Answer to Question #138546 in Electric Circuits for mungu

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.