Question

An electron traveling parallel to a uniform electric field increases its speed from $2.0\times {10}^{7}m/s\mathrm{to}4.0\times {10}^{7}m/s$ over a distance of $1.2cm$. What is the electric field strength?

Short answer

The electric field strength is$E=\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{N}\mathbf{/}\mathit{C}$

Step-by-step solution

Given information and formula used

Given :

The electron's speed ranges from : $2.0\times {10}^{7}m/s\mathrm{to}4.0\times {10}^{7}m/s$

Over a distance of : $1.2cm$

Theory used :

Newton's Second law states :

$$\begin{gathered} F=m·a \\ {F}_E=qE \\ \Rightarrow qE=m·a \end{gathered}$$

Formula for acceleration :${v_f}^2={v_i}^2+2ad$

Calculating the electric field strength 

Using the kinematics equation, we calculate acceleration :

$$\begin{gathered} {v_f}^2={v_i}^2+2ad \\ \Rightarrow {v_f}^2-{v_i}^2=2ad \\ \Rightarrow a=\frac{{v_f}^2-{v_i}^2}{2d} \\ =\frac{{(4.0\times {10}^7)}^2-{(2.0\times {10}^7)}^2}{2x0.012} \\ =\underline{5\times {10}^{16}m/s²} \end{gathered}$$

We can now use Newton's Second Law. The force in this situation is the electric force, which is given by $F=qE$.

$$\begin{gathered} qE=ma \\ \Rightarrow E=\frac{ma}{q} \\ =\frac{(9.1x{10}^{-31})\times (5\times {10}^{16})}{1.6x{10}^{-19}} \\ \Rightarrow E=\boxed{\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{N}\mathbf{/}\mathit{C}} \end{gathered}$$