If the electrical field intensity E in the region between two parallel plates is 3.0 x 104NC-1. Calculate: (a) The force F on the electron in the region. (b) The acceleration of an electron when acted on the force.

The force $F$ experienced by a charge $q$ in an electric field $E$ is given by:

$$F = qE$$

$$F = 4.806 \times 10^{-18} \text{ N}$$

The force on the electron is $\boxed{\text{4.806} \times 10^{-18} \text{ N}}$.

Newton's second law states that force equals mass times acceleration ($F = ma$). Therefore, acceleration $a$ can be found by:

$$a = \frac{F}{m_e}$$

$$a \approx 0.5276 \times 10^{13} \text{ m/s}^2$$ $$a \approx 5.276 \times 10^{12} \text{ m/s}^2$$

The acceleration of the electron is approximately $\boxed{\text{5.276} \times 10^{12} \text{ m/s}^2}$.