In figure, two equal positive point charges q₁ = q₂ = 2.0 $μ$ C interact with a third point charge Q = 4.0 $μ$ C. The magnitude, as well as direction, of the net force on Q is

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0.46 N in the +x-direction

0.46 N in the -x-direction

0.23 N in the +x-direction

0.23 N in the -x-direction

answer is B.

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Detailed Solution

$\begin{matrix} F_{net} = 2 |F_{31}| \cos α \\ = 2 \times \frac{1}{4 π ε_{0}} \times \frac{2 \times 4 \times 10^{- 12}}{(0.5)^{2}} \times \frac{4}{5} = 0.46 N \end{matrix}$

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