MHT CET · Physics · Electrostatics
Two charges \(\mathrm{q}_1=+6 \mathrm{q}\) and \(\mathrm{q}_2=-3 \mathrm{q}\) are placed as shown in figure. A proton is placed on x -axis away from \(\mathrm{q}_2\). To remain proton in equilibrium, the distance between \(\mathrm{q}_1\) and proton is
- A \(\left(\frac{\sqrt{2}}{\sqrt{2}-1}\right) \mathrm{L}\)
- B 2 L
- C \(\frac{L}{2}\)
- D \(\left(\frac{\sqrt{2}}{\sqrt{2}+1}\right) \mathrm{L}\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{\sqrt{2}}{\sqrt{2}-1}\right) \mathrm{L}\)
Step-by-step Solution
Detailed explanation
\(k \frac{|q_1| e}{x^2} = k \frac{|q_2| e}{(x-L)^2}\) \(\frac{6q}{x^2} = \frac{3q}{(x-L)^2}\)
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