Three point charges of magnitude 5, 0.16 and 0.3 are located at the vertices A, B, C of a right angled triangle whose sides are AB = 3 cm, BC = 3√(2) cm and CA = 3 cm and point A is the right angle corner. Charge at point A experiences ____________ N of electrostatic force due to the other two charges. Numerical answer type. Work it out, then open the answer.

Answer: 17. Right angle at A, so forces from B and C on A are perpendicular; AB = AC = 0.03 m. FB = 9 × 10⁹ × 5 × 10⁻⁶ × 0.16 × 10⁻⁶9 × 10⁻⁴ = 8 N FC = 9 × 10⁹ × 5 × 10⁻⁶ × 0.3 × 10⁻⁶9 × 10⁻⁴ = 15 N F = √(8² + 15²) = √(289) = 17 N.

Three point charges of magnitude 5, 0.16 and 0.3 are located at the… | JEE Main 2022 (26 Jul, Shift 2)

Q47JEE Main 2022 · 26 JUL · SHIFT 2Medium

Electric Charge & Coulomb's Law

Three point charges of magnitude $5\,\mu\text{C}, 0.16\,\mu\text{C}$ and $0.3\,\mu\text{C}$ are located at the vertices $A, B, C$ of a right angled triangle whose sides are $AB = 3\text{ cm}, BC = 3\sqrt{2}\text{ cm}$ and $CA = 3\text{ cm}$ and point $A$ is the right angle corner. Charge at point $A$ experiences ____________ $\text{N}$ of electrostatic force due to the other two charges.

Numerical answer type. Work it out, then open the answer.

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Answer: 17

Right angle at A, so forces from B and C on A are perpendicular; $AB = AC = 0.03\text{ m}$ .
$F_B = \dfrac{9 \times 10^9 \times 5 \times 10^{-6} \times 0.16 \times 10^{-6}}{9 \times 10^{-4}} = 8\text{ N}$
$F_C = \dfrac{9 \times 10^9 \times 5 \times 10^{-6} \times 0.3 \times 10^{-6}}{9 \times 10^{-4}} = 15\text{ N}$
$F = \sqrt{8^2 + 15^2} = \sqrt{289} = 17\text{ N}$ .

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