A uniform electric field exists in the region between two oppositely charged plane-parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of$\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{s}$. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

As given the distance between plates are r=1.60cm and the proton strikes the negative plate at$t=3.2\times {10}^{-6}s$

Using newton’s second law of motion for the determination of force (F)

$F=ma$

Now the electric force due to the electric field is

$F=qE$

Therefore, from the above two equations

$$\begin{gathered} qE=ma \\ a=\frac{qE}{a} \end{gathered}$$

Relation between the acceleration and the time using Newton’s Law of motion;

$$\begin{gathered} r-r_0=v_{0}t+\frac{1}{2}a{t}^2 \\ r=\frac{1}{2}a{t}^2 \\ r=\frac{1}{2}\left(\frac{qE}{m}\right)t^2 \\ E=2\frac{rm}{q{t}^2} \end{gathered}$$

Putting values;

$$\begin{gathered} E=2\frac{\left(0.016\right)\left(1.67\times {10}^{-27}\right)}{\left(1.6\times {10}^{-19}\right){\left(3.2\times {10}^{-6}\right)}^2} \\ E=33\mathrm{N}/\mathrm{C} \end{gathered}$$

Therefore, the magnitude of the electric field is 33N/C.

Speed can be determined by using Newton’s Law of motion;

$$\begin{gathered} v=v_0+at \\ v=v_0+\frac{qE}{m}t \end{gathered}$$

Putting values;

$$\begin{gathered} v=0+\frac{\left(1.6\times {10}^{-19}\right)\left(33\right)}{\left(1.67\times {10}^{-27}\right)} \\ v=1.00\times {10}^4\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the proton when it strikes a negatively charged plate is $1.00\times {10}^{4}m/s$.