Chapter 32, Problem 28A

Two equally charged particles exert equal forces on each other. Suppose that the charge on one of the particles is doubled. The charge on the other remains the same.

a. How much stronger is the force between them?

b. How does the force change if the charges of both particles are doubled?

To determine

The force between the charged particles for the given changes.

Answer to Problem 28A

Force becomes double.

Explanation of Solution

Given information:

Two equally charged particles exerts equal force on each other.

The charge on one is doubled, and the charge on other remains the same

Formula used:

The force is,

  $F=k\frac{q_1{q}_2}{r^2}$

Here,q₁and q₂are two charges and r is the separation between the two charges.

Calculation:

Here the charges are equal, so in place of q₁and q₂take q

  $F=k\frac{q^2}{r^2}$ (1)

When the charge on one is doubled then let q₁=2q and q₂=q then the force

  $F_1=k\frac{q^2}{r^2}=k\frac{2q\times q}{r^2}=k\frac{2q^2}{r^2}$ (2)

From equation (1) and (2),F₁=2F

Conclusion:

Therefore, the force is doubled.

To determine

The force between the charged particles for the given changes.

Answer to Problem 28A

Force becomes four times stronger

Explanation of Solution

Given:

Two equally charged particles exerts equal force on each other.

The charge on both particles is doubled.

Formula used:

The force is,

  $F=k\frac{q_1{q}_2}{r^2}$

Here,q₁and q₂are two charges and r is the separation between the two charges.

Calculation:

Here the charges are equal, so in place of q₁and q₂take q

  $F=k\frac{q^2}{r^2}$ (1)

When the charge is doubled then let q₁=2q and q₂= 2q then the force is,

  $F_1=k\frac{2q\times 2q}{r^2}=k\frac{4q^2}{r^2}$ (2)

From equation (1) and (2),F₁= 4F

Conclusion:

Therefore, the force will become 4 times.