pi \varepsilon_{0}r} \times \frac{1}{3} = \frac{9}{3} = 3J$ (Using (i))