Two charges 7 μc and –4 μc are placed at (–7 cm, 0, 0) and (7 cm, 0, 0) respectively

Question

Two charges \(7 \ \mu \mathrm{c}\) and \(-4 \ \mu \mathrm{c}\) are placed at (-7 cm, 0, 0) and (7 cm, 0, 0) respectively. Given, \(\epsilon_{0}=8.85 \times 10^{-12} \ \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\), the electrostatic potential energy of the charge configuration is :

(1) - 1.5 J

(2) -2.0 J

(3) -1.2 J

(4) -1.8 J

Answer 1

Correct option is : (4) – 1.8 J

P.E. of two charges

\( u=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}}\)

\( \mathrm{r}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}\) 

\( =14 \mathrm{~cm}\) 

\( \therefore \ \mathrm{u}=\frac{9 \times 10^{9} \times 7 \times 10^{-6} \times(-4) \times 10^{-6}}{14 \times 10^{-2}}\) 

\( =-1.8 \mathrm{~J}\)