Question

Four identical charges Q are placed at the corners of a square of side L. (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges

Short answer

(a) The free body diagram is:

(b) The force is and the direction of this force is away from the center of the square.

Step-by-step solution

Free body diagram

The free body diagram is as follows: charge 2 exerted force F₂ in y-direction only. While the charge 1 exerted a force F₁ which is divided into two forces F_1x,.and F_2y, as shown with the red dashed lines in the figure. Charge 3 exerted a force F3 which is divided into two forces F­_3x and F_3y as shown with the black dashed lines in the figure.

So the free-body diagram will be given as shown below:

Direction of the force

Calculate the magnitude and direction of the total force F_t; exerted on the charge 4 by the other three charges (1, 2, 3). The forces that exerted on the charge 4 are F₂, F_1x, F_1y, F_3x, and F_3y, Where F₂ is in the y-direction and F₁ and F₃, are with the same magnitude and in opposite directions, so their summation will be zero.

$F_{1x}+F_{3x}=0$

The total force exerted on charge 4 is:

$F_t=F_2+F_{1y}+F_3$

where

role="math" localid="1668404118501" $\begin{array}{r}{F}_2=\frac{1}{4\pi {\epsilon }_3}\frac{Q_3{Q}_4}{(\sqrt{2}L{)}^2}\\ F_{1y}=F_1\cos \theta =\frac{Q_1{Q}_4}{L^2}\cos \theta \\ F_{3y}=F_3\cos \theta =\frac{Q_3{Q}_4}{L^2}\cos \theta \\ \theta ={45}^{\circ }\end{array}$

Magnitude of the force

Substitute to find value of F­_t

_{$\begin{array}{r}{F}_t=F_2+F_{1y}+F_{3y}\\ =\frac{1}{4\pi {\epsilon }_y}\frac{Q^2}{(\sqrt{2}L{)}^2}+2\left(\frac{1}{4\pi {\epsilon }_2}\frac{Q^2}{(L{)}^2}\cos \theta \right)\\ =\frac{1}{8\pi {\epsilon }_y}\frac{Q^2}{L^2}(1+2\sqrt{2})\end{array}$}

Therefore, the force is $=\frac{1}{8\pi {\epsilon }_y}\frac{Q^2}{L^2}(1+2\sqrt{2})$and the direction of this force is away from the center of the square