Answer
The work done by the electric field is zero.
Work Step by Step
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.
