## College Physics (4th Edition)

Assuming that the electric potential at infinity is zero, we can find the electric potential at the point $c$: $V_c = k~(\frac{q_1}{r_1}+\frac{q_2}{r_2})$ $V_c = (9.0\times 10^9~N~m^2/C^2)~(\frac{8.00\times 10^{-9}~C}{0.12~m}+\frac{-8.00\times 10^{-9}~C}{0.12~m})$ $V_c = 0$ We can find the potential energy when $q_3$ is at point $c$: $U_c = q_3~V_a$ $U_c = (2.00\times 10^{-9}~C)(0)$ $U_c = 0$ Assuming that the electric potential at infinity is zero, we can find the electric potential at the point $b$: $V_b = k~(\frac{q_1}{r_1}+\frac{q_2}{r_2})$ $V_b = (9.0\times 10^9~N~m^2/C^2)~(\frac{8.00\times 10^{-9}~C}{0.040~m}+\frac{-8.00\times 10^{-9}~C}{0.040~m})$ $V_b = 0$ We can find the potential energy when $q_3$ is at point $b$: $U_b = q_3~V_b$ $U_b = (2.00\times 10^{-9}~C)(0)$ $U_b = 0$ We can find the change in potential energy from point $b$ to point $c$: $\Delta U = U_c-U_b = 0-0 = 0$ Since the work done by the electric field is $-\Delta U$, the work done by the electric field is zero.