Question

(II) A charge of 6.15 mC is placed at each corner of a square
0.100 m on a side. Determine the magnitude and direction
of the force on each charge.

Answer 1

Answer

All charges are same therefore force is equal but direction is different.

So diagram is as above

Force at above charge ( right above side corner)

$F=\frac{\sqrt{2}kq^2}{a^2}+\frac{kq^2}{\sqrt{2}a^2}$

With direction and side of square 0.1m and q=6.15mC

So force is

$F_R=728.12\times10^5\frac{(i+j)}{\sqrt{2}}$

Similiarly force for above left corner

$F'_R=728.12\times10^5\frac{(-i+j)}{\sqrt{2}}$

And now bottom left charge

$F''_R=728.12\times10^5\frac{(-i-j)}{\sqrt{2}}$

Now right bottom charge

$F_R=728.12\times10^5\frac{(i-j)}{\sqrt{2}}$