Two positive point charges of 12 $\mu$ C and 8 $\mu$ C (micro coulomb) respectively are 10 cm apart. The work done in bringing them 4 cm closer so that they are 6 cm apart is
(1 / (4 $\pi$\varepsilon_0) = $9 * 10^9$ mF^-1)

The correct answer is (A) 5.8 J.

• The work done $W$ in bringing two point charges closer is equal to the increase in electrostatic potential energy $\mathrm{\Delta }U$$\Delta$ U:
• $W = U_f - U_i =$$\frac{1}{4\pi\varepsilon_0}$ q_1 q_2 \left($\frac{1}{r_f}$ - $\frac{1}{r_i}$\right)
• Here $q_1 = 12$$\times$ 10^{-6} $\text{ C}$, $q_2 = 8$$\times$ 10^{-6} $\text{ C}$, $r_i = 10$$\text{ cm}$ = 0.10 $\text{ m}$, and $r_f = 10 - 4 = 6$$\text{ cm}$ = 0.06 $\text{ m}$:
• $\frac{1}{4\pi {\epsilon }_0}{q}_1{q}_2=(9\times {10}^9)\times (12\times {10}^{-6})\times (8\times {10}^{-6})=0.864\text{ N}\cdot {\text{m}}^2$$\frac{1}{4\pi\varepsilon_0}$ q_1 q_2 = (9 $\times$ 10^9) $\times$ (12 $\times$ 10^{-6}) $\times$ (8 $\times$ 10^{-6}) = 0.864 $\text{ N}$$\cdot$$\text{m}$^2
• $W = 0.864$$\times$ \left($\frac{1}{0.06}$ - $\frac{1}{0.10}$\right) = 0.864 $\times$ \left($\frac{50}{3}$ - 10\right) = 0.864 $\times$ $\frac{20}{3}$ = 5.76 $\text{ J}$ $\approx$ 5.8 $\text{ J}$