Chapter 32, Problem 51A

Proportional reasoning: Consider a pair of electrically charged coins suspended from insulating threads, a certain distance from each other. There is a specific amount of electrostatic force between them.

a. If the charge on one coin were halved, what would happen to the force between them?

b. If the charges on both coins were doubled, what would happen to the force between them?

c. If the distance between the coins were tripled, what would happen to the force between them?

d. If the distance between them were reduced to one-fourth the original distance, what would happen to the force between them?

e. If the charge on each object were doubled and the distance between them were doubled, what would happen to the force between them?

To determine

The effect on the force between two electrically charged coins.

Explanation of Solution

Introduction:

The electrostatic force depends on the amount of charges $\left(q_1,q_2\right)$ and distance between them $\left(r\right)$ .

Electrostatic force between two charges can be expressed as:

  $F=k\frac{q_1{q}_2}{r^2}$

Here, k is the Coulomb’s constant.

If the charge on one coin is halved, then the electrostatic force between charged coins would also halve.

  $F_{new}=k\frac{q_1/2\times q_2}{r^2}=\frac{F}{2}$

To determine

The effect on the force between two electrically charged coins.

Explanation of Solution

Introduction:

The electrostatic force depends on the amount of charges $\left(q_1,q_2\right)$ and distance between them $\left(r\right)$ .

Electrostatic force between two charges can be expressed as:

  $F=k\frac{q_1{q}_2}{r^2}$

Here, k is the Coulomb’s constant.

If the charge on both the coins is doubles, then the electrostatic force between charged coins would become four times the initial value.

  $F_{new}=k\frac{2q_1\times 2q_2}{r^2}=4F$

To determine

The effect on the force between two electrically charged coins.

Explanation of Solution

Introduction:

The electrostatic force depends on the amount of charges $\left(q_1,q_2\right)$ and distance between them $\left(r\right)$ .

Electrostatic force between two charges can be expressed as:

  $F=k\frac{q_1{q}_2}{r^2}$

Here, k is the Coulomb’s constant.

If the distance between the charged coins is tripled, then the electrostatic force between charged coins would become one-ninth times the initial value.

  $F_{new}=k\frac{q_1\times q_2}{{\left(3r\right)}^2}=\frac{F}{9}$

To determine

The effect on the force between two electrically charged coins.

Explanation of Solution

Introduction:

The electrostatic force depends on the amount of charges $\left(q_1,q_2\right)$ and distance between them $\left(r\right)$ .

Electrostatic force between two charges can be expressed as:

  $F=k\frac{q_1{q}_2}{r^2}$

Here, k is the Coulomb’s constant.

If the distance between the charged coins is decreased to one-fourth value, then the electrostatic force between charged coins would become sixteen times the initial value.

  $F_{new}=k\frac{q_1\times q_2}{{\left(r/4\right)}^2}=16F$

To determine

The effect on the force between two electrically charged coins.

Explanation of Solution

Introduction:

The electrostatic force depends on the amount of charges $\left(q_1,q_2\right)$ and distance between them $\left(r\right)$ .

Electrostatic force between two charges can be expressed as:

  $F=k\frac{q_1{q}_2}{r^2}$

Here, k is the Coulomb’s constant.

If the distance between the charged coins is increased two times and charge is doubled on each coin, then the electrostatic force between charged coins would remain same.

  $F_{new}=k\frac{2q_1\times 2q_2}{{\left(2r\right)}^2}=F$