Field as Force per Charge - Electric Fields, Electrostatics & Capacitors Step by Step

An electric field tells how much force a positive test charge would feel per coulomb at a point.

electrostatics A field vector is a local force-per-charge claim at one point.

Example

An electric field tells how much force a positive test charge would feel per coulomb at a point. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Use a test charge

Place a small positive test charge at the point where you want to know the electric field. Here the test charge is 3 coulombs.

q_{\text{test}} = 3\ \text{C}

Measure force on the test charge

The force on that test charge is 12 newtons. The electric field is force per coulomb, so divide the force by the test charge.

E = \frac{F}{q_{\text{test}}}

Compute force per charge

Divide 12 newtons by 3 coulombs. The result is 4 newtons per coulomb.

E = \frac{12\ \text{N}}{3\ \text{C}} = 4\ \text{N/C}

More force gives more field at fixed charge

Hold the test charge at 3 coulombs. The diagram shows the middle row; increasing force increases the field by the same factor.

\begin{array}{c|c|c}F&q_{\text{test}}&E\\6\ \text{N}&3\ \text{C}&2\ \text{N/C}\\12\ \text{N}&3\ \text{C}&4\ \text{N/C}\\18\ \text{N}&3\ \text{C}&6\ \text{N/C}\\\end{array}

More test charge lowers force per charge

Now hold the force at 12 newtons and change the test charge. A larger test charge spreads the same force over more coulombs, so the field value is smaller.

\begin{array}{c|c|c}F&q_{\text{test}}&E\\12\ \text{N}&2\ \text{C}&6\ \text{N/C}\\12\ \text{N}&3\ \text{C}&4\ \text{N/C}\\12\ \text{N}&6\ \text{C}&2\ \text{N/C}\\\end{array}