Two point charges q₁ = 3 and q₂ = -4 are placed at points (2i + 3j + 3k) and (i + j + k) respectively. Force on charge q₂ is ______ N. (Take (1)/(4πε₀) = 9 × 10⁹ SI Units) (a) (12i + 24j + 24k) × 10⁻³ (b) (4i + 8j + 8k) × 10⁻³ (c) (3i + 6j + 6k) × 10⁻³ (d) (-4i - 8j - 8k) × 10⁻³

Answer: (b). Vector from q₁ to q₂: r = ( i+ j+ k)-(2 i+3 j+3 k) = - i-2 j-2 k, | r| = 3 m. F₂ = (kq₁q₂)/(r³) r = 9×10⁹×3×10⁻⁶×(-4×10⁻⁶)27(- i-2 j-2 k) = -4×10⁻³(- i-2 j-2 k) = (4 i+8 j+8 k)×10⁻³ N So the answer is (b).

Two point charges q₁ = 3 and q₂ = -4 are placed at points (2i + 3j +… | JEE Main 2026 (8 Apr, Shift 2)

Q6JEE Main 2026 · 8 APR · SHIFT 2Easy

Electric Charge & Coulomb's Law

Two point charges $q_1 = 3\ \mu\text{C}$ and $q_2 = -4\ \mu\text{C}$ are placed at points $(2\hat{i} + 3\hat{j} + 3\hat{k})$ and $(\hat{i} + \hat{j} + \hat{k})$ respectively. Force on charge $q_2$ is ______ N. (Take $\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}$ SI Units)

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Answer: (b)

Vector from $q_1$ to $q_2$ : $\vec r = (\hat i+\hat j+\hat k)-(2\hat i+3\hat j+3\hat k) = -\hat i-2\hat j-2\hat k$ , $|\vec r| = 3\ \text{m}$ .
$\vec F_2 = \dfrac{kq_1q_2}{r^3}\vec r = \dfrac{9\times10^9\times3\times10^{-6}\times(-4\times10^{-6})}{27}(-\hat i-2\hat j-2\hat k)$
$= -4\times10^{-3}(-\hat i-2\hat j-2\hat k) = (4\hat i+8\hat j+8\hat k)\times10^{-3}\ \text{N}$
So the answer is (b).

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Electric Charge & Coulomb's Law question from JEE Main 2026 (8 Apr, Shift 2)

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