In Problems 74 through 76 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.

76.$$\begin{gathered} \frac{(9.0\times {10}^{9}N{m}^2/C^2)(3.0\times {10}^{-9}C)}{0.030m}+\frac{(9.0\times {10}^{9}N{m}^2/C^2)(3.0\times {10}^{-9}C)}{(0.030m)+d} \\ =1200V \end{gathered}$$

Given equation :

$$\begin{gathered} \frac{(9.0\times {10}^{9}N{m}^2/C^2)(3.0\times {10}^{-9}C)}{0.030m}+\frac{(9.0\times {10}^{9}N{m}^2/C^2)(3.0\times {10}^{-9}C)}{(0.030m)+d} \\ =1200V \end{gathered}$$

Theory used :

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

a. Real life problem :

"Two identical $3nC$point charges are arranged so that one is $3cm$distant and the other is $d$greater than the other from our point of potential measurement. Find $d$."

b. We may simplify the problem by doing numerical computations and finding

$\frac{27}{0.03}+\frac{27}{0.03+d}=1200$.

Now we can do some more calculations and remove the enumerator from the d parameter, as shown below.

$$\begin{gathered} 900(0.03+d)+27=1200(0.03+d) \\ \Rightarrow 300(0.03+d)=27 \\ \Rightarrow d=\frac{27}{300}-0.03=\boxed{\mathbf{0}\mathbf{.}\mathbf{06}\mathbf{}\mathit{m}} \end{gathered}$$

is now the solution to this problem.