Answer
(a) positive
(b) $-1.7\times 10^{-16}C$
Work Step by Step
(a) We know that the 4th charge is positive and therefore, there will be only one attractive force due to $q_3$ and the other two forces due to $q_1$ and $q_2$ will be repulsive. Thus, the net force acting on the 4th charge will be a repulsive force, which means the work done will be positive at point P.
(b) We know that
$W=Kq(\frac{q_1}{r_{14}}+\frac{q_2}{r_{24}}+\frac{q_3}{r_{34}})$
We plug in the known values to obtain:
$W=Kq(\frac{2.75\times 10^{-6}C}{0.625m}+\frac{7.45\times 10^{-6}C}{1.172}+\frac{(-1.72\times 10^{-6}C)}{0.625m})$
This can be rearranged as:
$q=\frac{W}{K}(\frac{2.75\times 10^{-6}C}{0.625m}+\frac{7.45\times 10^{-6}C}{1.172}+\frac{(-1.72\times 10^{-6}C)}{0.625m})$
$\implies q=\frac{-1.3\times 10^{-11}J}{(8.99\times 10^{9}Nm^2/C^2)(4.4\times 10^{-6}+6.898\times 10^{-6}-2.75\times 10^{-6})C/m}$
This simplifies to:
$q=-1.7\times 10^{-16}C$
