Two charges 7 and -4 are placed at (-7 cm, 0, 0) and (7 cm, 0, 0) respectively. Given, ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², the electrostatic potential energy of the charge configuration is: (a) -2.0 J (b) -1.5 J (c) -1.2 J (d) -1.8 J

Answer: (d). Separation r = 14 cm = 0.14 m. U = (kq₁q₂)/(r) = 9×10⁹ × 7×10⁻⁶ × (-4×10⁻⁶)0.14 U = (-0.252)/(0.14) = -1.8 J So the answer is (d) -1.8 J.

Two charges 7 and -4 are placed at (-7 cm, 0, 0) and (7 cm, 0, 0)… | JEE Main 2025 (23 Jan, Shift 2)

Q21JEE Main 2025 · 23 JAN · SHIFT 2Easy

Electrostatic Potential Energy

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

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Answer: (d)

Separation $r = 14$ cm $= 0.14$ m.
$U = \dfrac{kq_1q_2}{r} = \dfrac{9\times10^9 \times 7\times10^{-6} \times (-4\times10^{-6})}{0.14}$
$U = \dfrac{-0.252}{0.14} = -1.8$ J
So the answer is (d) $-1.8$ J.

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Electrostatic Potential Energy question from JEE Main 2025 (23 Jan, Shift 2)

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