Three point charges $+ 2 μC, + 2 μC,$ and $+ 5 μC$ are placed respectively at the vertices A, B, C of an equilateral triangle of side 0.2 m. The magnitude of the force experienced by the charge at C is $(\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} newton - m^{2} / C^{2})$

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$2.25 n e w t o n$

$2.25 \sqrt{3} n e w t o n$

$4.5 n e w t o n$

$4.5 \sqrt{3} n e w t o n$

answer is B.

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Detailed Solution

$F_{net} = \sqrt{F^{2} + F^{2} + 2 FFcos 60^{\circ}} = \sqrt{3} F$

Where, $F = \frac{9 \times 10^{9} \times 2 \times 10^{- 6} \times 5 \times 10^{- 6}}{0 . 2^{2}} N$

$\Rightarrow F_{n e t} = \sqrt{3} F = \frac{\sqrt{3} \times 9 \times 10^{9} \times 2 \times 5 \times 10^{- 12}}{0 \cdot 2^{2}} = 2 .25 \sqrt{3} N$

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