Question:medium

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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For equilibrium: equate magnitudes of forces and carefully choose correct region.
Updated On: May 14, 2026
  • \(\left(\frac{\sqrt{2}}{\sqrt{2}-1}\right)L\)
  • \(2L\)
  • \(\frac{L}{2}\)
  • \(\left(\frac{\sqrt{2}}{\sqrt{2}+1}\right)L\)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
For a charge to be in equilibrium, the net electrostatic force on it must be zero.
This occurs at a point where the electric fields produced by the two fixed charges are equal in magnitude and opposite in direction.
Since the charges have opposite signs, the equilibrium point must lie on the line joining them but outside the segment connecting them, closer to the smaller magnitude charge (\(q_2\)).
Step 2: Key Formula or Approach:
Coulomb's Law field formula: \(E = \frac{kq}{r^2}\).
Let the proton be at distance \(x\) from \(q_1\). Then its distance from \(q_2\) is \((x - L)\).
Step 3: Detailed Explanation:
Equating the magnitudes of the electric fields: \[ \frac{k(6q)}{x^2} = \frac{k(3q)}{(x - L)^2} \] Cancel common terms \(k\), \(q\), and simplify coefficients: \[ \frac{2}{x^2} = \frac{1}{(x - L)^2} \] Take the square root of both sides: \[ \frac{\sqrt{2}}{x} = \frac{1}{x - L} \] Cross-multiply to solve for \(x\): \[ \sqrt{2}(x - L) = x \] \[ \sqrt{2}x - \sqrt{2}L = x \] \[ x(\sqrt{2} - 1) = \sqrt{2}L \] \[ x = \frac{\sqrt{2}L}{\sqrt{2} - 1} = \left( \frac{\sqrt{2}}{\sqrt{2}-1} \right)L \] Step 4: Final Answer:
The distance from \(q_1\) to the proton for equilibrium is \( \left( \frac{\sqrt{2}}{\sqrt{2}-1} \right)L \).
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