IB Physics HLTopic 4 — Force FieldsPaper 1 & 2Ep = kq₁q₂/r~16 min read
Electric Potential Energy
Potential belonged to a point in space. Put an actual charge at that point and it now has energy — energy you could get back by letting go. Two positive charges held close together are a loaded spring. Two opposite charges are the reverse: they have already fallen together, and prising them apart costs you. That is the whole difference between a positive and a negative potential energy, and it is why electric potential energy, unlike gravitational, comes in both flavours.
📘 What you need to know
Electric potential energy of a system is the work done in bringing all the charges to their positions from infinity
Ep = kq1q2 / r — note 1/r, not 1/r2
Ep = 0 at infinity, just like potential
Like charges (repulsive force) give a positiveEp. Energy is released as they separate
Opposite charges (attractive force) give a negativeEp. Energy must be supplied to separate them
Gravitational potential energy is always negative. Electric potential energy can be either sign
Also Ep = qV: the charge multiplied by the potential it sits in
The area under a force–distance graph gives Ep (and the work done)
The steepness of the Ep–r graph gives the size of the force
ΔEp = Ep,final − Ep,initial. Positive means you did work; negative means the field did
What the energy actually is
Electric potential energy — definitionthe work done in bringing all the charges in a system to their positions from infinity
Start with the charges infinitely far apart, where they ignore each other completely and Ep = 0. Now walk them in to where you want them, and keep a careful account of the work.
Potential energy of two point chargesEp = kq1q2 / rq1, q2 = the two charges (C, with their signs) • r = separation of centres (m)
Look at the bottom. It is r, not r2. Do not square the distance. That is the single most punished slip in this section, and it happens because Coulomb’s law — which looks almost identical — does have the square. Energy goes as 1/r. Force goes as 1/r2. Say it twice.
There is a neat shortcut worth noticing. Potential was V = kQ/r, the energy per unit charge. So the energy of a charge q sitting in that potential is just:
Energy from potentialEp = qVcharge × the potential it sits in • the same thing, said two ways
Why the sign matters so much
Feed the charges in with their signs and the equation tells you the story.
Both curves obey 1/r. The only difference is which side of zero they sit on — and that is decided entirely by the product q1q2.
The two charges
The force
Sign of Ep
To separate them…
Like (same sign)
Repulsive
Positive
Energy is released. The field does the work
Opposite
Attractive
Negative
Energy must be supplied. You do the work
Now compare with gravity. Mass comes in one flavour, so the gravitational force is always attractive, so gravitational potential energy is always negative. Electric charge comes in two flavours — and the moment you allow repulsion, you allow a positive potential energy. Everything unusual about this page traces back to that one fact.
Moving a charge: which way does the energy go?
“Rising towards zero” still counts as an increase. −3.6 µJ is smaller than −0.9 µJ, and moving from one to the other means the energy has gone up.
Changes in energy
Almost every exam question asks not for Ep but for the change in it. There is a compact formula floating around, but it hides a sign trap. Do this instead, every time:
The safe wayΔEp = Ep,final − Ep,initialwork out kq1q2/r at each separation, then subtract. Signs included.
ΔEp is positive → the energy went up → work was done on the system, by you
ΔEp is negative → the energy went down → work was done by the field, and energy was released
You will meet the compact version kq1q2(1/r1 − 1/r2) in various books, and it is easy to get the wrong way round — with r1 as the starting separation it actually gives you minus the change in Ep (which is the work done by the field). If you insist on using it, test it first on a case you already know: two opposite charges moved apart must give an increase in Ep. Or just do it the safe way and never think about it again.
Reading it off a graph
Two graphs, two quantities, and it is easy to muddle which gives which.
Chasing that area out to infinity is awkward, which is exactly why Ep = kq1q2/r exists. Use the equation; understand the area.
And the reverse trip works too. On the Ep–r graph, the steepness of the curve at any separation gives the size of the force there. Strictly the force is the negative of that gradient, which is just physics saying that a charge always accelerates downhill in energy.
Force F = kq1q2/r2
area under F–r
Energy Ep = kq1q2/r
steepness of Ep–r
back to the force
🔌 Working an energy question
Divide by r, not r2. Write the equation out before you touch the calculator.
Put both charges in with their signs. The sign of Ep is the sign of the product.
Change in energy? Work out Ep at the start and at the end, then subtract: final − initial.
Positive ΔEp → you did work on the system. Negative → the field did work, energy released.
A charge fired at a nucleus? At the closest approach it has stopped, so all its kinetic energy has become Ep.
Given a potential instead?Ep = qV. Given a graph? Area under F–r.
WE 1
Two point charges of +3.0 nC and +5.0 nC are held 4.0 cm apart in a vacuum. (a) Calculate the electric potential energy of the system. (b) State whether work must be done on or by the field to bring the charges closer together. (c) State the potential energy if the 5.0 nC charge were negative.
(a) Step 1 — straight into the equation, dividing by rEp = kq₁q₂/r = (8.99 × 10⁹)(3.0 × 10⁻⁹)(5.0 × 10⁻⁹) / 0.040Ep = (1.3485 × 10⁻⁷) / 0.040Ep = +3.4 × 10⁻⁶ J = 3.4 μJ(b) Step 2 — they are like charges, so they repel
Pushing them together is pushing against the repulsion.
work must be done ON the field, by you(c) Step 3 — flip one sign, flip the answerEp = −3.4 × 10⁻⁶ JA lovely check: on the Coulomb’s law page these same magnitudes at the same separation gave F = 8.4 × 10⁻⁵ N. And Ep = F × r = (8.4 × 10⁻⁵)(0.040) = 3.4 × 10⁻⁶ J. It works because one has an r² and the other an r.
WE 2
A proton is fired directly at a stationary nucleus of charge +50e. It momentarily comes to rest at a distance of 6.0 × 10−14 m from the nucleus. (a) Calculate the electric potential energy at that instant. (b) Determine the initial kinetic energy of the proton. (c) Calculate its initial speed. (e = 1.60 × 10−19 C, mp = 1.673 × 10−27 kg)
(a) Step 1 — the two chargesq₁ = e = 1.60 × 10⁻¹⁹ C | q₂ = 50e = 8.0 × 10⁻¹⁸ CEp = (8.99 × 10⁹)(1.60 × 10⁻¹⁹)(8.0 × 10⁻¹⁸) / (6.0 × 10⁻¹⁴)Ep = 1.9 × 10⁻¹³ J(b) Step 2 — at rest, so all the kinetic energy has goneEk,initial = Ep at closest approachEk = 1.9 × 10⁻¹³ J(c) Step 3 — back out the speed½mv² = 1.9179 × 10⁻¹³ → v = √(2Ek/m)v = √(2 × 1.9179 × 10⁻¹³ / 1.673 × 10⁻²⁷)v = 1.5 × 10⁷ m s⁻¹That is about 5% of the speed of light, so we are just about safe ignoring relativity. Note the whole question hinges on one sentence: “momentarily comes to rest” means Ek → 0, so all of it has turned into Ep.
WE 3
A charge of −2.0 nC is moved from 2.0 cm to 8.0 cm away from a fixed charge of +4.0 nC. (a) Calculate the change in electric potential energy. (b) State whether work was done on or by the field, and explain.
(a) Step 1 — find Ep at the startEp,i = (8.99 × 10⁹)(4.0 × 10⁻⁹)(−2.0 × 10⁻⁹) / 0.020Ep,i = −3.60 × 10⁻⁶ JStep 2 — and at the endEp,f = −7.192 × 10⁻⁸ / 0.080 = −0.90 × 10⁻⁶ JStep 3 — subtract, final minus initialΔEp = (−0.90) − (−3.60) = +2.70 (× 10⁻⁶)ΔEp = +2.7 × 10⁻⁶ J(b) Step 4 — read the sign
The charges are opposite, so they attract.
Separating them means pulling against that attraction.
ΔEp is positive, so the energy went up.
work was done ON the field, by an external forceBoth energies are negative, and yet the change is positive. −0.90 μJ really is larger than −3.60 μJ. If you subtracted the other way and got −2.7 μJ, you have found the work done by the field — a perfectly good quantity, but not what was asked.
💡 Top tips
Do not square the distance.Ep goes as 1/r; only the force goes as 1/r2.
Put both charges in with their signs. Like charges → positive Ep; opposite → negative.
For a change, compute both energies and subtract final − initial. Never guess the formula’s order.
“Momentarily at rest” means all the kinetic energy has become potential energy.
Ep = qV is often the fastest route if you already know the potential.
Area under F–r gives energy. Steepness of Ep–r gives force. Don’t swap them.
Gravitational Ep is always negative; electric Ep takes the sign of q1q2.
⚠ Common mistakes
Writing Ep = kq1q2/r2. That is the force multiplied by r too many times
Dropping the signs, then claiming Ep is always positive (or always negative)
Thinking −0.9 µJ is less than −3.6 µJ. It is greater — less negative
Subtracting initial − final and reporting it as ΔEp
Assuming electric Ep must be negative “because gravitational potential energy is”. Charge has two signs
Using the gradient of the F–r graph, or the area under the Ep–r graph. Both back to front
Forgetting that r runs from the centres of the charges
Quick recap: Electric potential energy is the work done bringing the charges in from infinity: Ep = kq1q2/r — a 1/r law, and zero at infinity. Like charges give a positiveEp (energy released as they separate); opposite charges give a negative one (energy must be supplied). Gravitational Ep is always negative, because gravity only attracts. You can also write Ep = qV. The area under an F–r graph gives the energy; the steepness of the Ep–r graph gives the force. For changes, always take final minus initial.
That last sentence hides the next page. We have just said the gradient of an energy graph gives a force. Divide both sides by the charge and something rather beautiful drops out: the gradient of the potential gives the field strength. It even comes with a minus sign, for exactly the reason you would guess — the field points downhill, from high potential to low. Next page: Electric Potential Gradient.
Signs of Ep tripping you up?
Book a free meeting and we’ll drill the 1/r versus 1/r2, the “final minus initial” rule, and the on-or-by-the-field question.
