Understand electric fields the easy way, with plain English intuition, an interactive simulation, point charges and uniform fields, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.
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Bring two charges near each other and they feel a force without ever touching. Each charge fills the space around it with an invisible influence called an electric field, and any other charge dropped into that space gets pushed or pulled. The field is just a map that tells you, at every point, which way a charge would be shoved and how hard. Get the map right and the force falls straight out of it.
A field is a force waiting to happen
Here is the whole idea. An electric field is the region around a charge where another charge feels a force. The field has a strength, written E, measured in newtons per coulomb (which is the same as volts per metre).
The fieldE exists whether or not anything is sitting in it. It is a property of the space.
Drop a charge q into the field and it feels a forceF=qE, in the direction of the field for a positive charge.
So the field is the cause and the force is the effect. Find the field first, then multiply by the charge to get the force.
The two field shapes you must know
VCE gives you two field shapes, and almost every question is one of them.
The first is the field around a single point charge. It spreads out in all directions and gets weaker as you move away, following an inverse square law:
E=r2kQ,k=8.99×109.
Field lines point away from a positive charge and towards a negative charge. Double your distance and the field drops to a quarter, because of the r2 on the bottom.
The second is the uniform field between two parallel charged plates. Here the field is the same strength and direction everywhere between the plates, and the formula is beautifully simple:
E=dV.
V is the voltage across the plates and d is their separation. This is the setup in the diagram below, and the one the worked examples use.
The red arrow F is the force on the small positive charge, and it points the same way as the field, straight from the positive plate towards the negative plate. The light arrows are the field lines: evenly spaced and all the same length, which is exactly what a uniform field looks like. The separation between the plates is d.
Field gives force, every time
This is the line that ties the whole topic together. Once you know the field at a point, the force on any charge sitting there is one multiplication away.
See it for yourself
Drop positive and negative charges onto the canvas, then use the sensors to read the field strength and direction at any point. Place two opposite charges near each other and watch the field lines bend, or line up a row of charges to build something close to the uniform field between two plates.
Decide which field shape you have: a point charge or a uniform field between plates.
For a point charge, find the field with E=r2kQ using k=8.99×109.
For parallel plates, find the field with E=dV. Keep the separation in metres.
Once you have E, the force on a charge sitting in it is F=qE.
Watch the units. A plate separation given in centimetres or millimetres must be converted to metres first, and charges are almost always tiny powers of ten, so track the exponent carefully.
Lock it in with active recall
Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.
Active recall
Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).
What is an electric field, and what are its units?
The region around a charge where another charge feels a force. Field strength E is in newtons per coulomb (N/C), the same as volts per metre (V/m).
Write the field formula for a point charge and for parallel plates.
Point charge: E=r2kQ with k=8.99×109. Parallel plates: E=dV (uniform).
Which way do electric field lines point?
Out of positive charges and into negative ones — the direction a small positive test charge would be pushed. They never cross.
How do you find the force on a charge q sitting in a field E?
F=qE. A positive charge feels a force along the field; a negative charge feels it the opposite way.
Double the distance from a single point charge. What happens to E?
It drops to one quarter, because E=r2kQ is an inverse square law.
Recall · Gravitational Fields
How does the point-charge field compare with gravitational field strength?
Both are inverse square: E=r2kQ and g=r2GM. Gravity only ever attracts, while electric fields can push or pull.
Recall · Charged Particles in Fields
What path does a charge follow when it enters a uniform electric field sideways?
A parabola — the constant force F=qE acts like gravity on a projectile.
See the recipe in action in the Worked Examples tab, then test yourself in Try It.
Worked examples
Worked Example 1Field between two parallel plates
Two parallel plates are 0.020 m apart with a voltage of 100 V across them. Find the electric field strength between them.
1
Between parallel plates the field is uniform, so the strength is just the voltage divided by the separation.
E=dV
2
Substitute the voltage and the plate separation.
E=0.020100
3
Work out the division. The units are volts per metre, which is the same as newtons per coulomb.
E=5000 N/C (V/m)
Answer
E=dV=0.020100=5000 N/C (V/m)
Worked Example 2Force on a charge in a field
A charge of 2.0×10−6 C sits in a uniform field of 5000 N/C. Find the electric force on it.
1
The force on any charge in a field is the charge multiplied by the field strength.
F=qE
2
Substitute the charge and the field strength.
F=(2.0×10−6)(5000)
3
Multiply the two numbers to get the force in newtons.
F=0.010 N
Answer
F=qE=(2.0×10−6)(5000)=0.010 N
Practice questions
Practice test
Try it yourself
6 questions, 8 marks
Choose your answers, then submit to see your score and the full worked solutions.
Multiple choice is marked for you, just like Exam 2 Section A.
Q1.A small positive test charge is placed in an electric field. The direction of the force on it is:
1mark
Need a hint?
The field direction is defined as the direction of the force on a positive charge.
Show worked solution
The electric field direction is defined as the direction of the force on a small positive test charge, so a positive charge feels a force in the same direction as the field. A negative charge would feel a force in the opposite direction, but the field itself points the way a positive charge is pushed.
Q2.Two parallel plates are 0.010 m apart with 50 V across them. The uniform field strength between them is:
1mark
Need a hint?
Use E=dV with the numbers given.
Show worked solution
E=dV=0.01050=5000 N/C. Dividing the wrong way round, 500.010, gives the tiny distractor, a common slip when the separation is a small decimal.
Q3.A charge of 3.0×10−6 C sits in a uniform field of 2000 N/C. The electric force on it is:
1mark
Need a hint?
Use F=qE and keep track of the power of ten.
Show worked solution
F=qE=(3.0×10−6)(2000)=6.0×10−3 N. Multiplying 3.0×2000=6000=6.0×103, then carrying the 10−6 across gives 6.0×10−3 N. Losing the power of ten is the usual mistake here.
Q4.If you double the distance from a single point charge, the electric field strength at that point becomes:
1mark
Need a hint?
For a point charge E=r2kQ, so the field depends on the distance squared.
Show worked solution
For a point charge E=r2kQ. Doubling r means dividing by 22=4, so the field drops to a quarter of its value. This is the inverse square law, the same shape as gravity. Halving the field would only need a 2 increase in distance, not a doubling.
Q5.Two parallel plates are 0.040 m apart with 200 V across them. Find the uniform electric field strength between the plates, then find the force on a charge of 5.0×10−6 C placed between them. Show your working.
3marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
Between parallel plates the field is uniform, so
E=dV=0.040200=5000 N/C.
The force on the charge is
F=qE=(5.0×10−6)(5000)=0.025 N.
Q6.Two positive point charges, +Q and +9Q, are placed 20 cm apart. On the straight line between them, the electric field is:
1mark
Need a hint?
The field is zero where the two fields cancel: r2kQ=(0.20−r)2k(9Q).
Show worked solution
Setting the magnitudes equal, r2Q=(0.20−r)29Q, so (0.20−r)=3r, giving r=0.05 m =5.0 cm from +Q (option C). The null point sits closer to the weaker charge.
VCAA 2025 Physics Exam, Section A Q8
Frequently asked questions
What actually is an electric field?
It is the region of space around a charge where another charge would feel a push or a pull. The field strength tells you how big a force a one coulomb charge would feel at that spot, measured in newtons per coulomb.
Why is the field between two parallel plates uniform?
Because the field lines run straight from one plate to the other, evenly spaced and the same strength everywhere in between. That is why one simple formula, E equals V divided by d, works no matter where you are between the plates.
Which way do field lines point?
Field lines point away from positive charges and towards negative charges, always in the direction a small positive test charge would be pushed. They never cross and they are closest together where the field is strongest.
