Electric potential and potential energy — Physics

Before this

This leads to

Ohm's law and resistance
Motion of charges in electric fields · coming soon
Gravitational potential · coming soon

A voltage at a single point?

A voltmeter always has two leads. It measures a potential difference: the energy transferred to each coulomb of charge that moves from one point to the other. One volt is one joule per coulomb. So it seems that "voltage" only makes sense between two points.

Yet physicists talk about "the potential at this point" near a charged sphere, as if a single point had its own number of volts. How can that be?

The last lecture left two promises: to build on the fact that a charge crossing a p.d. V gains energy qV, whatever the gap, and to define the potential at every point, even near a single charge, where $E = V / d$ fails. Both are kept here. The key is an idea you already use for gravity: height.

Height needs a zero, and so does potential

A stone on a shelf has gravitational potential energy. But "how much?" has no answer until you choose a zero: the floor, the ground, or sea level. Only differences in height are physical. A climber gains the same energy going from 2000 m to 2100 m whether you measure from sea level or from the valley floor.

Electric potential works the same way. The physical thing is the potential difference between two points. To give each single point its own value, we choose a zero. For a charge on its own, the natural zero is infinity: a point so far away that the charge's force there is too small to matter.

Electric potential

The electric potential $V$ at a point is the work done per unit positive charge in bringing a small positive test charge from infinity to that point.

$V = \frac{W}{q}$ , measured in joules per coulomb, which is volts (V).

Three details in that definition matter. The work is done by you, the outside agent, against the field. You move the charge slowly, so it gains no kinetic energy on the way. And the test charge is small and positive, exactly as in the definition of field strength, so that it does not disturb the field and so that the sign is fixed.

A potential difference is now just a difference of two potentials. If point A is at 300 V and point B is at 100 V, the p.d. between them is 200 V, and moving 1 C from B to A takes 200 J.

Why the potential of a point charge is kQ/r

Now find the potential at a distance r from a point charge Q. This is the hard part of the lecture. Spend time here.

The obvious method fails. Work = force × distance needs a constant force. But the force on the test charge, $F = k Qq / r^{2}$ , grows as you come closer. Between parallel plates, the last lecture could use $W = qE d$ , because the field was uniform. Here it is not.

There is a rule that works for any force that changes: the work done is the area under the force–distance graph. You met it for a stretched spring. So the work to bring q in from infinity to r is the area under the curve $F = k Qq / r^{2}$ , from r out to the far right.

The force on a small test charge falls as 1/r². The work done in bringing it in from far away to r is the shaded area under the curve, cut here into strips.

That area stretches out forever, yet it is finite, because the curve falls so fast. Calculus finds it in one line at Pre-University level. But you can find it now, with a clever choice and some cancelling. Step through it below.

Step through — the work to bring a charge in from infinity

Press Next to add one step. Read the reason before you go on.

Each step gives one line of the argument and the reason it is allowed.

Cut the path from far away to r into thin steps. The force changes along the path, so force × distance cannot be used in one go, but over a thin step the force hardly changes.
For a step from a to b, use the force at the distance √(ab): F = kQq/(ab). Any distance inside a thin step gives almost the same force; this one is chosen because it makes the sum collapse.
Work for that step = kQq/(ab) × (b − a) = kQq(1/a − 1/b), because (b − a)/(ab) = 1/a − 1/b.
Add the steps: kQq(1/r − 1/r₂) + kQq(1/r₂ − 1/r₃) + kQq(1/r₃ − 1/r₄) + … Each step ends where the next one starts.
Everything in the middle cancels, leaving W = kQq(1/r − 1/(far away)). Far away, 1/distance is as close to 0 as you like, so W = kQq/r.
Divide by q: V = W/q = kQ/r. The test charge cancels, just as it did for E = kQ/r².

So the work done is $W = k Qq / r$ , and dividing by q gives the potential.

Potential near a point charge

At a distance $r$ from a point charge $Q$ , $V = \frac{k Q}{r}$ , with $k \approx 8.99 \times 1 0^{9} N m^{2} / C^{2}$ .

The same result holds outside a charged sphere, with $r$ measured from its centre.

Compare it with the field, $E = k Q / r^{2}$ . Double the distance and the potential halves, but the field falls to a quarter. The potential falls more slowly, because it adds up the work over the whole journey from infinity, not just the push at one place.

Top: potential falls as 1/r, to a half at 2r and a third at 3r. Bottom: field strength falls as 1/r², to a quarter at 2r and a ninth at 3r.

Positive and negative potentials

The formula keeps the sign of Q, and the sign has a meaning.

Near a positive charge, the field pushes a positive test charge away. You must push it in, so you do positive work, and V is positive. The closer you go, the more work, and the higher the potential.

Near a negative charge, the field pulls a positive test charge in. You must hold it back all the way, so the force you apply points outwards while the charge moves inwards. Your work is negative, and V is negative. A negative potential does not mean "less than nothing". It means the field itself did the work, and you would have to put energy in to take the charge back out to infinity.

Here is the picture to hold in your mind. A positive charge makes a hill of potential, and a negative charge makes a well. A positive test charge rolls down a hill, away from a positive charge, and falls into a well, towards a negative one.

Stop and think

The field gets stronger as you approach any charge. Does the potential always get bigger as you approach a charge?

Show the answer

No. Near a positive charge, V rises towards large positive values. Near a negative charge, V becomes more and more negative: it gets lower as you approach. The field strength is about how steep the hill or the well is, so it grows near both. The potential is about the height, so it goes up near one and down near the other.

Potential is a scalar: just add the numbers

Field strength is a vector. To combine the fields of two charges, you must add arrows, with their directions. Potential is energy per coulomb, and energy has no direction. So the potential of several charges is simply the sum of their separate potentials, each with its sign:

V = \frac{k Q _{1}}{r _{1}} + \frac{k Q _{2}}{r _{2}} + \frac{k Q _{3}}{r _{3}} + \dots

This is a large gift. Try it on two charges in the explorer below. Move the point P along the line, and then switch the second charge from negative to positive.

Explore — potential and field between two charges

A is +4.0 nC and B is 0.40 m away. Move the point P along the line between them. Then switch B between −4.0 nC and +4.0 nC, and compare V and E at the midpoint.

Charge B:

Distance of P from A: 0.20 m

This table gives the potential and field at points between A (+4.0 nC) and B, 0.40 m apart, with $k = 9.0 \times 1 0^{9}$ . Distances are from A. A field towards B is written "towards B".

B	P from A	V	E
−4.0 nC	0.10 m	240 V	4000 N/C towards B
−4.0 nC	0.20 m	0 V	1800 N/C towards B
+4.0 nC	0.10 m	480 V	3200 N/C towards B
+4.0 nC	0.20 m	360 V	0 N/C

At the midpoint, opposite charges give V = 0 with a strong field, and like charges give E = 0 with V = 360 V.

Look at the midpoint in each case. With opposite charges, the two potentials cancel, so V = 0, but the field is strong: both fields point towards the negative charge and add. With like charges, the fields cancel, so E = 0, but the potentials add to 360 V. A zero potential does not mean a zero field, and a zero field does not mean a zero potential. The two quantities answer different questions.

Potential energy and the work done between two points

If a charge q sits at a point where the potential is V, its electric potential energy is

U = q V

This is just the definition turned round: V is energy per coulomb, so q coulombs have q times as much. Put in $V = k q_{1} / r$ for the potential of a second charge, and you get the energy of a pair of charges a distance r apart:

U = \frac{k q _{1} q _{2}}{r}

It looks like Coulomb's law with r instead of r². The signs tell the story. For like charges, U is positive: you had to push them together, and they would fly apart if released. For unlike charges, U is negative: they attract, and you must supply energy to pull them apart.

Moving a charge q from one point to another changes its energy by

W = q Δ V

where $Δ V$ is the potential difference between the two points. The path does not matter, because the potential at each end is fixed. Go straight, go round in a loop, go far away and come back: the energy change is the same.

This is the idea the last lecture promised to build on. An electron accelerated through a p.d. of V volts gains energy $e V$ , whatever the shape of the field. That gives physicists a natural unit of energy. One electronvolt (eV) is the energy gained by one electron's charge crossing a p.d. of 1 V: $1 eV = 1.60 \times 1 0^{- 19}$ J.

Equipotentials, and the field as a slope

An equipotential is a surface on which every point has the same potential. Moving a charge along it takes no work, because $Δ V = 0$ .

Equipotentials always cross field lines at right angles. Here is the reason. Suppose the field had a part along an equipotential. Then moving a charge along that surface would mean the field pushing it along its path, which is work. But no work is done along an equipotential. So the field can have no part along it, and must be perpendicular to it.

Top: equipotentials (dashed) around a +10 nC charge, every 50 V, getting further apart outwards; field lines cross them at right angles. Bottom: between parallel plates the equipotentials are flat and equally spaced.

Around a point charge, the equipotentials are spheres, drawn as circles. In the figure they are drawn every 50 V, and they get further apart as you go out. That is the field getting weaker: you must travel further to drop the same number of volts. Between parallel plates, the equipotentials are flat, parallel and equally spaced, because the field is uniform.

This gives the exact meaning of the last lecture's $E = V / d$ . The field strength is how fast the potential falls with distance, the steepness of the hill, and the field points downhill, from high potential to low:

E = - \frac{Δ V}{Δ r}

The minus sign says that the field points in the direction in which V decreases. Over a short step, this works in any field. For plates, where V falls steadily across the gap, it becomes $E = V / d$ . That is why the last lecture's equation was a special case.

Worked examples

Worked Example 1 — Potential, and the energy of a charge placed there

A charge $Q = + 5.0$ nC is fixed in place. Find the potential 30 cm from it. Then find the potential energy of a charge $q = - 2.0$ nC placed at that point. Use $k = 9.0 \times 1 0^{9} N m^{2} / C^{2}$ .

Think first. Convert first: 5.0 nC = $5.0 \times 1 0^{- 9}$ C, and 30 cm = 0.30 m.

$V = \frac{k Q}{r} = \frac{9.0 \times 1 0 ^{9} \times 5.0 \times 1 0 ^{- 9}}{0.30} = \frac{45}{0.30} = 150$ V.

$U = q V = - 2.0 \times 1 0^{- 9} \times 150 = - 3.0 \times 1 0^{- 7}$ J.

Check. V is positive near a positive charge. U is negative because the charges are unlike and attract: energy would have to be supplied to pull q away to infinity. Using $k = 8.99 \times 1 0^{9}$ gives 149.8 V, which rounds to the same answer. ✓

Worked Example 2 — An electron through 500 V

An electron starts at rest and is accelerated through a p.d. of 500 V. Find the energy it gains in electronvolts and in joules, and its final speed. (Electron mass $9.11 \times 1 0^{- 31}$ kg.)

Think first. An electron is negative, so it moves towards higher potential. It gains energy $e V$ , where e is the size of its charge.

Energy gained = 500 eV $= 500 \times 1.60 \times 1 0^{- 19} = 8.0 \times 1 0^{- 17}$ J.

This becomes kinetic energy: $\frac{1}{2} m v^{2} = 8.0 \times 1 0^{- 17}$ , so $v = \frac{2 \times 8.0 \times 1 0 ^{- 17}}{9.11 \times 1 0 ^{- 31}} \approx 1.3 \times 1 0^{7}$ m/s.

Check. This is about 4% of the speed of light, so the ordinary formula for kinetic energy is still good enough. The answer does not depend on the shape of the field or the distance travelled, only on the p.d. ✓

Worked Example 3 — Adding potentials

A charge of +4.0 nC is at A and a charge of −2.0 nC is at B, 0.60 m away. Find the potential at the midpoint M, and the work needed to bring a +3.0 nC charge from infinity to M.

Think first. Potential is a scalar. Find each charge's potential at M, with its sign, and add. M is 0.30 m from each.

$V = \frac{9.0 \times 1 0 ^{9} \times 4.0 \times 1 0 ^{- 9}}{0.30} + \frac{9.0 \times 1 0 ^{9} \times ( - 2.0 \times 1 0 ^{- 9} )}{0.30} = 120 - 60 = 60$ V.

$W = q V = 3.0 \times 1 0^{- 9} \times 60 = 1.8 \times 1 0^{- 7}$ J.

Check. No directions were needed at any stage. The positive charge is bigger, so the total potential should be positive. ✓

Worked Example 4 — Moving between two points

A +8.0 nC charge is fixed. How much work is needed to move a +2.0 nC charge from 40 cm to 10 cm from it?

Think first. The force changes along the way, so do not use force × distance. Find the potential at each point and use $W = q Δ V$ .

$V_{10} = \frac{9.0 \times 1 0 ^{9} \times 8.0 \times 1 0 ^{- 9}}{0.10} = 720$ V and $V_{40} = \frac{72}{0.40} = 180$ V, so $Δ V = 540$ V.

$W = 2.0 \times 1 0^{- 9} \times 540 = 1.08 \times 1 0^{- 6}$ J $\approx 1.1 \times 1 0^{- 6}$ J.

Check. The work is positive, because two positive charges are being pushed together. Any path between the two points gives the same answer. ✓

Three traps

Trap 1 — Mixing 1/r and 1/r²

Potential falls as 1/r and field strength as 1/r². Doubling the distance halves V but quarters E. And $E = V / d$ is only for a uniform field: near a point charge, use $k Q / r$ and $k Q / r^{2}$ .

Trap 2 — Dropping the signs

In $V = k Q / r$ and $U = q V$ , the signs of the charges stay in. A negative charge gives a negative potential, and a negative charge at a positive potential has negative energy. This is the opposite of Coulomb's law, where the convention is to use sizes and find the direction separately.

Trap 3 — "Zero potential means zero field"

Midway between equal and opposite charges, V = 0 but the field is strong. Midway between equal like charges, E = 0 but V is not zero. V is a height and E is a slope. A height of zero can sit on a steep slope.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Use $k = 9.0 \times 1 0^{9} N m^{2} / C^{2}$ and $e = 1.60 \times 1 0^{- 19}$ C unless told otherwise.

Find the potential 20 cm from a −6.0 nC charge. How much work is done by an outside agent in bringing a +1.0 nC charge from infinity to that point? What does the sign mean?

Hint 1
Keep the sign of Q in $V = k Q / r$ , and convert 20 cm to metres.

Hint 2
The work by the outside agent is $W = q V$ .

Full answer
$V = 9.0 \times 1 0^{9} \times (- 6.0 \times 1 0^{- 9}) \div 0.20 = - 270$ V. $W = 1.0 \times 1 0^{- 9} \times (- 270) = - 2.7 \times 1 0^{- 7}$ J. The work is negative: the negative charge pulls the positive one in, so the outside agent only holds it back, and the field does the work. Check: a negative charge must give a negative potential. ✓
A proton (mass $1.67 \times 1 0^{- 27}$ kg) starts at rest and is accelerated through 2000 V. Find its energy gain in eV and in J, and its final speed.

Hint 1
A proton has charge +e, so it gains e × 2000 V.

Hint 2
Set the energy equal to $\frac{1}{2} m v^{2}$ and solve for v.

Full answer
Energy = 2000 eV = $2000 \times 1.60 \times 1 0^{- 19} = 3.2 \times 1 0^{- 16}$ J. Then $v = 2 \times 3.2 \times 1 0^{- 16} \div 1.67 \times 1 0^{- 27} \approx 6.2 \times 1 0^{5}$ m/s. Check: $\frac{1}{2} \times 1.67 \times 1 0^{- 27} \times (6.19 \times 1 0^{5})^{2} \approx 3.2 \times 1 0^{- 16}$ J, the energy we started with. ✓ The proton is much slower than the electron in Worked Example 2, even with four times the energy, because it is about 1800 times heavier.
A charge of +10 nC is alone in space. Find the radii of the equipotentials at 300 V, 200 V and 100 V. Are they equally spaced? Explain why, using the field.

Hint 1
Rearrange $V = k Q / r$ to $r = k Q / V$ .

Hint 2
kQ = 90 V m. Divide by each potential.

Full answer
$r = 90 \div V$ : 0.30 m, 0.45 m and 0.90 m. They are not equally spaced: the gaps are 0.15 m and then 0.45 m. Each gap is a 100 V drop, and further out the field is weaker, so you must travel further to drop 100 V. Check: $E \approx - Δ V /Δ r$ gives about 670 V/m in the inner gap and about 220 V/m in the outer one. The field is weaker further out. ✓
A stretch. In a hydrogen atom, picture the electron about $5.29 \times 1 0^{- 11}$ m from the proton. Find the potential due to the proton at that distance (use $k = 8.99 \times 1 0^{9}$ and $e = 1.602 \times 1 0^{- 19}$ C), and the electron's potential energy in joules and in eV.

Hint 1
The proton's charge is +e. Find V = ke/r first.

Hint 2
The electron's charge is −e, so U = −eV. In electronvolts, U is simply −(V in volts) eV.

Full answer
$V = 8.99 \times 1 0^{9} \times 1.602 \times 1 0^{- 19} \div 5.29 \times 1 0^{- 11} \approx 27.2$ V. $U = - e V \approx - 1.602 \times 1 0^{- 19} \times 27.2 \approx - 4.36 \times 1 0^{- 18}$ J, which is −27.2 eV. The negative sign means the electron is bound: energy must be supplied to pull it away. Check: the answers in J and eV agree, because 27.2 × 1.602 × 10⁻¹⁹ ≈ 4.36 × 10⁻¹⁸. ✓ The electron also has kinetic energy, so the energy needed to free it is less than 27.2 eV. That part of the story is for Pre-University atomic physics.

The bigger idea

The word "potential" was introduced in 1828 by George Green, a self-taught miller's son from Nottingham, in an essay on electricity and magnetism that only about fifty people subscribed to. His idea was the one in this lecture: replace a field of arrows by a single number at each point, found by adding up contributions divided by distance.

That swap, from a vector to a scalar, is one of the great simplifying moves in physics. Adding numbers is far easier than adding arrows, and once the potential is known, the field follows from its slope. The same move appears for gravity, where the Pre-University lectures meet a gravitational potential, and in circuits, where every point in a wire has a potential and a voltmeter reads the difference. The next topic, circuits, uses that last picture all the time.

Key concepts in this lesson

The electric potential at a point is the work done per unit positive charge in bringing a small positive test charge from infinity to that point, measured in volts.
Near a point charge, $V = k Q / r$ , keeping the sign of Q, so potential falls as 1/r while field strength falls as 1/r².
Potential is a scalar, so the potentials of several charges are simply added, with their signs.
A charge q at potential V has energy $U = q V$ , and moving it through a p.d. $Δ V$ takes $W = q Δ V$ by any path; 1 eV = $1.60 \times 1 0^{- 19}$ J.
Equipotentials cross field lines at right angles, and the field points from high to low potential with a strength equal to how fast the potential falls with distance.