Question

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?

Short answer

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

Step-by-step solution

Electrostatic force:

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

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 \(7.11{\rm{ }}m\).