IB Physics SLTopic 4 — Electric & Magnetic FieldsPaper 1 & 2E = F/q~9 min read
Electric Field Strength
A charge can feel a push or pull with nothing touching it — that’s the electric field at work. Field strength turns “there’s a force somewhere around here” into a precise number: how many newtons each coulomb feels at a point. Get E = F ÷ q and you can handle point charges, spheres and parallel plates alike.
📘 What you need to know
Electric field strength is the force per unit charge on a small positive test charge: E = F ÷ q, in N C⁻¹
Because we use a positive test charge, E points away from positive charges and towards negative ones — E is a vector
Around a point charge or charged sphere, E = kq ÷ r² — inverse-square; the sphere acts as a point charge at its centre
Inside a charged conducting sphere E = 0; outside it falls off as 1 ÷ r²
Fields add as vectors: along a line, add (same way) or subtract (opposite); at right angles, use Pythagoras
Between two parallel plates the field is uniform: E = V ÷ d, pointing from the + plate to the − plate
The units N C⁻¹ and V m⁻¹ are the same thing
What Electric Field Strength Means
An electric field is any region where a charge feels a force. To measure how strong the field is at a point, we drop in a tiny positive test charge and ask how much force each coulomb feels:
Electric field strength (definition)E = F ÷ q
Where E is the field strength (N C⁻¹), F is the force on the charge (N), and q is the charge (C). Because the test charge is positive, this also fixes a direction convention: the field points the way a positive charge would be pushed. That means field lines run away from positive charges and towards negative charges. Field strength is a vector — it has both size and direction.
positive test charge +q
feels force Eq along the field
away from + / towards −
Field Around a Point Charge
Combine the definition with Coulomb’s law and the field of a single point charge drops straight out. The force on a test charge q at distance r is kQq ÷ r², so dividing by q gives:
Field due to a point chargeE = kq ÷ r²
So the field around a point charge is another inverse-square quantity: it doesn’t depend on the test charge at all, only on the source charge q and the distance r from it. A charged sphere behaves exactly like a point charge sitting at its centre — but only outside the sphere. Inside a charged conductor the field is zero, because the forces on a test charge there cancel out.
Field strength for a charged sphere: zero everywhere inside, a maximum right at the surface (radius R), then the familiar inverse-square fall-off outside — where the sphere acts just like a point charge at its centre.
Adding Fields Together
When more than one charge is around, each creates its own field and the total field at a point is the vector sum of them. How you combine them depends on their directions:
fields point the same way
add
E1 + E2
fields point opposite ways
subtract
E1 − E2
And when the two fields meet at right angles, you combine them with Pythagoras:
Two fields meeting at a right angle at point X combine like the sides of a right triangle: the resultant is the diagonal, found with Pythagoras. Here E1 = 3 and E2 = 4 give a resultant of 5 (same units).
The Uniform Field Between Parallel Plates
Charge up two parallel plates with a p.d. and you get something different from a point charge: a uniform field — the same strength and direction everywhere between the plates. Its size is simply the voltage shared over the gap:
Uniform field between parallel platesE = V ÷ d
Where V is the p.d. between the plates (V) and d is their separation (m). So a bigger voltage or a smaller gap makes a stronger field. The field points from the positive plate to the negative plate. Handily, the units here (V m⁻¹) are exactly the same as N C⁻¹ from the definition.
Between parallel plates the field is uniform — equally spaced lines of equal length, running from the + plate to the − plate. Its strength is E = V ÷ d. (This can’t be used for a point charge, whose field is radial.)
One handy shortcut: if one plate is earthed, it sits at 0 V, so the full p.d. is just the voltage of the other plate. And when a point charge moves between the plates, you can equate the two field expressions, E = F ÷ q = V ÷ d, to get the force straight away.
🧭 Which E equation do I need?
Given a force on a known charge? Use E = F ÷ q (or F = Eq to go back)
A point charge or sphere, at distance r? Use E = kq ÷ r², with r from the centre
Between parallel plates? Use E = V ÷ d
More than one charge? Find each field, then add as vectors (Pythagoras at right angles)
Check the units — N C⁻¹ and V m⁻¹ are the same
Quick recap:E = F ÷ q (force per coulomb, a vector). For a point charge E = kq ÷ r² (zero inside a sphere); between plates E = V ÷ d. Combine several fields as vectors, and remember N C⁻¹ = V m⁻¹.
WE 1
A small sphere carries a charge of +8.0 nC. (a) Calculate the electric field strength at a point 0.12 m from its centre. (b) A −3.0 nC charge is placed at that point. Find the size and direction of the force on it.
Part (a) — point charge: E = kq ÷ r²E = (8.99 × 10⁹ × 8.0 × 10⁻⁹) ÷ (0.12)²E = 71.9 ÷ 0.0144E ≈ 5.0 × 10³ N C⁻¹directed away from the sphere (it’s positive)Part (b) — force from the field: F = EqF = (5.0 × 10³) × (3.0 × 10⁻⁹)F ≈ 1.5 × 10⁻⁵ Ndirected towards the sphere — the −3.0 nC charge is attracted to the positive sphere.
WE 2
Two parallel plates 4.0 cm apart have a potential difference of 6.0 kV across them. (a) Calculate the electric field strength between the plates. (b) Calculate the force on a +2.5 nC charge placed between them.
Part (a) — parallel plates: E = V ÷ d (convert kV → V, cm → m)E = 6000 ÷ 0.040E = 1.5 × 10⁵ V m⁻¹Part (b) — force on the charge: F = EqF = (1.5 × 10⁵) × (2.5 × 10⁻⁹)F ≈ 3.8 × 10⁻⁴ Ndirected towards the negative plate (the charge is positive, so it follows the field).
💡 Top tips
Two units, one quantity: N C⁻¹ = V m⁻¹. Don’t be thrown when a plate question quotes the field in V m⁻¹
Match the equation to the geometry:E = V ÷ d is only for parallel plates; E = kq ÷ r² is only for point charges / spheres
E is a vector. Combine two fields with directions — Pythagoras at right angles — not by just adding the numbers
Sphere = point charge at its centre for the field outside; inside a conductor, E = 0
⚠ Common mistakes
Using E = V ÷ d for a point charge, or E = kq ÷ r² between plates — wrong geometry
Forgetting E is a vector and adding two fields as plain numbers when they’re at an angle
Forgetting to square r, or measuring it from the surface instead of the centre
Not converting kV → V, cm → m, nC → C before substituting
You can now put a number on the field at any point. Up next: Electric Field Lines — the pictures behind these numbers: radial fields around point charges, the uniform field between plates, how line spacing shows strength, and why the lines always meet a conductor at right angles.
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