Chapter 32, Problem 47A

By how much is the electrical force between a pair of ions reduced when their separation distance is doubled? Tripled?

To determine

To calculate: The electric force between a pair of ions reduced for the given changes.

Answer to Problem 47A

Force becomes $\frac{1}{4}$ th of the original value when separation is doubled

Force becomes $\frac{1}{9}$ th of the original value when separation distance is tripled.

Explanation of Solution

Given: A pair of ions, separated by some distance.

The separation distance is doubled and if separation distance is tripled.

Formula used: $F=k\frac{q_1{q}_2}{r^2}$ here q₁ and q₂ are two charges and r is the separation between the two charges.

Calculation: From coulomb’s law $F=k\frac{q_1{q}_2}{r^2}$ ……………………….1

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

If the separation distance is doubled then let the force if F₁,So $F_1=k\frac{q_1{q}_2}{{(2r)}^2}=k\frac{q_1{q}_2}{4r^2}$ …………..2

Comparing equation 1 and 2 , the force reduced to $\frac{1}{4}$ th of the original force.

If the separation distance is tripled then let the force if F₂,So $F_2=k\frac{q_1{q}_2}{{(3r)}^2}=k\frac{q_1{q}_2}{9r^2}$ …………..3

Comparing equation 1 and 3 , the force reduced to $\frac{1}{9}$ th of the original force.