Force from Field - Electric Fields, Electrostatics & Capacitors Step by Step

Once the field at a point is known, multiplying by a test charge gives the force on that charge.

electrostatics Field is force per charge, so a larger test charge feels proportionally more force.

Example

Once the field at a point is known, multiplying by a test charge gives the force on that charge. 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 field value

At the test point, the electric field is 5 newtons per coulomb.

E = 5\ \text{N/C}

Multiply by the test charge

A field tells the force per coulomb. A positive 2 coulomb test charge feels 2 copies of that force-per-charge.

F = qE

Compute the force

Multiplying 2 coulombs by 5 newtons per coulomb gives 10 newtons.

F = 2\ \text{C}\cdot 5\ \text{N/C} = 10\ \text{N}

More charge means more force at fixed field

Hold the field at 5 newtons per coulomb. The diagram shows the middle row, and the force grows directly with test charge.

\begin{array}{c|c|c}q_{\text{test}}&E&F\\1\ \text{C}&5\ \text{N/C}&5\ \text{N}\\2\ \text{C}&5\ \text{N/C}&10\ \text{N}\\4\ \text{C}&5\ \text{N/C}&20\ \text{N}\\\end{array}

More field means more force at fixed charge

Now hold the test charge at 2 coulombs. A stronger field gives more force on the same charge.

\begin{array}{c|c|c}q_{\text{test}}&E&F\\2\ \text{C}&3\ \text{N/C}&6\ \text{N}\\2\ \text{C}&5\ \text{N/C}&10\ \text{N}\\2\ \text{C}&7\ \text{N/C}&14\ \text{N}\\\end{array}