Suggested languages for you:

Americas

Europe

Q14PE

Expert-verified
Found in: Page 664

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# How far apart must two point charges of $${\rm{75}}{\rm{.0 nC}}$$ (typical of static electricity) be to have a force of $${\rm{1}}{\rm{.00 N}}$$ between them?

The separation between the charges is $$7.11{\rm{ }}mm$$.

See the step by step solution

## Step 1: Electrostatic force:

When a charge enters the field of another charge, it experiences a force known as an electrostatic force.

## Step 2: Separation between charges

The electrostatic force between two point charges is given as,

$$F = \frac{{K{q_1}{q_2}}}{{{r^2}}}$$ ….. (1)

Here, $$F$$ is the electrostatic force, $$K$$ is the electrostatic force constant, $${q_1}$$is the charge on the first body, $${q_2}$$ is the charge on the second body, and $$r$$ is the separation between the charges.

Consider the given data as below.

The electrostatic force, $$F = 1.00{\rm{ }}N$$

The electrostatic force constant, $$K=9\times 10^{9} N.m^{2}/C^{2}$$

The charge on the first body, $${q_1} = 75.0{\rm{ }}nC$$

The charge on the second body, $${q_2} = 75.0{\rm{ }}nC$$

Rearranging equation (1) in order to get an expression for separation between the charges.

$$r = \sqrt {\frac{{K{q_1}{q_2}}}{F}}$$ ….. (2)

Substituting all known values into equation (2).

\begin{aligned} r &= \sqrt {\frac{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times \left( {75.0{\rm{ }}nC} \right) \times \left( {75.0{\rm{ }}nC} \right)}}{{\left( {1.00{\rm{ }}N} \right)}}} \\ &= \sqrt {\frac{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times \left( {75.0{\rm{ }}nC} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ }}C}}{{1{\rm{ }}nC}}} \right) \times \left( {75.0{\rm{ }}nC} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ }}C}}{{1{\rm{ }}nC}}} \right)}}{{\left( {1.00{\rm{ }}N} \right)}}} \\ &= 7.11 \times {10^{ - 3}}{\rm{ }}m \times \left( {\frac{{1000{\rm{ }}mm}}{{1{\rm{ }}m}}} \right)\\ &= 7.11{\rm{ }}m\end{aligned}

Hence, the separation between the charges is $$7.11{\rm{ }}m$$.