Question

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.

$\frac{(9.0\times {10}^9{Nm}^2/C^2)q^2}{(0.0150m{)}^2}=0.020N$

Short answer

1. If the electrostatic force acting between two identical charges separated by a distance of $1.50cm$is $0.020N$, calculate the magnitude of each charge.
2. The magnitude of each charge is calculated to be$22.3nC$.

Step-by-step solution

Part (a) Step 1: Given Information

Equation$=\frac{(9.0\times {10}^9{Nm}^2/C^2)q^2}{(0.0150m{)}^2}=0.020N$

Part (a) Step 2: Explanation

By comparing the given equation from the equation of the electrostatic force given by two identical point charges which is given as:

$E=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}$

We get,

Electrostatic force $=0.020N$

Separation of point charges $=0.0150m=1.50cm$

Constant role="math" localid="1648629164648" $=\frac{1}{4\pi {ϵ}_0}=9.0\times {10}^9{Nm}^2/C^2$

Hence, it can be concluded that the electrostatic force of $0.020N$is generated with two identical charges$\text{'}q\text{'}$at a distance of$1.50cm$.

Part (a) Step 3: Final answer

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

If the electrostatic force acting between two identical charges separated by a distance of $1.50cm$is $0.020N$, calculate the magnitude of each charge.

Part (b) Step 1: Given Information

Equation$=\frac{(9.0\times {10}^9{Nm}^2/C^2)q^2}{(0.0150m{)}^2}=0.020N$

Part (b) Step 2: Calculation

The given equation is:

$\frac{(9.0\times {10}^9{Nm}^2/C^2)q^2}{(0.0150m{)}^2}=0.020N$

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

If the electrostatic force acting between two identical charges separated by a distance of $1.50cm$is $0.020N$, calculate the magnitude of each charge.

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

$$\begin{gathered} q^2=\frac{0.020N\times (0.0150m{)}^2}{(9.0\times {10}^9{Nm}^2/C^2)} \\ q=\sqrt{\frac{0.020N\times (0.0150m{)}^2}{(9.0\times {10}^9{Nm}^2/C^2)}} \\ q=2.23\times {10}^{-8}C=22.3nC \end{gathered}$$

Part (b) Step 3: Final answer 

Hence, the magnitude of each charge is calculated to be$22.3nC$.