Answer
a) $v=4.55\times 10^5m/s$
b) $v=1.44\times 10^6m/s$
Work Step by Step
(a) We can find the required speed as follows
$F=qE=ma$
$\implies a=\frac{qE}{m}$
We know that
$v=\sqrt{v_{\circ}^2+2a\Delta x}$
$\implies v=\sqrt{v_{\circ}^2+\frac{2qE}{m}\Delta x}$
We plug in the known values to obtain:
$v=\sqrt{(0)^2+\frac{2(1.6\times 10^{-19})(1.08\times 10^5)(0.0100)}{1.673\times 10^{-27}}}$
$v=4.55\times 10^5m/s$
(b) We know that
$v=\sqrt{v_{\circ}^2+2a\Delta x}$
$\implies v=\sqrt{v_{\circ}^2+\frac{2qE}{m}\Delta x}$
We plug in the known values to obtain:
$v=\sqrt{(0)^2+\frac{2(1.6\times 10^{-19})(1.08\times 10^5)(0.100)}{1.673\times 10^{-27}}}$
$v=1.44\times 10^6m/s$
