Inverse-Square Distance - Electric Fields, Electrostatics & Capacitors Step by Step

Electric force falls with the square of distance. Doubling distance makes the force one quarter as large.

electrostatics Distance appears twice in the denominator, so spreading charges apart weakens the force quickly.

Example

Electric force falls with the square of distance. Doubling distance makes the force one quarter as large. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Start with a reference distance

Use the same charges and begin at a distance of 1 metre. In this reference case, the force size is 16 newtons.

r = 1\ \text{m}\qquad F = 16\ \text{N}

Double the distance

Now keep both charges the same but move the right charge farther away. The distance changes from 1 metre to 2 metres.

r_{\text{new}} = 2\ \text{m} \qquad \text{distance factor} = 2

Distance is an inverse-square input

Distance counts twice in the denominator. Doubling the distance makes the distance-square factor four, so the force becomes one quarter as large.

\text{force factor} = \frac{\text{one}}{2\cdot 2} = \frac{\text{one}}{4}

The force becomes one quarter

One quarter of 16 newtons is 4 newtons. The force shrinks fast because distance is squared.

F_{\text{new}} = 16\ \text{N}\cdot \frac{\text{one}}{4} = 4\ \text{N}

A three-row scan shows the square

Keep the charges fixed and compare three exact distances. The diagram shows the middle row. Doubling distance from there again would divide the force by another four.

\begin{array}{c|c|c}r&r\cdot r\text{ factor}&F\\1\ \text{m}&1&16\ \text{N}\\2\ \text{m}&4&4\ \text{N}\\4\ \text{m}&16&1\ \text{N}\\\end{array}