TASC TCE 2024-style practice: A +4.0\ μC charge and a +6.7\ μC charge are placed 10\ cm apart. A point P is located 7.0\ cm from both charges. With the aid of a suitable vector diagram, calculate the magnitude of the electric field strength at the point P.

Each charge sets up a radial field E = kqr^2 at P, pointing away from the (positive) charge along the line from the charge to P. From the +4.0\ μC charge: E_1 = (8.9910^9)(4.010^-6)(0.07)^2 = 7.34 × 10^6\ N C^-1. From the +6.7\ μC charge: E_2 = (8.9910^9)(6.710^-6)(0.07)^2 = 1.23 × 10^7\ N C^-1. Geometry: the triangle has sides 7, 7 and 10\ cm, so by the cosine rule the angle between the two field vectors at P is about 91^, essentially perpendicular. Resultant (treating them as perpendicular): $E = sqrt(E_1^2 + E_2^2) = sqrt((7.3410^6)^2 + (1.2310^7)^2) = 1.43 × 10^7\ N C^-1.$ The field at P is about 1.4 × 10^7\ N C^-1. Markers want each field from E = kqr^2 and the vector addition.

§-Syllabus dot point

TASPhysicsSyllabus dot point

How do we map the influence of a charge on the space around it?

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

Electric field strength as force per unit charge, the radial field of a point charge, the uniform field between parallel plates, and how field lines represent direction and strength.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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.

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.

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).

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 practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20245 marksA

+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.

TCE 20194 marksTwo 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.

TCE 20193 marksA 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.