Question

1. At each corner of a square of side L there are point charges of magnitudes Q, 2Q, 3Q, and 4Q. Determine the magnitude and direction of the force on the charge 2Q. 

Answer 1

The force exerted from Q to 2Q is $F_1 = \dfrac{kQ\cdot 2Q}{L^2} = \dfrac{2kQ^2}{L^2}$ ,

the force exerted from 4Q to 2Q is $F_4 = \dfrac{k4Q\cdot 2Q}{L^2} = \dfrac{8kQ^2}{L^2}$ .

The distance from 3Q to 2Q is $\sqrt2 L$ , so the force exerted from 3Q to 4Q is

$F_3 = \dfrac{k3Q\cdot 2Q}{2L^2} = \dfrac{3kQ^2}{L^2}$ .

Let x-axis be directed from 2Q to Q and y-axis be directed from 2Q to 4Q. Let us write the projections of forces on the axes:

$x: \; F_x = \dfrac{2kQ^2}{L^2} + \dfrac{3kQ^2}{L^2} \cos 45^\circ = \dfrac{kQ^2}{L^2}\cdot\left(2 + 3\dfrac{\sqrt2}{2}\right) = \dfrac{kQ^2}{L^2} \cdot4.12, \\ y: \; F_y = \dfrac{8kQ^2}{L^2} + \dfrac{3kQ^2}{L^2} \cos 45^\circ = \dfrac{kQ^2}{L^2}\cdot\left(8 + 3\dfrac{\sqrt2}{2}\right)= \dfrac{kQ^2}{L^2} \cdot10.12.$

The total force will be $F = \sqrt{F_x^2 + F_y^2} = \dfrac{kQ^2}{L^2} \cdot 10.9.$

The angle between F and x-axis is $\cos \alpha = F_x/F = 4.12/10.9 \approx 0.38, \; \alpha = 68^\circ.$