Question

Three point charges are arranged as shown in figure 40. Find(a) the magnitude and (b) the direction of the electric force on the particle at the origin.

Answer 1

The electrostatic force on the charge q due to q₁ is directed along negative x-axis

$F_x = -k\frac{qq_1}{r_1^2}$

k is the Coulomb’s constant.

$F_x = -(9 \times 10^9)\frac{(5 \times 10^{-9})(6 \times 10^{-9})}{0.3^2} = -3 \times 10^{-6} \;N$

The electrostatic force on the charge q due to q₂ is directed along negative y-axis

$F_y = -k\frac{qq_2}{r_2^2}$

$F_y = -(9 \times 10^9)\frac{(5 \times 10^{-9})(3 \times 10^{-9})}{0.1^2} = -1.35 \times 10^{-5} \;N$

a) The magnitude of the net force on the charge q is

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$F_{net} = \sqrt{(-3 \times 10^{-6})^2 + (-1.35 \times 10^{-5})^2} = 1.38 \times 10^{-5} \;N$

b) The direction of net force acting on the charge q is

$φ = tan^{-1}(\frac{F_y}{F_x})$

$φ = tan^{-1}(\frac{-1.35 \times 10^{-5}}{-3 \times 10^{-6}}) = 77.5$ º

Since both the components of electric force are negative, the direction of the net force is 180º + 77.5º = 257.5º counter clock wise from the positive x-axis.