## College Physics (4th Edition)

At the center of the square, the electric field is zero, and the electric potential is $5.7\times 10^6~V$
We can write an expression for the magnitude of the electric field at the center due to each charge: $E = \frac{k~q}{r^2}$ Since the electric field due to the charge at point $a$ is equal and opposite to the electric field due to the charge at point $c$, and the electric field due to the charge at point $b$ is equal and opposite to the electric field due to the charge at point $d$, the net electric field at the center of the square is zero. We can find the potential at the center of the square: $\Delta V = 4\times \frac{k~q}{r}$ $\Delta V = 4\times \frac{(9.0\times 10^9~N~m^2/C^2)(9.0\times 10^{-6}~C)}{\sqrt{2}\times 10^{-2}~m}$ $\Delta V = 5.7\times 10^6~V$ At the center of the square, the electric field is zero, and the electric potential is $5.7\times 10^6~V$