The problem describes two charged metal plates in a vacuum, 15 cm apart. The electric field between the plates is uniform and has a strength of 500 N/C. An electron (charge $e = 1.6 x

The electric force $F_e$ on a charge $q$ in an electric field $E$ is given by $F_e = qE$.

$$F_e = (1.6 \times 10^{-19} \, \text{C}) \times (500 \, \text{N/C})$$ $$F_e = 8.0 \times 10^{-17} \, \text{N}$$

According to Newton's second law, $F_e = m_e a$, where $a$ is the acceleration.

$$a = \frac{F_e}{m_e}$$ $$a = \frac{8.0 \times 10^{-17} \, \text{N}}{9.1 \times 10^{-31} \, \text{kg}}$$ $$a \approx 8.79 \times 10^{13} \, \text{m/s}^2$$

The distance $d$ between the plates is $15 \, \text{cm} = 0.15 \, \text{m}$. Since the electron starts from rest, we can use the kinematic equation $d = v_0 t + \frac{1}{2}at^2$. Here, $v_0 = 0$.

$$d = \frac{1}{2}at^2$$ $$t^2 = \frac{2d}{a}$$ $$t = \sqrt{\frac{2d}{a}}$$ $$t = \sqrt{\frac{2 \times 0.15 \, \text{m}}{8.79 \times 10^{13} \, \text{m/s}^2}}$$ $$t = \sqrt{\frac{0.30}{8.79 \times 10^{13}}} \, \text{s}$$ $$t = \sqrt{3.413 \times 10^{-15}} \, \text{s}$$ $$t \approx 5.84 \times 10^{-8} \, \text{s}$$

The time taken is $\boxed{\text{5.84} \times 10^{-8} \, \text{s}}$.

b) Final speed of the electron

We can use the equation $v = v_0 + at$. Since $v_0 = 0$:

$$v = at$$ $$v = (8.79 \times 10^{13} \, \text{m/s}^2) \times (5.84 \times 10^{-8} \, \text{s})$$ $$v \approx 5.13 \times 10^6 \, \text{m/s}$$

The final speed of the electron is $\boxed{\text{5.13} \times 10^6 \, \text{m/s}}$.