Question

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

• A. \(\left(\frac{\sqrt{2}}{\sqrt{2}-1}\right)L\)
• B. \(2L\)
• C. \(\frac{L}{2}\)
• D. \(\left(\frac{\sqrt{2}}{\sqrt{2}+1}\right)L\)

Hint:

For equilibrium: equate magnitudes of forces and carefully choose correct region.

Answer 1

The Correct Option is A

Concept:
For equilibrium of proton: \[ F_1 = F_2 \] (Coulomb's law) Step 1: Assume position. Let distance of proton from \(q_1\) be \(x\). Then distance from \(q_2\): \[ = x - L \] (Proton is placed beyond \(q_2\))
Step 2: Apply force balance. \[ \frac{k \cdot 6q \cdot e}{x^2} = \frac{k \cdot 3q \cdot e}{(x - L)^2} \] Cancel common terms: \[ \frac{6}{x^2} = \frac{3}{(x - L)^2} \]
Step 3: Simplify. \[ \frac{2}{x^2} = \frac{1}{(x - L)^2} \] \[ (x - L)^2 = \frac{x^2}{2} \]
Step 4: Take square root. \[ x - L = \frac{x}{\sqrt{2}} \]
Step 5: Solve for \(x\). \[ x - \frac{x}{\sqrt{2}} = L \] \[ x\left(1 - \frac{1}{\sqrt{2}}\right) = L \] \[ x = \frac{L}{1 - \frac{1}{\sqrt{2}}} \]
Step 6: Rationalize. \[ x = \frac{L\sqrt{2}}{\sqrt{2} - 1} \]
Step 7: Conclusion. \[ x = \left(\frac{\sqrt{2}}{\sqrt{2}-1}\right)L \]