Question

Three charges are arranged as shown in the figure below. Find the magnitude and direction of the electrostatic force on the charge q = 4.82 nC at the origin. (Let r₁₂ = 0.295m.)

magnitude	N
direction

Answer 1

X - component of the force on 4.82 nC due to 6.00 nC is

F_x = k * 4.82 * 10^-9 * 6.00 * 10^-9* cos 0⁰ / (0.295)²

    = 8.99 * 10⁹ *4.82 * 10^-9 * 6.00 * 10^-9* 1 / (0.295)²

F_x = 2.988 * 10^-6 N

Y - component of the force on 4.82 nC due to 6.00 nC is

F_y =k * 4.82 * 10^-9 * 6.00 * 10^-9* sin 0⁰ / (0.295)²

F_y = 0

Similarly,

X - component of the force on 4.82 nC due to -3.00 nC is

F_x = k * 4.82 * 10^-9 * 3.00 * 10^-9* cos 270⁰ / (0.100)²

F_x = 0

Y - component of the force on 4.82 nC due to -3.00 nC is

F_y =k * 4.82 * 10^-9 * 3.00 * 10^-9* sin 270⁰ / (0.100)²

    = 8.99 * 10⁹ *4.82 * 10^-9 * 3.00 * 10^-9* -1 / (0.01)

F_y = 12.9995 * 10^-6 N

Thus,

F_x = 2.988 * 10^-6 N

F_y = 12.9995 * 10^-6 N

Now, magnitude of the force is given by

F = [ (F_x)² + (F_y)² ]

F = 10^-6 * (8.928 +168.987)

F = 13.338 * 10^-6 N

Direction of the force is given by,

= tan^-1 ( F_{y /} F_x )

= 77.06⁰ in the third quadrant

i.e. = 257.06⁰ appr.