## College Physics (4th Edition)

The work done by the electric field is $2.40\times 10^{-6}~J$
Assuming that the electric potential at infinity is zero, we can find the electric potential at the point $a$: $V_a = k~(\frac{q_1}{r_1}+\frac{q_2}{r_2})$ $V_a = (9.0\times 10^9~N~m^2/C^2)~(\frac{8.00\times 10^{-9}~C}{0.040~m}+\frac{-8.00\times 10^{-9}~C}{0.12~m})$ $V_a = 1200~J/C$ We can find the potential energy when $q_3$ is at point $a$: $U_a = q_3~V_a$ $U_a = (2.00\times 10^{-9}~C)(1200~J/C)$ $U_a = 2.40\times 10^{-6}~J$ Assuming that the electric potential at infinity is zero, we can find the electric potential at the point $b$: $V_b = k~(\frac{q_1}{r_1}+\frac{q_2}{r_2})$ $V_b = (9.0\times 10^9~N~m^2/C^2)~(\frac{8.00\times 10^{-9}~C}{0.040~m}+\frac{-8.00\times 10^{-9}~C}{0.04~m})$ $V_b = 0$ 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 $a$ to point $b$: $\Delta U = U_b-U_a = 0-( 2.40\times 10^{-6}~J) = -2.40\times 10^{-6}~J$ Since the work done by the electric field is $-\Delta U$, the work done by the electric field is $2.40\times 10^{-6}~J$