Charges \(+1\ \mu\mathrm{C}\), \(-1\ \mu\mathrm{C}\), \(+1\ \mu\mathrm{C}\) and \(-1\ \mu\mathrm{C}\) are placed at the corners A, B, C and D of a square of side 9 cm, in that order. The work needed to put together this arrangement, starting with the charges very far apart, is about (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\) and \(\sqrt{2}=1.41\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-0.26\ \mathrm{J}\)
The work needed to put the charges together is the potential energy of the system. It is the sum of the energies of all six pairs of charges. The four sides join unlike charges at distance \( d \). The two diagonals join like charges at distance \( \sqrt{2}\,d \).
Given:
\( q=1\ \mu\mathrm{C} \)
\( d=0.09\ \mathrm{m} \)
\( \begin{aligned} \frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{d} &= \frac{9\times10^{9}\times(1\times10^{-6})^{2}}{0.09} \\ &= 0.1\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} U &= 4\left(-\frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{d}\right)+2\left(\frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{\sqrt{2}\,d}\right) \\ &= -\frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{d}\left(4-\sqrt{2}\right) \\ &= -0.1\times(4-1.41) \\ &\approx -0.26\ \mathrm{J} \end{aligned} \)
Therefore, the work needed is about \( -0.26\ \mathrm{J} \). It is negative because the attraction between the unlike neighbours is stronger than the repulsion along the diagonals.
NCERT (2026-27):
Example 2.4 – “Add the work done in steps (i), (ii), (iii) and (iv). The total work required is”
\( =\frac{-q^{2}}{4\pi\varepsilon_0 d}\left(4-\sqrt{2}\right) \)
“The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.”