Chapter 10, Problem 2TS

Make use of Coulomb’s law here. Suppose you have a pair of electrically charged metal spheres suspended a certain distance from each other by insulating threads. There is a specific amount of electrical forces between them,

(a) If the charge on the one sphere were doubled, what would happen to the force between them?

(b) If the charge on the two spheres were doubled, what would happen to the force between them?

(c) If the distance between the two spheres were tripled, what would happen to the force between them?

(d) If the distance between the two spheres were reduced to one fourth, what would happen to the force between them?

(e) If the charge on the one sphere were doubled and the distance between them were doubled, what would happen to the force between them?

To determine

The force between the charges.

Answer to Problem 2TS

The force between the charges doubles.

Explanation of Solution

Given info: The charge of one charge is $2q$.

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

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

Here,

$q_1$ is the first charge

$q_2$ is the second charge

$r$ is the distance between them

Substitute $2q$ for $q_1$ in the above expression to get $F^{\prime }$.

$\begin{array}{c}{F}^{\prime }=\frac{2q_1{q}_2}{r^2}\\ =2F\end{array}$

Conclusion:

Therefore, the force between the charges doubles.

To determine

The force between the charges.

Answer to Problem 2TS

The force between the charges becomes four times the initial force.

Explanation of Solution

Given info: The charge of one charge is $2q_1$ and the charge of second charge is $2q_2$.

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

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

Here,

$q_1$ is the first charge

$q_2$ is the second charge

$r$ is the distance between them

Substitute $2q_1$ for $q_1$ and $2q_2$ for $q_2$ in the above expression to get $F^{\prime }$.

$\begin{array}{c}{F}^{\prime }=\frac{2q_{1}2q_2}{r^2}\\ =4F\end{array}$

Conclusion:

Therefore, the force between the charges becomes four times the initial force.

To determine

The force between the charges.

Answer to Problem 2TS

The force between the charges becomes $\frac{1}{9}$ times the initial force.

Explanation of Solution

Given info: The distance between the charges is $3r$.

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

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

Here,

$q_1$ is the first charge

$q_2$ is the second charge

$r$ is the distance between them

Substitute $3r$ for $r$ in the above expression to get $F^{\prime }$.

$\begin{array}{c}{F}^{\prime }=\frac{q_1{q}_2}{{\left(3r\right)}^2}\\ =\frac{1}{9}F\end{array}$

Conclusion:

Therefore, the forces between the charges becomes $\frac{1}{9}$ times the initial force

To determine

The force between the charges.

Answer to Problem 2TS

The force between the charges becomes $16$ times the initial force

Explanation of Solution

Given info: The distance between the charges is $\frac{1}{4}r$

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

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

Here,

$q_1$ is the first charge

$q_2$ is the second charge

$r$ is the distance between them

Substitute $\frac{1}{4}r$ for $r$ in the above expression to get $F^{\prime }$.

$\begin{array}{c}{F}^{\prime }=\frac{q_1{q}_2}{{\left(\frac{1}{4}r\right)}^2}\\ =16F\end{array}$

Conclusion:

Therefore, the forces between the charges becomes $16$ times the initial force

To determine

The force between the charges.

Answer to Problem 2TS

The force between the charges becomes $\frac{1}{2}$ times the initial force

Explanation of Solution

Given info: The distance between the charges is $2r$ and the first charge is $2q_1$.

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

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

Here,

$q_1$ is the first charge

$q_2$ is the second charge

$r$ is the distance between them

Substitute $2r$ for $r$ and $2q_1$ for $q_1$ in the above expression to get $F^{\prime }$.

$\begin{array}{c}{F}^{\prime }=\frac{2q_1{q}_2}{{\left(2r\right)}^2}\\ =\frac{2q_1{q}_2}{4{\left(r\right)}^2}\\ =\frac{1}{2}F\end{array}$

Conclusion:

Therefore, the forces between the charges becomes $\frac{1}{2}$ times the initial force