91影视

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颅_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:

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" $\begin{array}{r}{F}_2=\frac{1}{4蟺{蔚}_3}\frac{Q_3{Q}_4}{(\sqrt{2}L{)}^2}\\ F_{1y}=F_1\cos 鈦�胃=\frac{Q_1{Q}_4}{L^2}\cos 鈦�胃\\ F_{3y}=F_3\cos 鈦�胃=\frac{Q_3{Q}_4}{L^2}\cos 鈦�胃\\ 胃={45}^{鈭�}\end{array}$

Substitute to find value of F颅_t

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

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