PrepRung
Sign inSign up free

JEE Main 2017 · Physics · Electrostatics · medium

Four point charges +q, -q, +q, -q are placed in that order at the successive corners of a square of side a. The work done in assembling this…

Single correct option

Four point charges +q,−q,+q,−q+q, -q, +q, -q are placed in that order at the successive corners of a square of side aa. The work done in assembling this arrangement from infinity is, with k=1/4πε0k = 1/4\pi\varepsilon_0,
  1. A. −4kq2a-4\dfrac{kq^2}{a}
  2. B. (22−4)kq2a(2\sqrt2 - 4)\dfrac{kq^2}{a}
  3. C. (2−4)kq2a(\sqrt2 - 4)\dfrac{kq^2}{a}
  4. D. (4−2)kq2a(4 - \sqrt2)\dfrac{kq^2}{a}
Show answer and solution

Answer: (C) (2−4)kq2a(\sqrt2 - 4)\dfrac{kq^2}{a}

Add up all six pair energies. Each side joins opposite charges: four pairs of energy −kq2/a-kq^2/a, a total of −4kq2/a-4kq^2/a. Each diagonal joins like charges at distance a2a\sqrt2: two pairs of energy +kq2/(a2)+kq^2/(a\sqrt2), a total of 2 kq2/a\sqrt2\,kq^2/a. So U=(2−4)kq2a≈−2.59 kq2aU = \left(\sqrt2 - 4\right)\dfrac{kq^2}{a} \approx -2.59\,\dfrac{kq^2}{a}. The work done by an external agent equals this potential energy, and it is negative, which means the configuration is bound.
Report a mistake

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake