# Two-point charges exert a 9.00 N force on each other. What will the force (in N) become if the...

## Question:

Two-point charges exert a 9.00 N force on each other. What will the force (in N) become if the distance between them is increased by a factor of 3?

## Coulomb's Law

The electrostatic force between two stationary point charges is described by Coulomb's law.

Coulomb's law states that the force obeys the equation:

{eq}\displaystyle F = \frac{Q_1 Q_2}{4 \pi \epsilon_0 r^2} {/eq}

Here:

{eq}\displaystyle Q_1 {/eq} and {eq}\displaystyle Q_2 {/eq} are the electric charges on the two point charges respectively,

{eq}\displaystyle \epsilon_0 \approx 8/854 \times 10^{-12} \: \text{F/m} {/eq} is the electric constant, and

{eq}\displaystyle r {/eq} is the distance separating the two charges.

## Answer and Explanation:

We are told that the two point charges exert a force (assumed to be electrostatic) of 9.00 N on one another.

We are then asked to find what the force becomes when the distance between the charges is increased by a factor of 3.

Using the fact that the force is 9.00 N we can write:

{eq}\displaystyle F_1=9.00 \: \text{N} = \frac{Q_1 Q_2}{4 \pi \epsilon_0 r_1^2} {/eq}

The distance between the two charges is then changed by a factor of 3 ({eq}\displaystyle r_2=3r_1 {/eq}).

{eq}\displaystyle F_2 = \frac{Q_1 Q_2}{4 \pi \epsilon_0 (3r_1)^2} = \frac{1}{9} \frac{Q_1 Q_2}{4 \pi \epsilon_0 r_1^2} = \frac{1}{9} (9.00 \: \text{N}) = 1.00 \: \text{N} {/eq}

When the distance between the charges is increased by a factor of 3 the force becomes 1.00 N.

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from UExcel Physics: Study Guide & Test Prep

Chapter 12 / Lesson 4