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Two charges of +200µC and -200µC are placed at the corners B and C of an equilateral triangle ABC of side 0.1 m. The force on a charge of 5µC placed A is

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600√3N

1200√3N

900N

1800 N

answer is D.

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Detailed Solution

$F_{1} = F_{2} = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r^{2}}; F_{r} = F_{1} = F_{2}$ because angle between then is 120°

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