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A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to

(a) -Q/4

(b) Q/4 

(c) -Q/2 

(d) Q/2

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Best answer

Correct option (a) -Q/4

Explanation:

The situation is as shown in the figure.

Let two equal charges Q each placed at points A and B at a distance r apart. C is the centre of AB where charge q is placed.

For equilibrium, net force on charge Q= 0

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Thanks dear
+4 votes
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The answer is: (a) -Q/4

Charge q is in equilibrium since charges A and B exert equal and opposite forces on it.

For equilibrium of charge Q at B;

\(F_{BC}\,+\,F_{AB}=0\)

\(\implies \frac{1}{4\pi\epsilon_o}\frac{qQ}{(\frac{1}{2})^2}\) + \(\frac{1}{4\pi\epsilon_o}\frac{Q.Q}{1^2}\) = 0

\(\implies \frac{1}{4\pi\epsilon_o}\frac{Q}{(1)^2}(4q+Q)\) = 0

\(\implies\) q = \(-\frac{Q}{4}\).

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