## 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 $a$: $V = k~(\frac{q_1}{r_1}+\frac{q_2}{r_2})$ $V = (9.0\times 10^9~N~m^2/C^2)~(\frac{8.00\times 10^{-9}~C}{0.040~m}+\frac{-8.00\times 10^{-9}~C}{0.12~m})$ $V = 1200~J/C$ We can find the change in potential energy when $q_3$ is moved to point $a$: $\Delta U = q_3~V$ $\Delta U = (2.00\times 10^{-9}~C)(1200~J/C)$ $\Delta U = 2.40\times 10^{-6}~J$ Since the work done by the electric field is $-\Delta U$, the work done by the electric field is $-2.40\times 10^{-6}~J$