Describe electric fields and potential and apply them to point charges and uniform fields between parallel plates

A focused answer to the WACE Year 12 Physics Unit 3 dot point on electric fields and potential. Coulomb's law, field strength for point charges and parallel plates, work, potential difference and the motion of a charge in a uniform field.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-29

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What this dot point is asking

WACE treats electric fields as a direct parallel to gravitational fields, with the key differences that charge comes in two signs and the force can attract or repel. You need to handle both the point-charge case and the uniform-field-between-plates case, and connect field, force, work and potential difference.

Coulomb's law and the field of a point charge

The force between two point charges is

F=k\frac{q_1q_2}{r^2},\qquad k=8.99\times10^{9}\ \text{N m}^2\,\text{C}^{-2}.

Like charges repel; opposite charges attract. The electric field strength is the force per unit positive charge,

E=\frac{F}{q}=k\frac{Q}{r^2}\quad(\text{N C}^{-1}).

Field is a vector, pointing away from positive source charges and toward negative ones. Field lines never cross, and their density represents field strength.

The uniform field between parallel plates

Two oppositely charged parallel plates produce a uniform field (except near the edges). Its magnitude is

E=\frac{V}{d}

where $V$ is the potential difference across the plates and $d$ their separation. Here the units $\text{V m}^{-1}$ and $\text{N C}^{-1}$ are equivalent. The field points from the positive plate to the negative plate, and the force $F=qE$ on a charge is constant throughout the gap.

Work and potential difference

Potential difference is the work done per unit charge moved between two points,

V=\frac{W}{q}\quad\Rightarrow\quad W=qV.

When a charge is accelerated from rest through a potential difference, that work becomes kinetic energy,

qV=\tfrac{1}{2}mv^2,

so $v=\sqrt{\dfrac{2qV}{m}}$ . This is the standard electron-gun calculation.

Motion of a charge in a uniform field

A charged particle entering a uniform field experiences a constant force, so its path is a parabola in exactly the same way a projectile is. The component along the field accelerates uniformly while the perpendicular component stays constant. This deflection of charges is the basis of cathode-ray tubes and is a favourite exam scenario.

Accelerating an electron and finding the field

Electron accelerated through 250 V between plates

An electron starts from rest and is accelerated through a potential difference of $250\ \text{V}$ . Take $q=1.60\times10^{-19}\ \text{C}$ and $m=9.11\times10^{-31}\ \text{kg}$ .

Work done on the electron:

W=qV=(1.60\times10^{-19})(250)=4.00\times10^{-17}\ \text{J}.

This equals its kinetic energy, so

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(4.00\times10^{-17})}{9.11\times10^{-31}}}=\sqrt{8.78\times10^{13}}=9.4\times10^{6}\ \text{m s}^{-1}.

If the plates are $5.0\ \text{cm}$ apart, the field between them is

E=\frac{V}{d}=\frac{250}{0.050}=5.0\times10^{3}\ \text{V m}^{-1}.

Linking the field and potential pictures

A graph of $E$ against $r$ for a point charge falls as $1/r^2$ , while a graph for parallel plates is flat. The work done moving a charge equals $q$ times the change in potential, which is the area interpretation you may be asked to use. Keep clear which scenario you are in before choosing a formula.

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