Answer
a) less than
b) $F_{net}=0.55N$
c) $F_{net}=0.96N$
Work Step by Step
(a) We know that the net force on $q_1$ is the vector sum of an attractive force and a repulsive force which partially cancel each other. On the other hand, the net force on charge $q_2$ is the vector sum of two attractive forces; therefore, the net force on $q_2$ is greater than that of $q_1$.
(b) We know that
$F_{net}=\frac{Kq^2}{a^2}$
We plug in the known values to obtain:
$F_{net}=\frac{(8.99\times 10^9Nm^2/C^2)(7.3\times 10^{-6}C)^2}{(0.93m)^2}$
$F_{net}=0.55N$
(c) We know that
$F_{net}=\frac{Kq^2}{2a^2}(2\sqrt 3)$
We plug in the known values to obtain:
$F_{net}=\frac{(8.99\times 10^9Nm^2/C^2)(7.3\times 10^{-6}C)^2(1.732)}{(0.93m)^2}$
$F_{net}=0.96N$
