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Q.

Electric charges of 1 μC, -1 μC and 2 μC are placed in air at the coners 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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a

3.6 N

b

2.7 N

c

1.8 N

d

0.9 N

answer is B.

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Detailed Solution

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$\begin{array}{l} F_{A} = Force on C due to charge placed at A \\ = 9 \times 10^{9} \times \frac{10^{- 6} \times 2 \times 10^{- 6}}{{(10 \times 10^{- 2})}^{2}} = 1 .8 N \\ F_{B} = Force on C due to charge placed at B \\ = 9 \times 10^{9} \times \frac{10^{- 6} \times 2 \times 10^{- 6}}{{(0 .1)}^{2}} = 1 .8 N \end{array}$

Question Image

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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