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Using Coulomb’s law to find the distance between charged objects · Physics Playground

Exercises Electric charge and fields

Example from Physics, 18.2 Coulomb's law

Using Coulomb’s law to find the distance between charged objects

An engineer measures the force between two ink drops by measuring their acceleration and their diameter. She finds that each member of a pair of ink drops exerts a repulsive force of $F = 5.5 mN$ on its partner. If each ink drop carries a charge $q_{ink drop} = −1 \times 10^{−10} C$ , how far apart are the ink drops?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
We know the force and the charge on each ink drop, so we can solve Coulomb’s law for the distance r between the ink drops. Do not forget to convert the force into SI units: $F = 5.5 mN = 5.5 \times 10^{−3} N.$
Solution
The charges in Coulomb’s law are $q_{1} = q_{2} = q_{ink drop}^{},$ so the numerator in Coulomb’s law takes the form $q_{1} q_{2} = q_{ink drop}^{2}$ . Inserting this into Coulomb’s law and solving for the distance r gives

$\begin{matrix} F & = & \frac{k q_{ink drop}^{2}}{r^{2}} \\ r & = & \pm \sqrt{\frac{k q_{ink drop}^{2}}{F}} \\ = & \pm \sqrt{\frac{(8.99 \times 10^{9} N \cdot m^{2} {/C}^{2}) {(−1 \times 10^{−10} C)}^{2}}{5.5 \times 10^{−3} N}} \\ = & \pm 1.3 \times 10^{−4} m \end{matrix}$
18.14

or 130 microns (about one-tenth of a millimeter).

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From Physics (1st edition) by OpenStax (Rice University), section 18.2, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.

Strategy

Solution