Electric fields and field lines - TCE Physics (Tasmania)

Syllabus dot point

“Define electric field strength and represent the fields of point charges and parallel plates with field lines.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min readUpdated 2 Jun 2026

Quick answer

The electric field strength at a point is the force per unit positive charge there, $E = \dfrac{F}{q}$ . A point charge makes a radial field $E = \dfrac{kQ}{r^2}$ , while two parallel charged plates make a uniform field $E = \dfrac{V}{d}$ . Field lines point the way a positive charge would be pushed.

What this dot point is asking

This dot point develops the field model for electricity, describing the influence a charge has on the space around it without needing a second charge to be present.

Electric field strength

The electric field strength $E$ at a point is defined as the force per unit charge that a small positive test charge would feel:

E = \frac{F}{q}

measured in newtons per coulomb. It is a vector pointing in the direction of the force on a positive charge. Rearranged, the force on any charge placed in a known field is $F = qE$ , which is the key to analysing charges in fields.

Electric field strength

The electric field strength at a point is the electric force experienced per unit positive charge placed at that point. It is a vector, measured in newtons per coulomb, pointing in the direction a positive charge would be pushed.

The field of a point charge

Around an isolated point charge the field is radial. Its strength follows the inverse-square law:

E = \frac{kQ}{r^2}

The field points away from a positive charge and toward a negative charge. Field lines for a point charge are straight, evenly spaced rays; their spacing widens with distance, showing the field weakening. To find the field at a point from two or more charges, compute each contribution with $E = \dfrac{kQ}{r^2}$ , draw its direction, then add the vectors, just as in the worked TCE questions above.

The uniform field between parallel plates

Two parallel plates with opposite charge and a potential difference $V$ across a separation $d$ produce a uniform field in the gap:

E = \frac{V}{d}

Here the field lines are straight, parallel and equally spaced, running from the positive plate to the negative plate. The field strength is the same everywhere between the plates (ignoring the edges), which makes this arrangement ideal for accelerating and deflecting charges in a controlled way.

Two equivalent units for field strength

Electric field strength can be written in newtons per coulomb or volts per metre, and these units are identical. The point-charge form $\dfrac{kQ}{r^2}$ naturally gives newtons per coulomb, while the parallel-plate form $\dfrac{V}{d}$ naturally gives volts per metre.

Reading field lines

Field lines obey simple rules: they start on positive charge and end on negative charge, they never cross, and their density shows the field strength, so closely packed lines mean a strong field. The line through a point gives the direction of the force on a positive charge placed there. For two charges, the pattern shows attraction (lines linking them) or repulsion (lines pushed apart). The field is zero at any point where contributions cancel; for two like charges this neutral point lies on the line between them, closer to the smaller charge.

Conductors and shielding

In a conductor at electrostatic equilibrium the field inside is zero, because free electrons rearrange until any internal field is cancelled. Excess charge sits only on the outer surface, and field lines meet a conductor's surface at right angles. This is the principle behind the Faraday cage: a hollow conductor shields its interior from external electrostatic fields, which is why a car or an aircraft offers some protection during a lightning strike.

Worked example: field between plates and force on an electron

Find the field strength

Two parallel plates are separated by $5.0\ \text{mm}$ with a potential difference of $200\ \text{V}$ :

E = \frac{V}{d} = \frac{200}{5.0\times10^{-3}} = 4.0\times10^{4}\ \text{V m}^{-1}.

Find the force on the electron

With $q = 1.6\times10^{-19}\ \text{C}$ :

F = qE = (1.6\times10^{-19})(4.0\times10^{4}) = 6.4\times10^{-15}\ \text{N}.

The electron feels $6.4 \times 10^{-15}\ \text{N}$ , directed toward the positive plate (opposite to the field, because the charge is negative).

Confusing the point-charge and parallel-plate formulas

A common error is using $E = \dfrac{kQ}{r^2}$ for the field between plates or $E = \dfrac{V}{d}$ for a point charge. The radial inverse-square form is for a single source charge; the $\dfrac{V}{d}$ form is only for the uniform field between parallel plates. Identify the geometry first, then pick the matching formula.

In the exam, decide whether the field is radial (point charge) or uniform (parallel plates) before choosing a formula. Remember the field direction is defined for a positive test charge, so a negative charge feels a force in the opposite direction to the field, and add multiple point-charge fields as vectors.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks

+4.0\ \mu\text{C}

charge and a

+6.7\ \mu\text{C}

charge are placed

10\ \text{cm}

apart. A point P is located

7.0\ \text{cm}

from both charges. With the aid of a suitable vector diagram, calculate the magnitude of the electric field strength at the point P.

Show worked answer →

Each charge sets up a radial field $E = \dfrac{kq}{r^2}$ at P, pointing away from the (positive) charge along the line from the charge to P.

From the $+4.0\ \mu\text{C}$ charge: $E_1 = \dfrac{(8.99\times10^9)(4.0\times10^{-6})}{(0.07)^2} = 7.34 \times 10^6\ \text{N C}^{-1}$ .

From the $+6.7\ \mu\text{C}$ charge: $E_2 = \dfrac{(8.99\times10^9)(6.7\times10^{-6})}{(0.07)^2} = 1.23 \times 10^7\ \text{N C}^{-1}$ .

Geometry: the triangle has sides $7$ , $7$ and $10\ \text{cm}$ , so by the cosine rule the angle between the two field vectors at P is about $91^\circ$ , essentially perpendicular.

Resultant (treating them as perpendicular):

E = \sqrt{E_1^2 + E_2^2} = \sqrt{(7.34\times10^6)^2 + (1.23\times10^7)^2} = 1.43 \times 10^7\ \text{N C}^{-1}.

The field at P is about $1.4 \times 10^7\ \text{N C}^{-1}$ . Markers want each field from $E = \dfrac{kq}{r^2}$ and the vector addition.

Practice4 marks

Two charges of

+2\ \text{nC}

and

-4\ \text{nC}

are arranged

3\ \text{cm}

apart. Find the magnitude of the total electric field strength at the vertex P of an equilateral triangle formed with the two charges, so P is

3\ \text{cm}

from each charge.

Show worked answer →

Each charge produces a field at P with $E = \dfrac{kq}{r^2}$ and $r = 0.03\ \text{m}$ .

From $+2\ \text{nC}$ : $E_1 = \dfrac{(8.99\times10^9)(2\times10^{-9})}{(0.03)^2} = 2.00 \times 10^4\ \text{N C}^{-1}$ , pointing away from the charge.

From $-4\ \text{nC}$ : $E_2 = \dfrac{(8.99\times10^9)(4\times10^{-9})}{(0.03)^2} = 4.00 \times 10^4\ \text{N C}^{-1}$ , pointing toward the charge.

The two field vectors are separated by $120^\circ$ at P. Combine using the cosine rule for vectors:

E = \sqrt{E_1^2 + E_2^2 + 2E_1 E_2 \cos 120^\circ} = \sqrt{(2.0\times10^4)^2 + (4.0\times10^4)^2 - (2.0\times10^4)(4.0\times10^4)} = 3.46 \times 10^4\ \text{N C}^{-1}.

The resultant field at P is about $3.5 \times 10^4\ \text{N C}^{-1}$ . Markers want both magnitudes, correct directions, and the $120^\circ$ vector sum.

Practice3 marks

A Faraday cage does not let external electrostatic effects into it. Describe why a closed, charged metal cage has no electric field inside due to an external charge.

Show worked answer →

A metal cage is a conductor, so its free electrons can move.

When an external charge is brought near, the free electrons redistribute over the surface until they reach equilibrium. They arrange themselves so that the field they create inside the cavity exactly cancels the external field at every interior point.

In electrostatic equilibrium the net electric field inside the conducting material is zero, and because there is no charge inside the cavity, the field throughout the enclosed space is also zero. Any excess charge resides only on the outer surface.

This is why sensitive electronics and people inside a metal enclosure are shielded from external electrostatic effects and even lightning. Markers want: conductor, free charges redistribute on the surface, induced field cancels the external field, so the interior field is zero.

Practise this

Exam-style questions 3 on this page
Flashcards Physics deck

Sources & how we know this

Physics Level 4 (PHY415115) course document (2015 onwards) — TASC