# A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges

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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

by (80.1k points)
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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

by (10 points)
Thanks dear
by (51.9k points)

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 = $-\frac{Q}{4}$.