Compare Two Changes - Electric Fields & Electrostatics Step by Step

When charge and distance both change, their exact factors combine. Double charge and double distance leave half the reference force.

Example

When charge and distance both change, their exact factors combine. Double charge and double distance leave half the reference force. 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 from the reference force

The reference force is 12 newtons for the stated charges and distance.

F_{\text{base}} = 12\ \text{N}

Change charge and distance together

Double the left charge from 2 coulombs to 4 coulombs. Also double the distance from 1 metre to 2 metres.

\text{charge factor} = 2\qquad \text{distance-square factor} = 2\cdot 2 = 4

The distance change wins partly

The charge change multiplies the force by two. The doubled distance divides the force by four. Together, the force is half as large.

\text{net factor} = \frac{2}{4}

Half of the reference force remains

Half of 12 newtons is 6 newtons. Changing more charge and more distance at the same time means the effects combine.

F_{\text{new}} = 12\ \text{N}\cdot \frac{2}{4} = 6\ \text{N}

The net factor is a ledger

Use the same reference force, then compare factor rows. The diagram shows the third row: charge factor 2, distance factor 2, net factor one half.

\begin{array}{c|c|c|c}q\text{ factor}&r\text{ factor}&\text{net factor}&F_{\text{new}}\\1&1&1&12\ \text{N}\\2&1&2&24\ \text{N}\\2&2&\tfrac{1}{2}&6\ \text{N}\\4&2&1&12\ \text{N}\\\end{array}

electrostatics Charge factors and distance factors combine; this lesson keeps the combined ratio exact instead of using decimals.