Two charges 7 C and -2 C are placed at (-9, 0, 0) cm and (9, 0, 0) cm respectively in an external field E = Ar^2 r, where A = 9 10^5 N/C.m^2. Considering the potential at infinity is 0, the…

Two charges 7 C and -2 C are placed at (-9, 0, 0) cm and (9, 0, 0… — Solution | JEE Physics

MCQMediumJEE 2026Electric Potential & Potential Energy

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
$1.4$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Charges are $q_1 = 7 \times 10^{-6} \, \text{C}$ and $q_2 = -2 \times 10^{-6} \, \text{C}$ . Their positions are $r_1 = 0.09 \, \text{m}$ and $r_2 = 0.09 \, \text{m}$ from the origin. The separation is $r_{12} = 0.18 \, \text{m}$ . The external field is $E = \frac{A}{r^2}\hat{r}$ with $A = 9 \times 10^5 \, \text{V m}$ .

Find: The total electrostatic energy of the configuration.

For charges placed in an external field, the total energy is

U = q_1 V(r_1) + q_2 V(r_2) + \frac{k q_1 q_2}{r_{12}}

First find the potential due to the external field:

V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{r} = -\int_{\infty}^{r} \frac{A}{r^2} \, dr = \frac{A}{r}

At $r = 0.09 \, \text{m}$ ,

V(0.09) = \frac{9 \times 10^5}{0.09} = 10^7 \, \text{V}

Energy of the first charge:

U_1 = q_1 V = (7 \times 10^{-6})(10^7) = 70 \, \text{J}

Energy of the second charge:

U_2 = q_2 V = (-2 \times 10^{-6})(10^7) = -20 \, \text{J}

Interaction energy between the two charges:

U_{12} = \frac{9 \times 10^9 (7 \times 10^{-6})(-2 \times 10^{-6})}{0.18} = -0.7 \, \text{J}

Therefore,

U = 70 - 20 - 0.7 = 49.3 \, \text{J}

Therefore, the total electrostatic energy is $49.3 \, \text{J}$ . The correct option is B.

Common mistakes

Using only the mutual interaction term $\frac{k q_1 q_2}{r_{12}}$ and ignoring the energies $q_1V(r_1)$ and $q_2V(r_2)$ due to the external field. This is wrong because the charges are placed in a pre-existing field. Always add both external-field contributions and pair interaction energy.
Taking the potential as negative, $V(r) = -\frac{A}{r}$ . This is wrong here because with $V(\infty)=0$ and $E = \frac{A}{r^2}\hat{r}$ , the integral gives $V(r)=\frac{A}{r}$ . Evaluate the definite integral carefully.
Using $9 \, \text{cm}$ and $18 \, \text{cm}$ directly without converting to metres. This gives incorrect numerical values because SI units are required in electrostatics formulas. Convert to $0.09 \, \text{m}$ and $0.18 \, \text{m}$ first.

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