Question

charges of 0.46 are placed at three corners of a square side length 3.4 cm. A charge of -0.64 is placed at the fourth corner. Find the net electric field and the total electric potential at the center of the square.

Answer 1

A q 0.46C B q side of cube a-3.4cm q-0.64C

distance $AC=\sqrt{(AB)^{2}+(AC)^{2}}=\sqrt{3.4^{2}+3.4^{2}}=4.80cm$

so $AO=2.40cm$

part a

net electric field at center O is

$V=\frac{kq}{AO}+\frac{kq}{BO}+\frac{kq}{CO}-\frac{kq}{DO}$   

as in cube shown above AO=BO=CO=DO

SO

$V=\frac{kq}{AO}+\frac{kq}{AO}+\frac{kq}{AO}-\frac{kq'}{AO}$

$V=3\frac{kq}{AO}-\frac{kq'}{AO}=\frac{2\times 9\times 10^{9}\times(3\times 0.46-0.64)}{2.40\times 10^{-2}}$

$V=2.775\times 10^{11}V$

PART B

due to 2 t 3 i duu to 1 at C ос Resultant ah. E, and E ki k1 Resultant ok Er and E