Question

In Problems 63 through 66 you are given the equation(s) used to solve a problem. For each of these

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem

$$\begin{gathered} 2.0\times {10}^{12}m/s^2=\frac{(1.60\times {10}^{-19}{C}^2)E}{(1.67\times {10}^{-27}kg)} \\ E=\frac{Q}{(8.85\times {10}^{-12}{C}^2/N{m}^2){(0.020m)}^2}Q \end{gathered}$$

Short answer

(a) What is the strength of the electric field between two parallel conducting planes when a proton is discharged in an area between planes with an acceleration of $2\times {10}^{12}m/s^2$?

(b) The solution is $\mathbf{7}\mathbf{.}\mathbf{36}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathit{C}$

Step-by-step solution

Given information and formula used

Given :

$$\begin{gathered} 2.0\times {10}^{12}m/s^2=\frac{(1.60\times {10}^{-19}{C}^2)E}{(1.67\times {10}^{-27}kg)} \\ E=\frac{Q}{(8.85\times {10}^{-12}{C}^2/N{m}^2){(0.020m)}^2}Q \end{gathered}$$

Theory used :

The electric field between two conducting parallel plates is the potential difference divided by the distance by which they are separated.

Writing a realistic problem and finding the solution of the problem 

(a) Realistic Problem :

What is the strength of the electric field between two parallel conducting planes when a proton is discharged in an area between planes with an acceleration of $2\times {10}^{12}m/s^2$?

(b) Solution :

$$\begin{gathered} 2.0\times {10}^{12}m/s^2=\frac{(1.60\times {10}^{-19}{C}^2)E}{(1.67\times {10}^{-27}kg)} \\ \Rightarrow E=\frac{(2.0\times {10}^{12}m/s^2)(1.67\times {10}^{-27}kg)}{(1.60\times {10}^{-19}{C}^2)}=\underline{2.08\times {10}^{4}N/C} \end{gathered}$$

Also,

$$\begin{gathered} E=\frac{Q}{(8.85\times {10}^{-12}{C}^2/N{m}^2){(0.020m)}^2} \\ \Rightarrow Q=(2.08\times {10}^{4}N/C)(8.85\times {10}^{-12}{C}^2/N{m}^2){(0.020m)}^2 \\ =\boxed{\mathbf{7}\mathbf{.}\mathbf{36}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathit{C}} \end{gathered}$$