Answer
(a) $20m/s$
(b) greater than
(c) $28.3m/s$
Work Step by Step
(a) We know that
$v_f=\sqrt{\frac{-2Kq_1q_2}{mr_i}}$
We plug in the known values to obtain:
$v_f=\sqrt{\frac{-2(8.99\times 10^9Nm^2/C^2)(20.2\times 10^{-6}C)(-5.25\times 10^{-6}C)}{(3.2\times 10^{-3}Kg)(1.49m)}}$
$v_f=20m/s$
(b) We know that the change in potential energy when the charge is at $r_i$ is given as
$\Delta U_1=\frac{-Kq_1q_2}{r_i}$
When the charge is at rest $\frac{r_i}{2}$, then
$\Delta U_2=\frac{-q_1q_2}{\frac{r_i}{2}}$
$\implies \frac{\Delta U_2}{\Delta U_1}=\frac{2}{1}\gt 1$
As $\Delta U_2\gt \Delta U_1$, therefore, the K.E will also be greater.
(c) We know that
$v_f=\sqrt{\frac{-2Kq_1q_2}{mr_i}}$
We plug in the known values to obtain:
$v_f=\sqrt{\frac{-2(8.99\times 10^9Nm^2/C^2)(20.2\times 10^{-6}C)(-5.25\times 10^{-6}C)}{(3.2\times 10^{-3}Kg)(0.7457m)}}$
This simplifies to:
$v_f=28.3m/s$
