Q. In figure, two equal positive point charges q_{1} = q_{2} = 2.0μC interact with a third point charge Q = 4.0 μC . The magnitude, as well as direction, of the net force on Q is

(a) 0.23 N in the +x-direction

(b) 0.46 N in the +x-direction

(c) 0.23 N in the -x-direction

(d) 0.46 N in the -x-direction

**Click to See Answer : **

_{1}is

$\displaystyle F_1 = \frac{1}{4\pi \epsilon_0}\frac{q_1 Q}{r^2} $

Force acting on Q due to q_{2} is

$ \displaystyle F_2 = \frac{1}{4\pi \epsilon_0}\frac{q_2 Q}{r^2} $

Here , F_{1} = F_{2}

Taking components along X & Y axis , the component along Y axis will be cancel out .

Net force on Q is

$ \displaystyle F_{net}= 2 F_1 cos\alpha $

$ \displaystyle F_{net}= 2(\frac{1}{4\pi \epsilon_0}\frac{q_1 Q}{r^2}) cos\alpha $

$ \displaystyle F_{net}= 2(9\times 10^9 \times \frac{2\times 10^{-6} \times 4\times 10^{-6}}{(0.5)^2}) (\frac{0.4}{0.5}) $

= 0.46 N in the +x-direction