Question

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

• A. 24.3
• B. 49.3
• C. -90.7
• D. 1.4

Hint:

For a system of charges in an external field, always calculate the self-energy $qV$ for each charge and add the pair-wise interaction energies $\frac{kq_iq_j}{r_{ij}}$.

Answer 1

The Correct Option is B

The total potential energy of the system is given by the sum of the potential energies of individual charges in the external field and the interaction energy between the charges.
$U = q_1 V(r_1) + q_2 V(r_2) + \frac{k q_1 q_2}{r_{12}}$.
First, determine the electric potential $V(r)$ from the external field $E = \frac{A}{r^2}$.
$V(r) = -\int_{\infty}^r E \cdot dr = -\int_{\infty}^r \frac{A}{r^2} dr = \left[ \frac{A}{r} \right]_{\infty}^r = \frac{A}{r}$.
Given $A = 9 \times 10^5 \text{ Vm}$. The positions are $r_1 = 9 \text{ cm} = 0.09 \text{ m}$ and $r_2 = 9 \text{ cm} = 0.09 \text{ m}$. Separation $r_{12} = 18 \text{ cm} = 0.18 \text{ m}$.
Calculate potential at the locations: $V(0.09) = \frac{9 \times 10^5}{0.09} = 10^7 \text{ V}$.
Energy of $q_1 = 7 \mu C$: $U_1 = (7 \times 10^{-6})(10^7) = 70 \text{ J}$.
Energy of $q_2 = -2 \mu C$: $U_2 = (-2 \times 10^{-6})(10^7) = -20 \text{ J}$.
Interaction energy: $U_{12} = \frac{9 \times 10^9 (7 \times 10^{-6})(-2 \times 10^{-6})}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 \text{ J}$.
Total Energy $U = U_1 + U_2 + U_{12} = 70 - 20 - 0.7 = 49.3 \text{ J}$.