Answer
(a) $0.86J$
(b) less than
(c) $-0.54J$
Work Step by Step
(a) We know that
$r_{32}=\sqrt{(0.25m)^2+(0.16m)^2}$
$r_{32}=0.296m$
$W=-Kq_2(\frac{q_1}{r_{12}}+\frac{q_2}{r_{32}})$
We plug in the known values to obtain:
$W=-(8.99\times 10^{9}Nm^2/C^2)(+2.7\times 10^{-6}C)(\frac{-6.1\times 10^{-6}C}{0.25m}+\frac{-3.3\times 10^{-6}C}{0.296m})$
$\implies W=0.86J$
(b) We know that the repulsive force for the charges $-6.1\mu C$ and $-3.3\mu C$ is of greater magnitude when compared to the attractive force between the charges $-6.1\mu C$ and $+2.7\mu C$. Thus, we conclude that the work done in the given case is less than that determined in part (a).
(c) We can determine the required work done as follows:
$W=-Kq_1(\frac{q_3}{r_{13}}+\frac{q_2}{r_{12}})$
We plug in the known values to obtain:
$W=-(8.99\times 10^{9}Nm^2/C^2)(-6.1\times 10^{-6}C)(\frac{-3.3\times 10^{-6}C}{0.16m}+\frac{2.7\times 10^{-6}C}{0.25m})$
This simplifies to:
$W=-0.54J$
