The parallel-plate field connects circuit voltage to electrostatic field. 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 voltage per distance

For ideal parallel plates, field strength is voltage divided by gap distance.

E=VdE = \frac{V}{d}

Substitute the values

Use 12 volts across 3 metres.

E=12 V3 mE = \frac{12\ \text{V}}{3\ \text{m}}

Use the field unit from electrostatics

Volts per metre is the same field dimension as newtons per coulomb.

V/m=N/C\text{V}/\text{m} = \text{N/C}

Compute the field

The field is 4 newtons per coulomb.

E=4 N/CE = 4\ \text{N/C}
Field from voltageEvery displayed field vector is checked against the same value.4 N/C4 N/C4 N/C

Voltage is the numerator

With gap fixed at 3 metres, more voltage gives a stronger field.

VdE6 V3 m2 N/C12 V3 m4 N/C18 V3 m6 N/C\begin{array}{c|c|c}V&d&E\\6\ \text{V}&3\ \text{m}&2\ \text{N/C}\\12\ \text{V}&3\ \text{m}&4\ \text{N/C}\\18\ \text{V}&3\ \text{m}&6\ \text{N/C}\\\end{array}
Field from voltageThe middle table row is the checked diagram.4 N/C4 N/C4 N/C

Distance is the denominator

With voltage fixed at 12 volts, a wider gap weakens the field because the same voltage is spread over more distance.

VdE12 V2 m6 N/C12 V3 m4 N/C12 V6 m2 N/C\begin{array}{c|c|c}V&d&E\\12\ \text{V}&2\ \text{m}&6\ \text{N/C}\\12\ \text{V}&3\ \text{m}&4\ \text{N/C}\\12\ \text{V}&6\ \text{m}&2\ \text{N/C}\\\end{array}
Field from voltageThe middle table row is the checked diagram.4 N/C4 N/C4 N/C