Potential of charge systems

A spark from a door handle

Walk across a carpet and touch a metal door handle. A small spark jumps to your hand. Your body may have been at several thousand volts. Yet the spark is harmless, because it carries very little charge.

Here we learn how to find the potential of a charge, and of many charges together. The good news is that potential is a number, so adding potentials is much easier than adding fields.

The potential of a point charge

Bring a test charge $q_0$ from infinity to a distance $r$ from a fixed point charge $q$. The force on it at a distance $x$ is $\dfrac{kqq_0}{x^2}$. It grows as the charge comes closer. So we cannot simply multiply force by distance. We cut the path into thin slices, find the small work in each slice and add them all. Calculus does this sum, and the result is

The work for each unit of charge is the potential:

Here $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)$. The sign of $q$ goes into the answer.

• A positive charge has a positive potential all round it. Bringing a positive test charge close needs work.
• A negative charge has a negative potential. The test charge is pulled in, and the field does the work.
• Compare the field, $E = kq/r^2$, with the potential, $V = kq/r$. The potential falls more slowly with distance.

A worked example. A point charge of $+5.0$ nC is in air.

1. At $20$ cm. First $kq = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2 \times 5.0 \times 10^{-9}\ \text{C} = 45\ \text{V m}$. Then $V = \dfrac{kq}{r} = \dfrac{45\ \text{V m}}{0.20\ \text{m}} = 225\ \text{V}$.
2. At $50$ cm. $V = \dfrac{45\ \text{V m}}{0.50\ \text{m}} = 90\ \text{V}$.
3. The work to carry $+1.0$ nC from $50$ cm to $20$ cm. $W = q(V_{20} - V_{50}) = 1.0 \times 10^{-9}\ \text{C} \times 135\ \text{V} = 1.35 \times 10^{-7}\ \text{J}$.

Many charges: add the numbers

Suppose there are several point charges $q_1, q_2, q_3, \dots$ at the distances $r_1, r_2, r_3, \dots$ from a point P. Each one makes its own potential at P, and the potentials add:

There are no angles to worry about. Only the signs of the charges and the distances matter.

A worked example. A charge of $+8.0$ nC is at one end of a line. A charge of $-2.0$ nC is $20$ cm away.

1. At the midpoint. Both charges are $0.10$ m away. The sum of the charges is $(8.0 - 2.0)\ \text{nC} = 6.0\ \text{nC}$, so $k(q_1 + q_2) = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2 \times 6.0 \times 10^{-9}\ \text{C} = 54\ \text{V m}$. Then $V = \dfrac{54\ \text{V m}}{0.10\ \text{m}} = 540\ \text{V}$.
2. At a point P on the line, $10$ cm beyond the $-2.0$ nC charge. The distances are $0.30$ m and $0.10$ m. The charges give $kq_1 = 72\ \text{V m}$ and $kq_2 = -18\ \text{V m}$. So $V = \dfrac{72\ \text{V m}}{0.30\ \text{m}} + \dfrac{-18\ \text{V m}}{0.10\ \text{m}} = 240\ \text{V} - 180\ \text{V} = 60\ \text{V}$.
3. The work to bring $+1.0$ nC from infinity to the midpoint. $W = qV = 1.0 \times 10^{-9}\ \text{C} \times 540\ \text{V} = 5.4 \times 10^{-7}\ \text{J}$.

Look at the two graphs. For two equal charges the potential has its lowest point at the midpoint, and the field is zero there. For equal and opposite charges, the potential is exactly zero at the midpoint. Yet the field is not zero there, because the two fields point the same way. A point of zero potential is not a point of zero field.

A ring and a thin shell

Take a thin ring of radius $R$ with a total charge $Q$ spread evenly round it. Pick a point on its axis, a distance $x$ from the centre. Every bit of the ring is at the same distance $\sqrt{R^2 + x^2}$ from this point. So all the bits can be added at once:

For a ring of radius $6.0$ cm with $10$ nC, the centre has $V = \dfrac{kQ}{R} = 1500$ V. At $x = 8.0$ cm the distance is $10$ cm, and $V = 900$ V.

A thin spherical shell of charge $Q$ and radius $R$ gives $V = kQ/r$ at every point outside it, as if all the charge were at the centre. Inside the shell the field is zero, so the potential cannot change. It has the same value everywhere inside: $V = kQ/R$.

The potential of a dipole

A dipole is a pair of charges $+q$ and $-q$ a small distance $2a$ apart. Its dipole moment is $p = q \times 2a$, and it points from the negative charge to the positive charge.

Take a point on the axis, a distance $r$ from the centre, on the side of the $+q$ charge. The distances are $r - a$ and $r + a$:

Far from the dipole, $r$ is much larger than $a$, and $V = \dfrac{kp}{r^2}$. In general, for a point at an angle $\theta$ from the dipole axis:

• On the axis, $\cos\theta = \pm 1$, and the potential is largest. It is positive on the side of the positive charge.
• On the equatorial line, $\theta = 90^\circ$, so $V = 0$. Each point there is the same distance from both charges, and the two potentials cancel.
• The potential falls as $1/r^2$, faster than the $1/r$ of a single charge.

A worked example. A water molecule has a dipole moment of about $6.2 \times 10^{-30}\ \text{C m}$. On its axis, $2.0$ nm away, the potential is $V = \dfrac{kp}{r^2} = \dfrac{9.0 \times 10^{9} \times 6.2 \times 10^{-30}}{(2.0 \times 10^{-9})^2}\ \text{V} = 0.014\ \text{V}$, that is $14$ mV. The two charges of the molecule are only a few hundredths of a nanometre apart, so the far-away formula is a good one here.

The potential energy of a group of charges

Bring a charge $q_2$ from infinity to a distance $r$ from a fixed charge $q_1$. The potential of $q_1$ at that point is $kq_1/r$, so the work done is $q_2 \times kq_1/r$. This is the energy stored in the pair:

• For like charges, $U$ is positive. We had to do work to bring them together, and they would move apart if released.
• For unlike charges, $U$ is negative. The pair is bound. We would have to do work to pull them apart to infinity.

For three or more charges, bring them in one at a time. The total energy is the sum of the energy of every pair. Each pair is counted once.

A worked example. Three charges sit at the corners of an equilateral triangle of side $10$ cm. They are $+3.0$ nC, $+3.0$ nC and $-3.0$ nC. Each pair has a size $\dfrac{kq^2}{a} = \dfrac{9.0 \times 10^{9} \times (3.0 \times 10^{-9})^2}{0.10}\ \text{J} = 8.1 \times 10^{-7}\ \text{J}$. The two unlike pairs are negative, and the like pair is positive:

The group is bound. It takes $8.1 \times 10^{-7}$ J of work to pull the three charges far apart.

A dipole in a uniform field

Place a dipole in a uniform field $E$, with its axis at an angle $\theta$ to the field. The two forces, $qE$ and $-qE$, are equal and opposite, so there is no net force. But they act along different lines and twist the dipole. The torque is $\tau = qE \times 2a\sin\theta$:

The twist turns the dipole towards the field. The work done by an outside torque in turning it from $\theta_1$ to $\theta_2$ is $pE(\cos\theta_1 - \cos\theta_2)$. We choose the potential energy to be zero at $\theta = 90^\circ$. Then

At $\theta = 0$ the dipole points along the field. The energy is $-pE$, the lowest possible, and this position is stable. At $\theta = 180^\circ$ the energy is $+pE$, and the position is unstable. Turning the dipole through half a turn takes $2pE$.

For a water molecule in a field of $2.0 \times 10^{5}$ V/m, $pE = 1.2 \times 10^{-24}$ J. The thermal energy of a molecule at room temperature is about $4 \times 10^{-21}$ J, which is much larger. This is why the molecules of water in a glass are not all lined up by ordinary fields.

Practise with new numbers

Try it in Python

The program adds the potentials of the $+8.0$ nC and $-2.0$ nC charges at several points of the line. Then it walks along the line between them and looks for the place where the potential changes sign.

Common mistakes

• Using $r^2$ in the potential. The potential of a point charge has $r$. The field has $r^2$.
• Adding potentials as vectors. Potential is a scalar. Add the numbers with their signs.
• Dropping the sign of a negative charge. A negative charge makes a negative potential.
• Saying zero potential means zero field. On the equatorial line of a dipole, $V = 0$ but the field is not zero.
• Forgetting to count every pair in the energy of a group. Three charges make three pairs.
• Mixing nC, μC and C. Change to coulomb before you calculate, or keep a clear rule such as nC with metres.

Remember these

• $V = \dfrac{kq}{r}$, with $k = \dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2$. The sign of $q$ is kept.
• For many charges, $V = k\sum \dfrac{q_i}{r_i}$. Potentials add as numbers.
• A ring on its axis: $V = \dfrac{kQ}{\sqrt{R^2 + x^2}}$. A shell: $\dfrac{kQ}{r}$ outside, and $\dfrac{kQ}{R}$ everywhere inside.
• A dipole: $V = \dfrac{kp\cos\theta}{r^2}$ far away, and $V = 0$ on the equatorial line.
• A pair has $U = \dfrac{kq_1q_2}{r}$. A group has the sum over all pairs.
• A dipole in a field has $\tau = pE\sin\theta$ and $U = -pE\cos\theta$.

Check yourself

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

1. What is the potential at $45$ cm from a point charge of $-3.0$ nC in air?

2. Two equal positive charges $+q$ are $2a$ apart. What are the potential and the field at the midpoint? Is the potential zero? Is the field zero?

3. On the axis of a small dipole, a point is moved to twice the distance from the centre. What happens to the potential?

4. Three equal charges of $2.0$ nC sit at the corners of an equilateral triangle of side $5.0$ cm. Find the potential energy of the group.

5. The energy of a pair of charges is negative. What does the minus sign tell us?