Question

Three 3.0μC charges are at the three corners of a square of side 0.50m. The last corner is occupied by a -3.0μC charge. Find the electric field at the center of the square.

Answer 1

Electric field due to charge

$q_1$ And $q_3$

$E_1=E_3=\frac{kq_1}{r^2}=\frac{kq_3}{r^2}$

$$E_1=E_3=\frac{9\times10^9\times3\times10^{-6}}{0.25^2}=432000N/C$$

At symmetry $E_1,E_3$ Are equal and opposite direction

Now net electric field due to charge

$q_2,q_4$

$E_{net}=E_2+E_4$

$$E_{net}=\frac{9\times10^9\times3\times10^{-6}}{0.25^2}+\frac{9\times10^9\times3\times10^{-6}}{0.25^2}=432000+432000=864000N/C$$