Explanation

(c) When charges are placed at vertices of an equilateral triangle of side 1m, then potential energy of combination is
[image:2_image119.emf]
$U_{1} = \frac{1}{4\pi\varepsilon_{0}}.\frac{1 \times 2 \times 10^{- 12}}{(1)} + \frac{1}{4\pi\varepsilon_{0}}.\frac{2 \times 3 \times 10^{- 12}}{(1)}$
$+ \frac{1}{4\pi\varepsilon_{0}}.\frac{3 \times 1 \times 10^{- 12}}{(1)}$
$= 11 \times \frac{1}{4\pi\varepsilon_{0}} \times 10^{- 12}J$
When charges are placed at vertices of an equilateral triangle of side 0.5m, then potential energy of combination is
[image:2_image120.emf]
$U_{1} = \frac{1}{4\pi\varepsilon_{0}}.\frac{1 \times 2 \times 10^{- 12}}{(0.5)} + \frac{1}{4\pi\varepsilon_{0}}.\frac{2 \times 3 \times 10^{- 12}}{(0.5)}$
$+ \frac{1}{4\pi\varepsilon_{0}}.\frac{3 \times 1 \times 10^{- 12}}{(0.5)}$
$= 22 \times \frac{1}{4\pi\varepsilon_{0}} \times 10^{- 12}J$
$\therefore Work\ done = \mathrm{\Delta}U = U_{2} - U_{1}$
$= 22 \times \frac{1}{4\pi\varepsilon_{0}} \times 10^{- 12} - 11 \times \frac{1}{4\pi\varepsilon_{0}} \times 10^{- 12}$
$= 11 \times \frac{1}{4\pi\varepsilon_{0}} \times 10^{- 12}$
$= 11 \times 9 \times 10^{9} \times 10^{- 12} = 99 \times 10^{- 3}$
$= 0.099J \approx 0.01J$