In Problems 69 through 72 you are given the equation(s) used to solve a problem. For each of these,

1. Write a realistic problem for which this is the correct equation(s).
2. Finish the solution of the problem.

$\frac{(9.0\times {10}^9{Nm}^2/C^2)(15\times {10}^{-9}C)}{r^2}=54,000N/C$

Equation$=\frac{(9.0\times {10}^9{Nm}^2/C^2)(15\times {10}^{-9}C)}{r^2}=54,000N/C$

By comparing the given equation from the equation of the magnitude of the electric field by a point charge which is given as:

$E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$

We get,

Electric field $=54,000N/C$

Amount of charge role="math" localid="1648631611790" $=15\times {10}^{-9}C=15nC$

Constant role="math" localid="1648631387865" $=\frac{1}{4\pi {ϵ}_0}=9.0\times {10}^9{Nm}^2/C^2$

Hence, it can be concluded that an electric field of$54,000N/C$is generated by a point charge ofrole="math" localid="1648631579441" $15nC$at a distance$\text{'}r\text{'}$.

Hence, the realistic problem that can be generated for the given question is:

Find the distance where the electric field is $54,000N/C$from the charge having a magnitude of $15nC$.

Equation$=\frac{(9.0\times {10}^9{Nm}^2/C^2)(15\times {10}^{-9}C)}{r^2}=54,000N/C$

The given equation is:

$\frac{(9.0\times {10}^9{Nm}^2/C^2)(15\times {10}^{-9}C)}{r^2}=54,000N/C$

Based on the equation, a realistic problem can be made as:

Find the distance where the electric field is $54,000N/C$from the charge having a magnitude of $15nC$.

Hence, by rearranging the terms to get the value of r,

$$\begin{gathered} r^2=\frac{(9.0\times {10}^9{Nm}^2/C^2)(15\times {10}^{-9}C)}{54,000N/C} \\ r=\sqrt{\frac{(9.0\times {10}^9{Nm}^2/C^2)(15\times {10}^{-9}C)}{54,000N/C}} \\ r=0.05m=5.0cm \end{gathered}$$

Hence, the distance from the charge is calculated to be$5.0cm$.