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Expert-verified Found in: Page 664 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three?

If the separation is increased to three times of the original, force will become 0.556N .

See the step by step solution

## Step 1: Given Data

• Value of point charges is 5.00 N

## Step 2: Electrostatic force

Coulomb stated that when two-point charges are separated by some distance, they experience some force of attraction or repulsion. This force of attraction or repulsion is known as electrostatic force.

## Step 3: New force

Initially, when the charges q and Q are separated by some distance r the electrostatic force between them is,

$F=\frac{KqQ}{{r}^{2}}..........\left(1.1\right)$

Here, F is the electrostatic force (F = 5.00 N) , and K is the electrostatic force constant.

When the separation between the charges increased by a factor of three $\left({r}^{\mathbf{\text{'}}}=3r\right)$. The new electrostatic force (F') between the point charges is,

${F}^{\text{'}}=\frac{KqQ}{{r}^{\text{'}2}}\phantom{\rule{0ex}{0ex}}=\frac{KqQ}{{\left(3r\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{KqQ}{9{r}^{2}}$

Using equation (1.1),

${F}^{\text{'}}=\frac{F}{9}$

Substituting 5.00 N for F,

${F}^{\text{'}}=\frac{5.00N}{9}\phantom{\rule{0ex}{0ex}}=0.556N$

Hence, when the separation becomes three times of the original, the new force will be 0.556 N. ### Want to see more solutions like these? 