Question 19: Electrostats & Capacitance

Question

Two charges $7\mu	ext{C}$ and $-2\mu	ext{C}$ are placed at $(-9, 0, 0)	ext{ cm}$ and $(9, 0, 0)	ext{ cm}$ respectively in an external field $E = \frac{A}{r^2}\hat{r}$, where $A = 9 	imes 10^5	ext{ N/C}\cdot	ext{m}^2$. Considering the potential at infinity is $0$, the electrostatic energy of the configuration is ______ $	ext{J}$.

Accepted Answer

Step 1: Energy of individual charges in the external fieldFor charge $q_1 = 7\mu	ext{C}$ at $r_1 = 9	ext{ cm} = 0.09	ext{ m}$:$U_1 = q_1 V(r_1) = \frac{q_1 A}{r_1} = \frac{(7 	imes 10^{-6}) 	imes (9 	imes 10^5)}{0.09} = 70	ext{ J}$$For charge $q_2 = -2\mu	ext{C}$ at $r_2 = 9	ext{ cm} = 0.09	ext{ m}$:$$U_2 = q_2 V(r_2) = \frac{q_2 A}{r_2} = \frac{(-2 	imes 10^{-6}) 	imes (9 	imes 10^5)}{0.09} = -20	ext{ J}$$ $\$Step 2: Interaction energy between the two chargesThe distance between the charges is $r_{12} = 9 - (-9) = 18	ext{ cm} = 0.18	ext{ m}$:$$U_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}} = \frac{(9 	imes 10^9) 	imes (7 	imes 10^{-6}) 	imes (-2 	imes 10^{-6})}{0.18}$$$$U_{12} = \frac{-126 	imes 10^{-3}}{0.18} = -0.7	ext{ J}$$ $\$Step 3: Total Electrostatic EnergyThe total energy $U$ is the sum of the work done to assemble the configuration:$$U = U_1 + U_2 + U_{12}$$ $$U = 70 - 20 - 0.7 = 49.3	ext{ J}$$ $$\$Hence, the correct option is (A).

Answer

A. $49.3$

Answer

B. $1.4$

Answer

C. $24.3$

Answer

D. $-90.7$

Electrostats & Capacitance

Choose the correct answer: