IB Physics HLTopic 4 — Force FieldsPaper 1 & 2E = F/q~19 min read
Electric Field Strength
Coulomb’s law needs two charges. But a single charge, alone in empty space, is plainly doing something — it is waiting, ready, and the moment another charge arrives it pushes. So stop asking “what force do these two exert on each other” and ask instead: what would this charge do to a test charge placed here? Divide the force by the test charge, the second charge vanishes from the equation, and what is left behind belongs to space itself. That is a field.
📘 What you need to know
An electric field is a region of space in which a charge experiences a force
Electric field strength is the force per unit charge on a small positive test charge at that point
E = F/q, measured in N C−1 — which is identical to V m−1
E is a vector: directed away from a positive charge, towards a negative charge
A charge +q feels a force Eqalong the field; a charge −q feels Eq in the opposite direction
Around a point charge: E = kq/r2 — another inverse square law
A charged sphere behaves exactly like a point charge at its centre — but E = 0 everywhere inside it
Between parallel plates the field is uniform: E = V/d, directed from the positive plate to the negative plate
Fields from several charges add as vectors: along a line, add or subtract; at right angles, use Pythagoras
What a field is, and how we measure it
An electric field is simply a region of space in which an electric charge experiences a force. Every charged particle makes one, and the field is what reaches out and does the pushing.
To measure how strong a field is at some point, drop a charge there and see how hard it is shoved. But a big charge would disturb the very field you are trying to measure, and a negative one would be pulled backwards. So we agree on a convention: a small, positivetest charge.
Electric field strength — definitionthe force per unit charge experienced by a small positive test charge placed at that point
Electric field strengthE = F / qE = field strength (N C−1) • F = force on the charge (N) • q = charge (C)
The word positive in that definition is not decoration — it is what fixes the direction of every field arrow you will ever draw. And a small aside worth banking now: N C−1 and V m−1 are the same unit. Follow it through: a volt is a joule per coulomb, and a joule is a newton-metre, so V m−1 = N m / (C m) = N C−1. Use whichever the question uses, and never let the swap alarm you.
Which way does the field point?
Because the test charge is positive, the field arrow points wherever a positive charge would be pushed. That gives one rule with two halves:
Field strength is a vector. A charge of −q placed in either field feels a force of the same size Eq, but pointing the other way along the arrow.
A positive charge +q experiences a force Eqin the direction of the field
A negative charge −q experiences a force Eq in the opposite direction
The field around a point charge
Now put Coulomb’s law and the definition together. The force on a test charge q sitting a distance r from a source charge Q is F = kQq/r2. Divide by q, and the test charge disappears:
Field strength due to a point chargeE = F/q = kQ / r2Q = the charge making the field • r = distance from its centre
Look at what just happened. The test charge cancelled. The field at that point does not care what you put there, or whether you put anything there at all — it belongs to Q and to the empty space around it. This is exactly the same vanishing act as g = GM/r2, and for exactly the same reason.
Charged spheres, inside and out
A uniformly charged sphere behaves, from the outside, exactly like a point charge with the same total charge, sitting at the sphere’s centre. Outside it, E = kQ/r2 and nothing changes.
Inside is another story. Every scrap of charge on the shell pulls on a test charge in a different direction, and the whole lot cancels exactly. The field inside a charged sphere is zero, everywhere.
Zero inside, maximum at the surface, then an inverse-square fall. Note the field jumps at r = R — it does not climb gently from the centre.
The uniform field between parallel plates
Take two flat metal plates, park them parallel, and connect a potential difference across them. The field between them is uniform — the same strength and the same direction at every point. And its size is beautifully simple:
Uniform field between parallel platesE = V / dV = potential difference between the plates (V) • d = plate separation (m)
The field runs from the positive plate to the negative plate. Between them it is uniform; beyond the edges it bulges and is non-uniform.
The equation earns its keep by telling you two things at a glance:
The greater the voltage between the plates, the stronger the field
The greater the separation, the weaker the field
Two traps live here. First: E = V/d is only for parallel plates. The field around a point charge is radial, not uniform, and this equation is meaningless there. Second: if a question tells you one plate is earthed, that plate is at 0 V, so V is simply the potential of the other one. Free mark, if you are watching for it.
And when a charged particle sits between the plates, you have two expressions for the same E. Equate them:
A charge between parallel platesE = F/q = V/d → F = qV / dthe same balance Millikan used to hold his oil drop still
Radial field (point charge)
Uniform field (parallel plates)
Equation
E = kQ/r2
E = V/d
Field lines
Radial, spreading out
Parallel, equally spaced
Strength
Decreases with distance
Same everywhere between the plates
Direction
Away from +, towards −
From the + plate to the − plate
Combining fields
Electric force is a vector, so electric field strength is a vector too. Two charges, two fields at a point — and you add them the way you add any vectors.
To get the direction of each contribution, remember the point behaves as a positive test charge: it is pushed away from positive charges and pulled towards negative ones.
Force on a test charge q
divide by q
E = F/q
substitute Coulomb’s law
E = kQ/r2
⚡ Working a field strength question
Radial or uniform? Point charge or sphere → E = kQ/r2. Parallel plates → E = V/d. Never swap them.
Inside a charged sphere? Stop. E = 0. No calculation.
Force on a charge?F = Eq, using the magnitude of q, then read the direction from the sign.
Convert. cm and mm → metres; nC and µC → coulombs. Halve any diameter.
More than one charge? Find each E separately, then add as vectors: same line → add or subtract; right angle → Pythagoras.
Units. N C−1 and V m−1 are the same thing. Either is correct.
WE 1
A charge of 5.0 nC placed at a point in an electric field experiences a force of 2.4 × 10−4 N. (a) Calculate the electric field strength at that point. (b) State the direction of the force if the charge were replaced by an electron.
(a) Step 1 — convert the chargeq = 5.0 nC = 5.0 × 10⁻⁹ CStep 2 — use the definitionE = F / q = (2.4 × 10⁻⁴) / (5.0 × 10⁻⁹)E = 4.8 × 10⁴ N C⁻¹(b) Step 3 — think about the sign
The field is defined by the force on a positive test charge.
An electron is negative.
the force is opposite to the fieldThe field strength does not change one bit when you swap the charge — that’s the whole point of dividing by q. Only the force changes, in size and in direction.
WE 2
An isolated metal sphere of radius 12 cm carries a uniformly distributed charge of +40 nC. Calculate the electric field strength (a) at the surface of the sphere, (b) at a point 30 cm from its centre, and (c) at the centre of the sphere. (k = 8.99 × 109 N m2 C−2)
(a) Step 1 — outside, treat it as a point charge at the centreE = kQ / r² = (8.99 × 10⁹)(40 × 10⁻⁹) / (0.12)²E = 359.6 / 0.0144E = 2.5 × 10⁴ N C⁻¹, directed away from the sphere(b) Step 2 — same equation, bigger rE = 359.6 / (0.30)² = 359.6 / 0.090E = 4.0 × 10³ N C⁻¹(c) Step 3 — inside the sphereE = 0Check (b) against (a) without a calculator: r went up by a factor of 0.30/0.12 = 2.5, so E must fall by 2.5² = 6.25. And 2.5 × 10⁴ ÷ 6.25 = 4.0 × 10³. It agrees. Part (c) needs no working at all — the charges on the shell cancel exactly.
WE 3
Two parallel plates are 8.0 mm apart. The upper plate is held at 2.4 kV and the lower plate is earthed. (a) Calculate the electric field strength between the plates. (b) Calculate the force on a charge of 3.2 × 10−19 C placed between them. (c) State the direction of that force if the charge is negative.
(a) Step 1 — the earthed plate is at 0 V, so V is just 2.4 kVV = 2400 V | d = 8.0 mm = 8.0 × 10⁻³ mE = V / d = 2400 / 0.0080E = 3.0 × 10⁵ V m⁻¹(b) Step 2 — force on a charge in a fieldF = Eq = (3.0 × 10⁵) × (3.2 × 10⁻¹⁹)F = 9.6 × 10⁻¹⁴ N(c) Step 3 — the field runs from + plate to − plate, so downwards
A negative charge feels a force opposite to the field.
upwards, towards the positive plateThat charge is 3.2 × 10⁻¹⁹ C, which is exactly 2e — quantised, as always. And notice (a) and (b) together are just F = qV/d, the Millikan balance in disguise.
WE 4
At a point X, one charge produces a field of 3.0 × 104 N C−1 and a second charge produces a field of 4.0 × 104 N C−1. Determine the resultant field strength at X if the two fields are (a) in the same direction, (b) in opposite directions, (c) at right angles to one another.
(a) same direction — simply addE = 3.0 + 4.0 (× 10⁴)E = 7.0 × 10⁴ N C⁻¹(b) opposite directions — subtractE = 4.0 − 3.0 (× 10⁴)E = 1.0 × 10⁴ N C⁻¹, in the direction of the 4.0 × 10⁴ field(c) right angles — PythagorasE = √(3.0² + 4.0²) = √25 (× 10⁴)E = 5.0 × 10⁴ N C⁻¹A 3–4–5 triangle, so the resultant sits at tan⁻¹(4/3) = 53° to the smaller field. In part (b), always say which way the resultant points — a bare magnitude will not get the mark.
💡 Top tips
Define E with all three phrases: force per unit charge, on a small, positivetest charge.
N C−1 = V m−1. Both are correct; use whichever the question uses.
E = V/d is only for parallel plates. E = kQ/r2 is only for radial fields.
One plate earthed? It is at 0 V, so V is the potential of the other plate.
Inside a charged sphere, E = 0. Outside, all the charge acts from the centre.
In E = kQ/r2, Q is the charge making the field, not the one feeling it.
Adding fields? Get each direction first by asking what a positive test charge would do.
⚠ Common mistakes
Using E = V/d for a point charge. That field is radial, not uniform
Forgetting E = 0 inside a charged sphere, and grinding out a number anyway
Putting the test charge into E = kQ/r2 instead of the source charge
Squaring r in E = V/d, or forgetting to square it in E = kQ/r2
Saying the field around a negative charge is “negative”. The field points inwards — direction, not sign
Adding field magnitudes when the two fields are not along the same line
Leaving d in mm, or a radius as a diameter
Thinking a bigger test charge means a bigger field. The test charge cancels
Quick recap: Electric field strength is the force per unit charge on a small positive test charge: E = F/q, in N C−1 (= V m−1). It is a vector, pointing away from positive charges and towards negative ones. Around a point charge or a sphere, E = kQ/r2 — and E = 0 inside a charged sphere. Between parallel plates the field is uniform, E = V/d, running from the positive to the negative plate. Fields from several charges add as vectors.
We have been drawing arrows all page — radiating out of a positive charge, marching between two plates. Those arrows are not just decoration. They are field lines, and they carry information we have not yet cashed in: their direction tells you where a positive charge would go, and their spacing tells you how strong the field is. Close together means strong. Spread apart means weak. There are rules for drawing them, and marks for drawing them properly. Next page: Electric Field Lines.
Radial and uniform fields blurring together?
Book a free meeting and we’ll separate E = kQ/r2 from E = V/d, and drill the vector addition until it’s automatic.
