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.
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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 at a point is defined as the force per unit charge that a small positive test charge would feel:
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 , 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:
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.
The uniform field between parallel plates
Two parallel plates with opposite charge and a potential difference across a separation produce a uniform field in the gap:
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).
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.
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.
2024 TASC5 marksA +4.0 microC charge and a +6.7 microC 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.Show worked answer →
Each charge sets up a radial field E = k q / r^2 at P, pointing away from the (positive) charge along the line from the charge to P.
From the +4.0 microC charge: E1 = 8.99 x 10^9 x 4.0 x 10^-6 / (0.07)^2 = 35960 / 0.0049 = 7.34 x 10^6 N C-1.
From the +6.7 microC charge: E2 = 8.99 x 10^9 x 6.7 x 10^-6 / (0.07)^2 = 60233 / 0.0049 = 1.23 x 10^7 N C-1.
Geometry: the triangle has sides 7, 7 and 10 cm, so the angle between the two field vectors at P is found from the cosine rule (10^2 = 7^2 + 7^2 - 2 x 7 x 7 cos(angle)), giving about 91 degrees, essentially perpendicular.
Resultant (treating them as perpendicular): E = sqrt(E1^2 + E2^2) = sqrt((7.34 x 10^6)^2 + (1.23 x 10^7)^2) = sqrt(5.39 x 10^13 + 1.51 x 10^14) = 1.43 x 10^7 N C-1.
The field at P is about 1.4 x 10^7 N C-1. Markers want each field from E = kq/r^2 and the vector (near-perpendicular) addition.
2019 TASC4 marksTwo charges of +2 nC and -4 nC are arranged 3 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 cm from each charge).Show worked answer →
Each charge produces a field at P: E = k q / r^2 with r = 0.03 m.
From +2 nC: E1 = 8.99 x 10^9 x 2 x 10^-9 / (0.03)^2 = 17.98 / 9 x 10^-4 = 2.00 x 10^4 N C-1, pointing away from the +2 nC charge.
From -4 nC: E2 = 8.99 x 10^9 x 4 x 10^-9 / (0.03)^2 = 35.96 / 9 x 10^-4 = 4.00 x 10^4 N C-1, pointing toward the -4 nC charge.
The two field vectors are separated by 120 degrees at P (equilateral triangle). Combine using the cosine rule for vectors:
E = sqrt(E1^2 + E2^2 + 2 E1 E2 cos120)
= sqrt((2.0 x 10^4)^2 + (4.0 x 10^4)^2 + 2(2.0 x 10^4)(4.0 x 10^4)(-0.5))
= sqrt(4 x 10^8 + 16 x 10^8 - 8 x 10^8) = sqrt(12 x 10^8) = 3.46 x 10^4 N C-1.
The resultant field at P is about 3.5 x 10^4 N C-1. Markers want both field magnitudes, correct directions (away from positive, toward negative), and the 120 degree vector sum.
2019 TASC3 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 charges (free electrons) are free to move.
When an external charge is brought near, the free electrons in the conductor redistribute over the surface until they are in 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.