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Electric charges of $1 μC, - 1 μC$ and $2 μC$ are placed in air at the corners A, B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is

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3.6 N

2.7 N

0.9 N

1.8 N

answer is B.

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Detailed Solution

F_A = force on C due to charge placed at A

$= 9 \times 10^{9} \times \frac{10^{- 6} \times 2 \times 10^{- 6}}{{(10 \times 10^{- 2})}^{2}} = 1 .8 N$

F_B = force on C due to charge placed at B

$= 9 \times 10^{9} \times \frac{10^{- 6} \times 2 \times 10^{- 6}}{(0 .1)^{2}} = 1 .8 N$

Net force on C

$F_{net} = \sqrt{{(F_{A})}^{2} + {(F_{B})}^{2} + 2 F_{A} F_{B} \cos 120^{o}} = 1 .8 N$

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