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

Given: q1 = 7 10^-6 C, q2 = -4 10^-6 C, 0 = 8.85 10^-12 C^2 N^-1 m^-2. The charges are at (-7 cm, 0, 0) and (7 cm, 0, 0), so the separation is r = 14 cm = 0.14 m. Find: The electrostatic potential energy of the two-charge configuration. For two point charges, U = k q1 q2r where k = 14 0 Now, k = 14 8.85 10^-12 8.99 10^9 N m^2 C^-2 Substituting the values, U = 8.99 10^9 7 10^-6 (-4) 10^-60.14 U = -8.99 10^9 28 10^-120.14 U = -251.72 10^-30.14 U = -1.8 J The potential energy is negative because the charges are of opposite signs. Therefore, the correct option is D. The solution states: The Correct Option is D. Given: q1 = 7 10^-6 C, q2 = -4 10^-6 C, r = 0.14 m. Find: U. Using the relation directly, U = q1 q24 0 r Substitute: U = (7 10^-6)(-4 10^-6)4 (8.85 10^-12)(0.14) U = -1.8 J Hence, the electrostatic potential energy is -1.8 J, so the correct option is D.

Two charges 7 C and -4 C are placed at (-7 cm, 0, 0) and (7 cm, 0… — Solution | JEE Physics

MCQEasyJEE 2025Electric Potential & Potential Energy

Two charges $7 \, \mu C$ and $-4 \, \mu C$ are placed at $(-7 \, cm, 0, 0)$ and $(7 \, cm, 0, 0)$ respectively. Given, $\epsilon_0 = 8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}$ , the electrostatic potential energy of the charge configuration is:

A
$-1.5 \, J$
B
$-2.0 \, J$
C
$-1.2 \, J$
D
$-1.8 \, J$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $q_1 = 7 \times 10^{-6} \, \text{C}$ , $q_2 = -4 \times 10^{-6} \, \text{C}$ , $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2}$ . The charges are at $(-7 \, \text{cm}, 0, 0)$ and $(7 \, \text{cm}, 0, 0)$ , so the separation is $r = 14 \, \text{cm} = 0.14 \, \text{m}$ .

Find: The electrostatic potential energy of the two-charge configuration.

For two point charges,

U = \frac{k q_1 q_2}{r}

where

k = \frac{1}{4\pi \epsilon_0}

Now,

k = \frac{1}{4\pi \times 8.85 \times 10^{-12}} \approx 8.99 \times 10^9 \, \text{N m}^2 \text{C}^{-2}

Substituting the values,

U = \frac{8.99 \times 10^9 \times 7 \times 10^{-6} \times (-4) \times 10^{-6}}{0.14}

U = \frac{-8.99 \times 10^9 \times 28 \times 10^{-12}}{0.14}

U = \frac{-251.72 \times 10^{-3}}{0.14}

U = -1.8 \, \text{J}

The potential energy is negative because the charges are of opposite signs. Therefore, the correct option is D.

The solution states: The Correct Option is D.

Direct Formula Form

Given: $q_1 = 7 \times 10^{-6} \, \text{C}$ , $q_2 = -4 \times 10^{-6} \, \text{C}$ , $r = 0.14 \, \text{m}$ .

Find: $U$ .

Using the relation directly,

U = \frac{q_1 q_2}{4 \pi \epsilon_0 r}

Substitute:

U = \frac{(7 \times 10^{-6})(-4 \times 10^{-6})}{4 \pi (8.85 \times 10^{-12})(0.14)}

U = -1.8 \, \text{J}

Hence, the electrostatic potential energy is $-1.8 \, \text{J}$ , so the correct option is D.

Common mistakes

Using $r = 7 \, \text{cm}$ instead of $14 \, \text{cm}$ . The charges are on opposite sides of the origin, so the separation is the total distance between them. Add the magnitudes of the coordinates to get $r = 14 \, \text{cm}$ .
Forgetting to convert microcoulomb and centimetre into SI units. Using $\mu \text{C}$ and $\text{cm}$ directly gives an incorrect numerical value. First convert to $\text{C}$ and $\text{m}$ .
Missing the negative sign in potential energy. Since the two charges have opposite signs, $q_1 q_2 < 0$ and the electrostatic potential energy must be negative, indicating attraction.

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