Potential energy and potential

Lifting a book

Lift a book from the floor to a shelf. You push up, gravity pulls down, and you do work. That work is not lost. It is stored as gravitational potential energy. Let the book fall, and the energy comes back as motion.

An electric field works in the same way. Put a small positive charge near a large positive charge. They repel. To push the small charge closer, you must do work against the field. Let go, and the small charge moves away. The energy you gave it comes back as kinetic energy.

Now we put numbers on this idea. It gives us two new quantities: the electric potential energy of a charge, and the electric potential of a point.

A field that stores energy

The force between two point charges depends only on the distance between them, and it acts along the line joining them. A force like this is called conservative. The work it does on a charge that goes from A to B depends only on A and on B. It does not depend on the path. If the charge goes round a loop and comes back to A, the net work is zero.

Because of this, every point in the field can be given a definite potential energy, just as every point in a room has a height.

Potential energy of a charge

Take a small positive test charge $q_0$. It must be so small that it does not disturb the charges that make the field. Carry it slowly from A to B, so that it has no kinetic energy at either end. The field pushes on it with the force $q_0\vec{E}$. So we must push with an equal and opposite force. The work that we do is the gain in potential energy:

The field itself does the work $-W_{AB}$ on the charge. If $W_{AB}$ is positive, the charge went against the field and stored energy. If it is negative, the field did the pushing.

Potential difference

Divide the work by the test charge. The test charge drops out, and we get a number that belongs to the points A and B alone:

This is the potential difference between B and A. In words: it is the work done by an outside agent in carrying one unit of positive charge slowly from A to B. Its unit is the volt, and $1\ \text{V} = 1\ \text{J/C}$.

Only differences matter, just as only differences of height matter on a hill. We are free to choose where the potential is zero. For charges in a small region, the usual choice is the far-away region, which we call infinity. Then:

The potential at a point is the work done in bringing one unit of positive charge from infinity to that point.

Potential has a size and a sign, but no direction. It is a scalar.

A charge $q$ at a point where the potential is $V$ has the potential energy

So when $q$ moves from A to B, its potential energy changes by $q(V_B - V_A)$.

Which way do charges go?

• A positive charge, released from rest, moves towards lower potential. Its potential energy $qV$ falls.
• A negative charge, released from rest, moves towards higher potential. Its $qV$ is negative, and it falls when $V$ rises.
• In both cases the potential energy falls and the kinetic energy rises. Their sum stays the same.

The field between two plates

Take two large parallel plates a distance $d$ apart, with a potential difference $V$ across them. Between them the field is uniform, and it points from the positive plate to the negative plate.

Carry a positive charge $q$ from the negative plate to the positive plate. The field pushes it back with the force $qE$, so we push with $qE$ over the distance $d$. The work is $W = qEd$. Dividing by $q$ gives $V = Ed$, so

The unit of the field is V/m, and it is the same as N/C. This gives us a second way to measure a field: as the fall of potential in each metre.

The potential rises steadily from the negative plate to the positive plate. If the negative plate is at 0 V, the potential at a distance $x$ from it is $V(x) = Ex$.

A worked example. Two plates are $8.0$ cm apart, with $240$ V across them. A charge of $+5.0$ nC is carried slowly from the negative plate.

1. The field. $E = \dfrac{V}{d} = \dfrac{240\ \text{V}}{0.080\ \text{m}} = 3000\ \text{V/m}$.
2. The potential at $2.0$ cm. $V = Ex = 3000\ \text{V/m} \times 0.020\ \text{m} = 60\ \text{V}$.
3. The work to reach that point. $W = qV = 5.0 \times 10^{-9}\ \text{C} \times 60\ \text{V} = 3.0 \times 10^{-7}\ \text{J}$.
4. The work to reach the positive plate. $W = 5.0 \times 10^{-9}\ \text{C} \times 240\ \text{V} = 1.2 \times 10^{-6}\ \text{J}$.

Step 3 is a quarter of step 4, because $2.0$ cm is a quarter of the gap.

Make the gap smaller and keep the voltage. What happens to the field? Then turn the charge negative. What happens to its potential energy at the same point?

The electronvolt

The joule is far too large for one atom or one electron. We use the electronvolt (eV) instead. It is the energy gained by a particle of charge $e$ that falls through $1$ V:

A particle of charge $e$ that falls through $500$ V gains $500$ eV. A particle of charge $2e$ gains $1000$ eV. The eV is a unit of energy, not of potential.

The speed after a fall through a potential difference

A particle of charge $q$ and mass $m$ starts from rest and falls through a potential difference $V$. The field does the work $qV$, and this becomes kinetic energy:

We want $v$, so $v = \sqrt{\dfrac{2qV}{m}}$.

A worked example. An electron starts from rest and falls through $150$ V.

1. The energy. $K = eV = 150\ \text{eV} = 150 \times 1.6 \times 10^{-19}\ \text{J} = 2.4 \times 10^{-17}\ \text{J}$.
2. The speed. $v = \sqrt{\dfrac{2 \times 2.4 \times 10^{-17}\ \text{J}}{9.1 \times 10^{-31}\ \text{kg}}} = 7.3 \times 10^{6}\ \text{m/s}$.
3. In words. That is about $2$ per cent of the speed of light. The formula is good while the speed is far below $3.0 \times 10^8$ m/s.

The mass does not appear in the energy. Two particles of the same charge gain the same energy. The lighter one goes faster.

Practise with new numbers

Try it in Python

The program gives an electron, a proton and an alpha particle the same potential difference. It prints the energy and the speed of each.

Common mistakes

• Mixing up potential and potential energy. Potential is in volt, and it belongs to a point. Potential energy is in joule, and it belongs to a charge at that point: $U = qV$.
• Getting the direction wrong. Going along the field, the potential falls. Going against it, the potential rises.
• Forgetting the sign of the charge. The work to carry a negative charge to a higher potential is negative.
• Using $E = V/d$ when the field is not uniform. The simple form holds only for the field between large parallel plates.
• Not changing units. Change cm to m, and nC or μC to C, before you calculate.
• Dropping the factor 2 in $\tfrac{1}{2}mv^2 = qV$.

Remember these

• $W_{AB} = U_B - U_A$, and $V_B - V_A = W_{AB}/q_0$. The unit is the volt, $1\ \text{V} = 1\ \text{J/C}$.
• The potential at a point is the work to bring a unit positive charge from infinity.
• $U = qV$. Positive charges fall to lower potential, and negative charges to higher potential.
• Between plates, $E = V/d$ in V/m, and the potential changes by $Ed$ in a distance $d$ along the field.
• $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$. A fall through $V$ gives $\tfrac{1}{2}mv^2 = qV$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. An outside agent carries a charge of $3.0\ \mu\text{C}$ from a point at $20$ V to a point at $85$ V. How much work does the agent do? What is the change in the potential energy of the charge?

2. A field of $4000$ V/m acts between two parallel plates that are $2.0$ cm apart. What is the potential difference between the plates?

3. An ion of charge $+2e$ starts from rest and falls through $350$ V. How much kinetic energy does it gain, in eV and in J?

4. Point A is at $50$ V and point B is at $20$ V. A positive charge and a negative charge are each released from rest at A. Which way does each one move?

5. A point has a potential of zero. Does that mean the electric field is zero there?