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Physics Question on electrostatic potential and capacitance

The work done by electric field during the displacement of a negatively charged particle towards a fixed positively charged particle is 9 J9\, 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

3 J

Explanation

Solution

Here, U1=Q(−q)4πε0r;U2=Q(−q)4πε0(r/2)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)}
U1−U2=Q(−q)4πε0[1r−2r]U_1 - U_2 = \frac{Q\left(-q\right)}{4 \pi \varepsilon_{0}} \left[ \frac{1}{r} - \frac{2}{r} \right]
= Qq4πε0r=9 \frac{Qq}{4 \pi \varepsilon_{0}r} = 9 ..(i)
When negative charge travels first half of distance, i.e.,r/4i.e., r/4, potential energy of the system
U3=Q(−q)4πε0(3r/4)=−Qq4πε0r×43U_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 = U1−U3U_1 - U_3
=Q(−q)4πε0r+Q r4πε0r×43= \frac{Q\left(-q\right)}{4 \pi \varepsilon_{0}r} + \frac{Q\,r}{4 \pi \varepsilon_{0}r} \times \frac{4}{3}
=Qq4πε0r×13=93=3J= \frac{Qq}{4 \pi \varepsilon_{0}r} \times \frac{1}{3} = \frac{9}{3} = 3J (Using (i))