Question

A charge of 4.10mC is placed at each corner of a square 0.100 m on a side.

A) Determine the magnitude of the force on each charge.

B) Determine the direction of the force on each charge. Assume that the positive x-axis is directed to the right.

Answer 1

First let's calculate the magnitude and direction of force due to other charges on charge at the upper right hand corner of the square (Q₃)

X - component of the forces due to other charges

F_x = F₃₁ cos 45 + F₃₂cos 90 + F₃₄ cos 0

        = 8.99 * 10⁹ * 4.10 * 10^-3 * 4.10 * 10^-3 *[(0.707)/(0.1414)² + (1 + 0) / (0.100)²]

F_x = 20.454 * 10⁶ N

Y - component of the forces due to other charges

F_y = F₃₁ sin 45 + F₃₂sin 90 + F₃₄ sin 0

        = 8.99 * 10⁹ * 4.10 * 10^-3 * 4.10 * 10^-3 *[(0.707)/(0.1414)² + (1 + 0) / (0.100)²]

F_y = 20.454 * 10⁶ N

Magnitude of the force on Q₃ due to other charges is,

F = [ ( Fx)² +( F_y)² ]

F = 28.922 * 10⁶ N

Direction of the force is given by,

= tan^-1 ( F_y / F_x)

= 45⁰

Since, Q₁ = Q₂ = Q₃ = Q₄, magnitude of the force on each charge due to other charges is the same and the directions are,

For F₁, = 225⁰

For F₂, = 315⁰

For F₃, = 45⁰

For F₄, = 135⁰, all with respect to positive x-axis.

(Image for representation only, not drawn to the scale.)