Electric charges of $1 μ C, - 1 μ C$ and $2 μ C$ 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

0.9 N

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1.8 N

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2.7 N

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3.6 N

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Solution

The correct option is B 1.8 N
$F_{A}$ = 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
$F_{n e t} = \sqrt{(F_{A})^{2} + (F_{B})^{2} + 2 F_{A} F_{B} c o s 120^{\circ}} = 1.8 N$

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