Answer
(a) $Q = 3.98 \times 10^{-11} \mathrm{C}$
(b) $E = 5 \times 10^{3} \mathrm{V} / \mathrm{m}$
(c) $v = 2.96 \times 10^{6} \mathrm{m} / \mathrm{s}$
Work Step by Step
(a) The charge for the two parallel plates is related to the potential and the capacitance between the two plates by
\begin{align}
Q&= CV\\
&= \dfrac{\epsilon_{0} A}{ d} (V)\\
& =\frac{\left(8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}\right)(0.03 \mathrm{m})^{2}}{0.005 \mathrm{m}} (25 \mathrm{V})\\
&= \boxed{3.98 \times 10^{-11} \mathrm{C}}
\end{align}
(b) It is easy to get the electric field by knowing the value of the potential difference between the two plates in the form
$$E=\dfrac{V }{ d}=\dfrac{(25.0 \mathrm{V})}{ (0.0050 \mathrm{m})}=\boxed{5 \times 10^{3} \mathrm{V} / \mathrm{m}}$$
(c) The speed of the electron could be found using the conservation law by
$$v=\sqrt{\frac{2 e V}{m}}=\sqrt{\frac{2\left(1.60 \times 10^{-19} \mathrm{C}\right)(25 \mathrm{V})}{9.11 \times 10^{-31} \mathrm{kg}}}=\boxed{2.96 \times 10^{6} \mathrm{m} / \mathrm{s}}$$
