Answer
(a) $-6.96J$
(b) $20.1m/s$
Work Step by Step
(a) We can find the electric potential energy as follows:
$U=\frac{1}{4\pi \epsilon_{\circ}}.\frac{Qq}{r}$
We plug in the known values to obtain:
$U=\frac{(8.99\times 10^9Nm^2/C^2)(87.1\times 10^{-6}C)(-2.87\times 10^{-6}C)}{0.323m}$
$U=-6.96J$
(b) We can find the required speed as follows:
$U_f=\frac{1}{4\pi \epsilon_{\circ}}.\frac{q_1Q}{r}$
We plug in the known values to obtain:
$U_f=\frac{(8.99\times 10^9Nm^2/C^2)(87.1\times 10^{-6}C)(-2.87\times 10^{-6}C)}{0.121m}$
$U_f=-18.57J$
According to law of conservation of energy
$K.E_i+U_i=K.E_f+U_f$
We plug in the known values to obtain:
$0-6.96J=\frac{1}{2}mv^2-18.57J$
$\frac{1}{2}mv^2=18.57J-6.96J$
$\frac{1}{2}mv^2=11.61J$
$\implies v^2=\frac{(2)(11.61J)}{m}$
$\implies v=\sqrt{\frac{(2)(11.61J)}{0.0576Kg}}$
$v=20.1m/s$
