Chapter 10, Problem 1TS

Two point charges are separated by $6\text{ cm}$. The attractive force between them is $20\text{ N}$. Find the force between them when the distance between them is $12\text{ cm}$. (Why can you solve this problem without knowing the magnitudes of the charges?)

To determine

The force at a distance $12\text{cm}$.

Answer to Problem 1TS

The force at $12\text{cm}$ is $5\text{N}$.

Explanation of Solution

Given info: Distance between the two charges is $6\text{cm}$ and the force of attraction is $20\text{N}$.

According to Coulomb’s law, write the force between two charges.

$F=\frac{q_1{q}_2}{r^2}$

Here,

$q_1$ is the first charge,

$q_2$ is the second charge

$r$ is the distance between them

Substitute $6\text{cm}$ for $r$ in the above expression to get $F_6$.

$F_6=\frac{q_1{q}_2}{6^2}$

Substitute $12\text{cm}$ for $r$ in the above in the above expression to get $F_{12}$.

$F_{12}=\frac{q_1{q}_2}{{12}^2}$

Taking the ratio of the forces at the two given distances.

$\begin{array}{c}\frac{F_6}{F_{12}}=\frac{\frac{q_1{q}_2}{6^2}}{\frac{q_1{q}_2}{{12}^2}}\\ =\frac{{12}^2}{6^2}\\ =4\end{array}$

The force at $6\text{cm}$ is four times larger as at $12\text{cm}$.

Write the expression to calculate the force at $12\text{cm}$.

$\begin{array}{c}{F}_{12}=\frac{20\text{N}}{4}\\ =5\text{N}\end{array}$

Conclusion:

Therefore, the force at $12\text{cm}$ is $5\text{N}$.