IB Physics HLTopic 4 — Force FieldsPaper 1 & 2V = kQ/r~16 min read
Electric Potential
Field lines told you which way a charge would be pushed. But shoving a positive charge against those arrows takes work — and the harder you shove, the more energy you bank. Divide that energy by the charge you shoved and you get a number that belongs to the point in space, not to the charge. It is called potential. Best of all, it is a scalar. No arrows. No triangles. Just add the numbers up, signs and all.
📘 What you need to know
Electric potential at a point is the work done per unit charge in bringing a small positive test charge from infinity to that point
Measured in J C−1, which is the volt (V)
It is a scalar — but it carries a sign, and the sign matches the sign of the charge making it
Potential is zero at infinity
Ve = kQ / r — note 1/r, not 1/r2
Around a positive charge V is positive and increases as you get closer
Around a negative charge V is negative and decreases as you get closer
Potential always decreases in the direction of the field lines
Unlike gravitational potential (always negative), electric potential takes the sign of the charge
Inside a charged conducting sphere V is constant and equal to its surface value — it is not zero, even though E is
Potentials from several charges add as scalars: V = kQ1/r1 + kQ2/r2
The definition, unpacked
Electric potential — definitionthe work done per unit charge in taking a small positive test charge from infinity to that point
Every phrase is load-bearing, and the marks are hidden in them:
Work done per unit charge. That is why the unit is the joule per coulomb, the volt
Positive test charge. The same convention as field strength, and it is what fixes the sign
From infinity. Infinitely far away, the charge feels nothing, so we define V = 0 there and measure everything relative to it
Zero at infinity is a choice, not a discovery. We could have put the zero anywhere — but infinity is the one place every charge in the universe agrees is uninteresting, so it makes the bookkeeping clean. It is exactly the same convention you met for gravitational potential.
The equation, and the sign
Potential due to a point chargeVe = kQ / rQ = the charge making the potential (C, with its sign) • r = distance from its centre (m)
Look hard at the bottom of that fraction. It is r, not r2. Potential falls off much more slowly than field strength. Double the distance and the field drops to a quarter, but the potential only halves.
Gravitational potential is always negative, because mass comes in one flavour. Electric potential is negative around a negative charge and positive around a positive one.
Moving a test charge around
Think about what your hand has to do, and the signs stop being arbitrary.
Near a positive charge, a positive test charge is repelled. To push it closer you must do work on it, so its stored energy — and the potential — increases. V is positive, and it grows as r shrinks
Near a negative charge, a positive test charge is attracted. It falls in on its own, and energy is released. To drag it back out to infinity you must do work. V is negative, and it grows (towards zero) as r increases
Here is the one-line check that never fails: potential always decreases in the direction of the field lines. Field lines point away from a positive charge, and sure enough V falls as you move away. Field lines point into a negative charge, and sure enough V falls as you move towards it — from a small negative number to a big one. Draw the arrows, and the signs follow.
Charged spheres: the trap
The equation V = kQ/r works for a conducting sphere too, treating all the charge as if it sat at the centre. But inside the sphere something happens that catches almost everyone.
Inside, the field is zero, so no work is done moving a charge about — and if no work is done, the potential cannot change. It stays pinned at its surface value all the way to the centre.
Put these two graphs side by side in your head. E inside a sphere is zero. V inside a sphere is not. They are not the same statement, and a question will test whether you know it. The logic runs one way only: no field means no force, so no work, so no change in potential. Constant is not the same as zero.
Adding potentials
Here is where potential repays all the effort. Field strength is a vector: two charges mean two arrows, and you are into Pythagoras. Potential is a scalar. Two charges mean two numbers, and you simply add them — taking care with the signs.
Combining potentialsV = kQ1/r1 + kQ2/r2 + …signed scalar addition • no directions, no components, no triangles
In the right-hand panel the two potentials cancel exactly. The field at X does not — both charges push a positive test charge the same way, to the right.
Two situations, mirror images of each other, and both are examiner favourites. Midway between two like charges: the fields cancel, so E = 0, but V is not. On the bisector of a dipole: the potentials cancel, so V = 0, but E is not. If you can say which is which, you have understood the difference between a vector and a scalar better than most.
Work done from infinity
divide by the charge
V = W/q
for a point charge
V = kQ/r
Field strength E
Potential V
Type of quantity
Vector
Scalar
Point charge
E = kQ/r2
V = kQ/r
Double the distance
Falls to a quarter
Falls to a half
Combining
Vector addition (Pythagoras)
Signed scalar addition
Inside a charged sphere
Zero
Constant, at the surface value
Unit
N C−1 or V m−1
J C−1 or V
🔋 Working a potential question
Put Q in with its sign. A negative charge gives a negative potential. Always.
Divide by r, not r2. This is the single commonest error in the topic.
Several charges? Work out each kQ/r separately, then add them like numbers.
Distances run from the centre of each charge. Add the radius if you are given a distance from a surface.
Inside a conducting sphere?V is constant at its surface value kQ/R. Not zero.
Work done bringing a charge q from infinity to a point of potential V is simply W = qV.
WE 1
A point charge of +5.0 nC sits in a vacuum. Calculate the electric potential (a) at a distance of 20 cm from it, and (b) at 40 cm from it. (c) State the potential at 20 cm if the charge were −5.0 nC instead. (k = 8.99 × 109 N m2 C−2)
(a) Step 1 — straight into V = kQ/rV = (8.99 × 10⁹)(5.0 × 10⁻⁹) / 0.20 = 44.95 / 0.20V = 2.2 × 10² V(b) Step 2 — double the distanceV = 44.95 / 0.40V = 1.1 × 10² V(c) Step 3 — the sign of Q carries straight throughV = −2.2 × 10² VDoubling r halved the potential, because V ∝ 1/r. Had this been field strength, it would have dropped to a quarter. That one difference is worth more marks than any calculation on this page.
WE 2
An isolated metal sphere of radius 20 cm is charged until its surface is at a potential of 150 kV. Determine (a) the charge on the sphere, (b) the potential at a point 30 cm from its surface, and (c) the potential at the centre of the sphere.
(a) Step 1 — the sphere acts like a point charge at its centreV = kQ/R → Q = RV / kQ = (0.20 × 150 × 10³) / (8.99 × 10⁹)Q = 3.3 × 10⁻⁶ C = 3.3 μC(b) Step 2 — careful: r is from the CENTREr = 0.20 + 0.30 = 0.50 mV = kQ / r = 30 000 / 0.50V = 60 kV(c) Step 3 — inside the sphere
V is constant inside, equal to its surface value.
V = 150 kVTwo traps, both sprung. In (b), “30 cm from the surface” means 0.50 m from the centre — and a quick check confirms it: V should fall by the factor R/r = 0.20/0.50 = 0.40, and 150 × 0.40 = 60 kV exactly. In (c), the field inside is zero but the potential most certainly is not.
WE 3
A point P lies 3.0 cm from a charge of +4.0 nC and 5.0 cm from a charge of −6.0 nC. (a) Calculate the electric potential at P. (b) Calculate the work done in bringing a charge of +2.0 nC from infinity to P. (c) Explain why the potentials were added as scalars, and not as vectors.
(a) Step 1 — one potential at a time, signs includedV₁ = (8.99 × 10⁹)(+4.0 × 10⁻⁹) / 0.030 = +1199 VV₂ = (8.99 × 10⁹)(−6.0 × 10⁻⁹) / 0.050 = −1079 VStep 2 — add them like numbersV = 1199 − 1079V = +1.2 × 10² V(b) Step 3 — potential is work per unit chargeW = qV = (2.0 × 10⁻⁹) × 120W = 2.4 × 10⁻⁷ J(c) Step 4 — why scalars?
Potential is work done per unit charge, and work is energy.
Energy is a scalar, so it has no direction to resolve.
just add the values, keeping their signsNotice the directions from P to each charge never entered the calculation — only the two distances. Had this asked for the field at P, you would have needed the angle between them, and Pythagoras. Potential is the easy one. Use it when you can.
💡 Top tips
Define it with all three phrases: work done per unit charge, from infinity, on a small positive test charge.
V = kQ/r, not r2. Double the distance and Vhalves.
Put Q in with its sign. Electric potential can be positive or negative — gravitational potential never can.
Potential decreases in the direction of the field lines. Use it to check every sign you write.
Inside a charged sphere V is constant, not zero. It is E that is zero.
Adding potentials is scalar addition. Signs, yes. Triangles, no.
Reach for potential rather than field strength whenever a question lets you. It is far less work.
⚠ Common mistakes
Using r2 in V = kQ/r. That is the field, not the potential
Saying V = 0 inside a charged sphere. V is constant; E is zero
Dropping the sign of Q, then wondering why the answer is positive
Assuming electric potential is always negative, “like gravitational potential”. It follows the sign of the charge
Combining potentials with Pythagoras. They are scalars — just add them
Measuring r from a sphere’s surface instead of its centre
Assuming that if E = 0 at a point then V = 0 there, or the other way round. Neither follows
Quick recap: Electric potential is the work done per unit charge in bringing a small positive test charge from infinity to a point, measured in volts. It is a scalar whose sign matches the charge: V = kQ/r, an inverse relationship, not inverse-square. Potential always falls in the direction of the field lines. Inside a charged conducting sphere V is constant at kQ/R — not zero, even though E is. Several charges? Add their potentials as signed scalars. And the work done bringing a charge q in from infinity is W = qV.
We have been talking about the energy of a test charge without quite saying so. Multiply the potential at a point by the charge you actually put there, and you get the energy stored in that arrangement — the energy you would get back if you let go. For two charges it turns out to be kq1q2/r, and unlike gravitational potential energy it can be positive. Next page: Electric Potential Energy.
Mixing up potential and field strength?
Book a free meeting and we’ll separate the 1/r from the 1/r2, and settle the “inside a sphere” question for good.
