Two charges q₁ = +6q and q₂ = -3q placed as shown in figure. A proton is placed on x -axis away from q₂. To remain proton in equilibrium, the distance between q₁ and proton is (a) ((√(2))/(√(2)-1))L (b) 2L (c) L/2 (d) ((√(2))/(√(2)+1))L

Answer: (a). Proton beyond q₂ at distance x from q₁: forces from q₁ (repulsion) and q₂ (attraction) must balance. 6q/x² = (3q)/((x - L)²) ⇒ (x)/(x - L) = √(2) x(√(2) - 1) = √(2)L ⇒ x = (√(2))/(√(2) - 1)L So the answer is (a).

Two charges q₁ = +6q and q₂ = -3q placed as shown in figure. A proton… | MHT CET 2025 (19 Apr, Shift 1)

Q24MHT CET 2025 · 19 APR · SHIFT 1Medium

Electric Charge & Coulomb's Law

Two charges $q_1 = +6q$ and $q_2 = -3q$ placed as shown in figure. A proton is placed on x -axis away from $q_2$ . To remain proton in equilibrium, the distance between $q_1$ and proton is

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Answer: (a)

Proton beyond $q_2$ at distance $x$ from $q_1$ : forces from $q_1$ (repulsion) and $q_2$ (attraction) must balance.
$\dfrac{6q}{x^2} = \dfrac{3q}{(x - L)^2} \Rightarrow \dfrac{x}{x - L} = \sqrt{2}$
$x(\sqrt{2} - 1) = \sqrt{2}L \Rightarrow x = \dfrac{\sqrt{2}}{\sqrt{2} - 1}L$
So the answer is (a).

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Electric Charge & Coulomb's Law question from MHT CET 2025 (19 Apr, Shift 1)

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