Question: An electron is accelerated eastward at $1.80\times {10}^{9}m/s^2$by an electric field. Determine the field (a) magnitude and (b) direction.

Acceleration of the electron in eastward direction,$1.80\times {10}^{9}m/s^2$

Using the concept of force due to Newton’s second law on a body and relating the value to the electric field produced by the particle, we can get the magnitude of the electric field with its direction.

Formulae:

The force due to Newton’s second law, F = mg (i)

The force and electric field relation, F = qE (ii)

With east being the$\overbrace{i}$ direction, we have the electric field by using equation (i) in equation (ii) as follows:

data-custom-editor="chemistry" $$\begin{gathered} \overrightarrow{\mathrm{E}}=-\left|\frac{m}{e}\right|\overrightarrow{a} \\ =\frac{\left(9.11\times {10}^{-31}kg\right)}{\left(1.60\times {10}^{-19}C\right)}\left(1.80\times \frac{{10}^{9}m}{s^2}\overbrace{i}\right) \\ =\left(-0.0102N/C\right)\stackrel{⏜}{i} \end{gathered}$$

Hence, the magnitude of the electric field is 0.0102 N/C

From the results of part (a), it can be seen that the field E is directed in the –x direction, or westward direction.