Question

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first$\mathbf{3}\mathbf{.}\mathbf{00}\mathit{\mu }\mathit{s}$after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

Short answer

1. Magnitude and direction of the electric field is 5.69N/C and the direction is downwards.
2. Influence of gravity can be neglected.

Step-by-step solution

Step 1:

(E) Electric field is a vector quantity and is the force unit per charge exerted on any test charge at any point.

A charge in an electric field is accelerated by a net force.

Then force is$\overrightarrow{F}=m\overrightarrow{a}=q\overrightarrow{E}$

So let +y be upwards

Electron charge q=-e

Step 2:

As$v_{oy}=0,a_y=a$

$$\begin{gathered} y-y_0=v_{oy}t+\frac{1}{2}{a}_y{t}^2 \\ y-y_0=\frac{1}{2}a{t}^2 \end{gathered}$$

Putting all the values,

$$\begin{gathered} \begin{array}{rl}a& =\frac{2\left(y-y_0\right)}{t^2}\end{array} \\ =\frac{2\left(4.50\right)}{{\left(3\times {10}^{-6}\right)}^2} \\ =1\times {10}^{12}\mathrm{m}/s^2 \end{gathered}$$

Now, the magnitude of the electric field is

$$\begin{gathered} \begin{array}{rl}E& =\frac{F}{q}=\frac{ma}{q}\end{array} \\ =\frac{\left(9.11\times {10}^{31}\right)\left(1\times {10}^{12}\right)}{1.6\times {10}^{-19}} \\ =5.69\mathrm{N}/\mathrm{C} \end{gathered}$$

Therefore, the Magnitude and direction of the electric field is 5.69N/C and the direction is downwards.

Step 3:

As, the net force acting on the electron is an electrical force because, as electron acceleration is much higher, therefore the influence of gravity can be neglected.