IB Physics HLTopic 4 — Force FieldsPaper 1 & 2E = −ΔV/Δr~17 min read
Electric Potential Gradient
Last page ended on a promise. The gradient of an energy graph gives a force — so divide both sides by the charge, and the gradient of the potential must give the field strength. It does. And it arrives with a minus sign that is not a nuisance but a statement: the electric field always points downhill, from high potential to low. Get that one idea and the whole topic finally locks together.
📘 What you need to know
Moving a charge through an electric field does work: W = qΔV
Potential difference is ΔV = Vf − Vi — final minus initial
Due to a point charge: ΔV = kQ(1/rf − 1/ri)
In W = qΔV, the q is the charge being moved; the Q is the charge producing the potential. Never swap them
The potential gradient is the rate of change of electric potential with displacement in the direction of the field
E = −ΔV / Δr — the field strength is the negative gradient of the V–r graph
The minus sign means E points opposite to the direction of increasing potential — the field points downhill
The area under an E–r graph gives the potential difference ΔV
The V–r curve is shallower than the E–r curve, because V ∝ 1/r while E ∝ 1/r2
Work done on a charge
Potential was defined as the work done per unit charge. So if you actually move a charge q between two points whose potentials differ by ΔV, the work involved is simply that potential difference multiplied back up:
Work done moving a chargeW = q ΔVW = work done (J) • q = the charge being moved (C) • ΔV = potential difference (V)
Potential differenceΔV = Vf − Viand for a point charge ΔV = kQ ( 1/rf − 1/ri )final minus initial • the same convention as ΔEp
ΔVpositive → W is positive → you did work on the charge, pushing it uphill
ΔVnegative → W is negative → the field did the work, and energy was released
There are two different charges in these questions and examiners know it. The little q is the charge you are carrying around. The big Q is the charge sitting there making the field. A question will hand you both, precisely to see whether you can tell them apart. Circle them on the paper before you start.
Notice too that W = qΔV is just last page’s ΔEp = qΔV wearing a different hat. Work done on the charge is the change in its potential energy. Same equation, same sign convention, nothing new to learn.
The gradient of the potential
Potential gradient — definitionthe rate of change of electric potential with respect to displacement in the direction of the field
Draw V against r for a positive charge and you get a falling curve. Take the gradient at any point on it — the steepness of the tangent — and you have the field strength there.
Field strength from the potential gradientE = − ΔV / ΔrE in V m−1 • Δr = displacement in the direction of the field
The tangent slopes down, so ΔV/Δr is negative. The minus sign in front flips it, and E comes out positive — pointing outwards, exactly as it should for a positive charge.
Why the minus sign, really? Because a positive charge released in a field rolls downhill. It accelerates towards lower potential, and the field arrow shows it the way. So E must point opposite to the direction in which V increases. Every hill you have ever walked down obeys the same rule: the force is minus the gradient of the potential energy. This is that, divided by charge.
The uniform-field check
Try it on the parallel plates. The potential falls steadily from V at the positive plate to 0 at the earthed one, over a separation d. So the gradient is constant, equal to −V/d, and:
A familiar faceE = −ΔV/Δr = −(0 − V)/d = V/dthe parallel-plate formula, falling straight out of the general one
Going the other way: area under the E–r graph
Gradient takes you from V to E. To go back, you need the inverse operation — and the inverse of a gradient is an area.
Chase the shaded strip from r1 all the way out to infinity and its area becomes the potential at r1 itself — because V = 0 out there.
Which curve is steeper? Always the E–r one. It falls as 1/r2, while V–r only falls as 1/r. If an exam hands you two unlabelled curves and asks which is which, the steeper one is the field. Every time.
The whole topic on one page
Four quantities, and they are not four separate things to memorise. They are one thing, seen from four angles.
Learn this square and you have learned the topic. Area always moves you from the 1/r2 column to the 1/r column; gradient always moves you back.
V–r graph
take the gradient
E = −ΔV/Δr
take the area
back to ΔV
📐 Working a potential gradient question
Identify the two charges.q is the one being moved; Q is the one making the potential.
ΔV = final − initial. Always that order, so the sign means something.
Work done?W = qΔV. Positive → you did it. Negative → the field did it.
Given a V–r graph? Draw a tangent, find its gradient, then E = −gradient.
Given an E–r graph? The area under it between two points is ΔV.
Parallel plates? The gradient is constant, so E = V/d is just the special case.
WE 1
A charge of +6.0 nC is moved from a point at a potential of 250 V to a point at a potential of 90 V. (a) Calculate the potential difference. (b) Calculate the work done. (c) State whether the work was done on or by the field, and explain.
(a) Step 1 — final minus initialΔV = 90 − 250ΔV = −160 V(b) Step 2 — multiply by the charge being movedW = qΔV = (6.0 × 10⁻⁹)(−160)W = −9.6 × 10⁻⁷ J(c) Step 3 — read the sign
The charge moved to a region of lower potential.
A positive charge does that on its own — it rolls downhill.
the field did the work; 9.6 × 10⁻⁷ J was releasedIf nothing held the charge back, that released energy would appear as kinetic energy — the charge speeds up. A negative answer for W is never a mistake; it is information.
WE 2
A charge of +8.0 nC is fixed in place. Point A lies 20 cm from it and point B lies 50 cm from it. (a) Calculate the potential at A and at B. (b) Determine the work done in moving a charge of +2.0 nC from A to B. (c) State whether this work is done on or by the field.
(a) Step 1 — V = kQ/r, using the charge that MAKES the potentialVA = (8.99 × 10⁹)(8.0 × 10⁻⁹) / 0.20 = 71.92 / 0.20VA = 360 VVB = 71.92 / 0.50VB = 144 V(b) Step 2 — potential difference, then workΔV = VB − VA = 143.84 − 359.60 = −215.8 VW = qΔV = (2.0 × 10⁻⁹)(−215.8)W = −4.3 × 10⁻⁷ J(c) Step 3 — does that make sense?
A positive charge moving away from a positive charge is being repelled.
the field does the workTwo charges, two jobs. The 8.0 nC made the potential and went into V = kQ/r. The 2.0 nC was the one carried about, and went into W = qΔV. Swap them and every number is wrong.
WE 3
(a) Between two parallel plates 4.0 cm apart, the potential falls from 600 V to 0 V. Use the potential gradient to calculate the field strength. (b) A point charge of +5.0 nC produces a field. Determine the potential difference between points 10 cm and 25 cm from it, and state what this quantity represents on the E–r graph.
(a) Step 1 — the gradient is constant, so use it directlyΔV = 0 − 600 = −600 V | Δr = 0.040 mE = −ΔV/Δr = −(−600) / 0.040E = 1.5 × 10⁴ V m⁻¹Step 2 — sanity check with the plate formulaE = V/d = 600 / 0.040 = 1.5 × 10⁴ V m⁻¹
Same answer — because E = V/d is E = −ΔV/Δr for a uniform field.
(b) Step 3 — potentials firstV(0.10) = 44.95 / 0.10 = 449.5 VV(0.25) = 44.95 / 0.25 = 179.8 V|ΔV| = 449.5 − 179.8ΔV = 2.7 × 10² Vthis is the AREA under the E–r graph between r = 0.10 m and r = 0.25 mPart (a) is the reassuring one: the general law collapses to the plate formula the moment the field stops changing. If a derivation ever gives you something you already knew, you did it right.
💡 Top tips
q is the charge moved; Q is the charge that made the potential. Label them before you begin.
ΔV = final − initial. The sign is doing real work — don’t discard it.
Negative W means the field did the work and energy was released. It is not an error.
Quote the definition in full: rate of change of potential with displacement in the direction of the field.
The minus sign in E = −ΔV/Δr means the field points down the potential hill. Say so if asked.
Gradient of V–r → E. Area under E–r → ΔV. Never the other way round.
The E–r curve is always the steeper of the two. That’s how you identify it.
⚠ Common mistakes
Using Q (the source) in W = qΔV instead of q (the charge being moved)
Computing ΔV as initial − final, then reporting the wrong sign for the work
Dropping the minus sign in E = −ΔV/Δr, or treating it as a typo
Taking the gradient of the E–r graph, or the area under V–r. Both are meaningless
Treating a negative W as an arithmetic slip. It means the field did the work
Saying the V–r curve is steeper. It is shallower — 1/r against 1/r2
Using E = V/d for a radial field. It only works where the gradient is constant
Quick recap: Moving a charge through a field does work: W = qΔV, with ΔV = Vf − Vi — and remember q is the charge moved, Q the charge that made the potential. The potential gradient is the rate of change of V with displacement along the field, and E = −ΔV/Δr: the field points downhill, towards lower potential. The gradient of the V–r graph gives E; the area under the E–r graph gives ΔV. And E = V/d is nothing more than this rule applied to a uniform field.
One consequence has been sitting quietly in the corner. If E = −ΔV/Δr, then wherever you move without changing the potential, ΔV = 0 and no work is done. Join up all the points at the same potential and you get a surface — a contour line on the electric landscape — along which a charge can be slid for free. They are always perpendicular to the field lines, and they turn every field diagram into a map. Next page: Electric Equipotential Surfaces.
That minus sign still bothering you?
Book a free meeting and we’ll settle E = −ΔV/Δr, the q-versus-Q trap and the gradient-or-area question for good.
