Answer
The work done by the electric field is $-2.40\times 10^{-6}~J$
Work Step by Step
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 $-2.40\times 10^{-6}~J$
