Question

Three charged particles are arranged in a line as shown in figure below. Charge A = -5 μC, charge B = +10 μC and charge C = -12 μC. Calculate the net electrostatic force on particle B due to the other two charges

Answer 1

Electrostatic force between two charges q1 and q2

$F = \frac{kq_1q_2}{d^2}$

d = distance between them

Forces applied on particle B will be attractive in nature.

$F_1 = \frac{k(10 \times 10^{-6})(12 \times 10^{-6})}{(4 \times 10^{-2})^2} \\ = \frac{9 \times 10^9 \times 120 \times 10^{-12}}{16 \times 10^{-4}} \\ = \frac{2700}{4} \;N \\ = 675 \;N$

$F_2 = \frac{9 \times 10^9 \times (5 \times 10^{-6})(10 \times 10^{-6})}{(6 \times 10^{-2})^2} \\ = \frac{4500}{36} \\ = 125 \;N$

Net force $= F_1 -F_2$

$= 675 -125 \\ = 550 \;N$

Answer: 550 N