Question

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.

Short answer

1. Magnitude of the electric field is 33 N/C .
2. Speed of proton when it strikes negatively charged plate is $1.00\times {10}^{4}m/s$.

Step-by-step solution

Step 1:

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

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

Step 2:

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.

Step 3:

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 $1.00\times {10}^{4}m/s$.