91影视

a)The net force on 3q at the center of the square is only $F_3$because$F_1$and$F_2$are equal in magnitude and opposite in direction so$F_{net}$

$F_{net}=\frac{3k{q}^2}{r^2}=\frac{3k{q}^2}{(\sqrt{2L/2}{)}^2}=\frac{6k{q}^2}{L^2}$

Hence the net force is $\frac{6k{q}^2}{L^2}$at the angle 45deg. Below the horizontal.

The force acting on 3q at the corner has two-component

${\mathrm{F}}_{\mathrm{x}}=鈭�\mathrm{F}={\mathrm{F}}_1+{\mathrm{F}}_3\sin 鈦�\left({45}^{鈭�}\right)鈥�鈭�>\left(1\right)$

The magnitude of $F_1$

$F_1=\frac{3k{q}^2}{L^2}$

And $F_1$is

$F_3=\frac{3k{q}^2}{(\sqrt{2}L{)}^2}$

Substitute the above values in equation(1)

$F_x=\frac{3k{q}^2}{L^2}+\frac{3k{q}^2}{(\sqrt{2}L{)}^2}\left(\frac{\sqrt{2}}{2}\right)=\frac{3k{q}^2}{L^2}\left(1+\frac{\sqrt{2}}{4}\right)=\left(1.35\right)\frac{3k{q}^2}{L^2}$

Similarly, the force in the y-direction is

$F_y=鈭�F=F_2+F_3\cos 鈦�\left({45}^{鈭�}\right)$

The magnitude of

$F_2=\frac{3k{q}^2}{L^2}$

The net force in the y-direction is

$F_y=\frac{3k{q}^2}{L^2}+\frac{3k{q}^2}{(\sqrt{2}L{)}^2}\left(\frac{\sqrt{2}}{2}\right)=\frac{3k{q}^2}{L^2}\left(1+\frac{\sqrt{2}}{4}\right)=\left(1.35\right)\frac{3k{q}^2}{L^2}$

The magnitude of the field

$F=\sqrt{\left(F_x^2+F_y^2\right)}=\left(1.35\right)\left(\sqrt{2}\right)\left(\frac{3k{q}^2}{L^2}\right)=\left(1.9\right)\left(\frac{3k{q}^2}{L^2}\right)$

Hence the net force is $\left(1.9\right)\left(\frac{3k{q}^2}{L^2}\right)$and the angle is below the horizontal by 45deg.