Fields of charge distributions

Charge that is not a point

A real charged object is not a point. A plastic rod, a metal ring and a sheet of foil all carry charge spread over them. How do we find the field of such a body?

We use the same idea as before. Cut the body into pieces so small that each one is a point charge $dq$. Find the field $dE$ of each piece. Then add all the $dE$ with care for their directions. The adding is done with an integral, but the steps in each case are short, and the results are worth knowing.

First we need a way to say how the charge is spread. There are three densities.

• Linear charge density, for a wire or a rod: $\lambda = \dfrac{Q}{L}$, in C/m.
• Surface charge density, for a sheet or a plate: $\sigma = \dfrac{Q}{A}$, in C/m².
• Volume charge density, for a solid body: $\rho = \dfrac{Q}{V}$, in C/m³.

When the charge is spread evenly, these are just the total charge divided by the length, area or volume. A rod of length $0.50$ m with $2.0\ \mu$C on it has $\lambda = 4.0\ \mu\text{C/m}$. A square plate of side $0.10$ m with $3.6$ nC on it has $\sigma = \dfrac{3.6 \times 10^{-9}\ \text{C}}{0.010\ \text{m}^2} = 3.6 \times 10^{-7}\ \text{C/m}^2$.

A small piece of a rod of length $dx$ carries $dq = \lambda\,dx$. A small piece of a sheet of area $dA$ carries $dq = \sigma\,dA$.

A ring, on its axis

A thin ring of radius $R$ carries a charge $Q$ spread evenly round it. We want the field at a point $P$ on the axis, a distance $z$ from the centre.

Take a small piece $dq$. Its distance from $P$ is $r = \sqrt{R^2 + z^2}$, and its field has the size $dE = \dfrac{k\,dq}{r^2}$. This field points away from the piece. Split it into two parts. One part is along the axis, $dE\cos\theta$. The other part is across the axis, $dE\sin\theta$.

Now take the piece on the opposite side of the ring. It makes the same $dE$, but its across part points the other way. So the across parts cancel in pairs all round the ring. Only the parts along the axis are left, and they add:

From the picture, $\cos\theta = \dfrac{z}{r}$. So

The field is along the axis and away from a positive ring. Let us test this in three places.

• At the centre, $z = 0$ and $E = 0$. All round the ring, the fields cancel in pairs.
• Far away, $z \gg R$, so $R^2 + z^2 \approx z^2$ and $E \approx \dfrac{kQ}{z^2}$. From far away the ring looks like a point charge.
• In between, the field is not zero and not a point-charge field. It rises from zero, reaches a greatest value, and then falls.

Where is it greatest? Set the slope to zero. For $E = kQ\,z\,(R^2 + z^2)^{-3/2}$ the slope is $kQ\,(R^2 + z^2)^{-5/2}(R^2 - 2z^2)$. It is zero when $z = \dfrac{R}{\sqrt{2}}$. At that place, $E_{\max} = \dfrac{2kQ}{3\sqrt{3}\,R^2} = 0.385\,\dfrac{kQ}{R^2}$.

A worked example. A ring of radius $0.20$ m carries $+5.0$ nC. Find the field on the axis $0.30$ m from the centre, and compare it with the field of a point charge of $5.0$ nC at the centre.

1. The ring. $R^2 + z^2 = 0.040 + 0.090 = 0.13\ \text{m}^2$, and $(0.13)^{3/2} = 0.0469\ \text{m}^3$. So $E = \dfrac{9.0 \times 10^{9} \times 5.0 \times 10^{-9} \times 0.30}{0.0469} = 288\ \text{N/C}$.
2. The point charge. $E = \dfrac{kQ}{z^2} = \dfrac{45\ \text{N m}^2/\text{C}}{0.090\ \text{m}^2} = 500\ \text{N/C}$.

Close to the ring, the point-charge idea is far off. At $z = 2.0$ m, the ring gives $11.08$ N/C and the point charge gives $11.25$ N/C, only $1.5\ \%$ more.

Make the radius smaller and watch the peak move in and become higher. Move the point far out along the axis. Do the two lines meet?

A straight line of charge

Now a straight rod of length $L$ with a charge $Q$ spread evenly along it, so that $\lambda = Q/L$. Take a point on the line at right angles to the rod through its middle, at a distance $r$ from the rod.

A piece of the rod at a distance $x$ from the middle has $dq = \lambda\,dx$. Its distance from the point is $\sqrt{r^2 + x^2}$. Its field has a part along the rod and a part across it. For each piece on one side there is one on the other side, and the parts along the rod cancel. The parts across the rod add:

The integral is $\dfrac{x}{r^2\sqrt{r^2 + x^2}}$ between the limits. This gives

The field is directed away from a positive rod, at right angles to it. Two limits are easy to see.

• Far from the rod, $r \gg L$. Then $\sqrt{r^2 + (L/2)^2} \approx r$, and $E \approx \dfrac{kQ}{r^2}$. The rod looks like a point charge.
• Close to a long rod, $r \ll L$. Then $\sqrt{r^2 + (L/2)^2} \approx \dfrac{L}{2}$, and $E \approx \dfrac{2k\lambda}{r}$. This is the field of a very long wire. It falls as $1/r$, not $1/r^2$.

In the next course we find the same result for a long wire with Gauss's law, in a few lines.

A worked example. A rod $0.60$ m long carries $12$ nC. Find the field at $0.40$ m from the rod, on the line at right angles through its middle. Compare with a very long wire with the same charge on each metre.

1. The charge on each metre. $\lambda = \dfrac{12\ \text{nC}}{0.60\ \text{m}} = 20\ \text{nC/m}$.
2. The rod. $\sqrt{r^2 + (L/2)^2} = \sqrt{0.16 + 0.09}\ \text{m} = 0.50\ \text{m}$. So $E = \dfrac{9.0 \times 10^{9} \times 12 \times 10^{-9}}{0.40 \times 0.50} = 540\ \text{N/C}$.
3. The long wire. $E = \dfrac{2k\lambda}{r} = \dfrac{2 \times 9.0 \times 10^{9} \times 20 \times 10^{-9}}{0.40} = 900\ \text{N/C}$.
4. A point charge of $12$ nC. $E = \dfrac{kQ}{r^2} = 675\ \text{N/C}$.

The rod lies between the two. At $r = 0.05$ m, close to the rod, the formula gives $7102$ N/C and the long wire gives $7200$ N/C, only $1.4\ \%$ more.

Make the rod long and bring the point close to it. Which dashed line does the solid line follow? Now make the rod short and move away. Which one now?

A flat disc

A thin disc of radius $R$ carries a charge spread evenly with a surface density $\sigma$. Cut it into thin rings. A ring of radius $s$ and width $ds$ carries $dq = \sigma\,2\pi s\,ds$, and its field on the axis is the ring field above. Adding the rings from $s = 0$ to $s = R$ gives, at a distance $z$ along the axis,

Two things are worth seeing. When $z \gg R$, the bracket becomes $\dfrac{R^2}{2z^2}$ and $E \approx \dfrac{kQ}{z^2}$, a point charge. When $z \ll R$, the bracket is nearly $1$, and $E \approx \dfrac{\sigma}{2\varepsilon_0}$. The field close to a large flat sheet does not depend on the distance. We use this result a great deal in the next course.

A worked example. A disc of radius $0.50$ m carries $\sigma = 4.0\ \text{nC/m}^2$. Find the field $0.10$ m from its centre on the axis. Use $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)$.

1. The sheet value. $\dfrac{\sigma}{2\varepsilon_0} = \dfrac{4.0 \times 10^{-9}\ \text{C/m}^2}{2 \times 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 226\ \text{N/C}$.
2. The bracket. $1 - \dfrac{0.10}{\sqrt{0.010 + 0.25}} = 1 - 0.196 = 0.804$.
3. The field. $E = 226\ \text{N/C} \times 0.804 = 182\ \text{N/C}$, away from the disc.

Practise with new numbers

Try it in Python

The program cuts a rod into $2000$ small pieces. It adds the part of each field that is across the rod. Then it compares the sum with the formula, and with the field of a very long wire.

Common mistakes

• Adding all the pieces as if they pointed the same way. Only the parts along the axis of a ring survive. The parts across it cancel.
• Using the point-charge formula for a ring close to its centre. The ring has no field at its centre, and a point charge has an infinite one.
• Mixing up $R$ and $z$. In the ring formula, $R$ is the radius of the ring and $z$ is the distance along the axis.
• Taking the long-wire result for a short rod. The result $2k\lambda/r$ needs $r$ much smaller than $L$.
• Mixing up the densities. $\lambda$ is for a length, $\sigma$ for an area and $\rho$ for a volume.

Remember these

• Cut the body into pieces $dq = \lambda\,dx$, $\sigma\,dA$ or $\rho\,dV$. Add the fields as vectors, using symmetry to cancel parts.
• Ring, on the axis: $E = \dfrac{kQz}{(R^2 + z^2)^{3/2}}$. It is zero at the centre, greatest at $z = R/\sqrt{2}$, and $kQ/z^2$ far away.
• Rod, on the perpendicular bisector: $E = \dfrac{kQ}{r\sqrt{r^2 + (L/2)^2}}$. For a very long wire, $E = \dfrac{2k\lambda}{r}$.
• Disc, on the axis: $E = \dfrac{\sigma}{2\varepsilon_0}\left(1 - \dfrac{z}{\sqrt{z^2 + R^2}}\right)$. Near a large sheet, $E = \dfrac{\sigma}{2\varepsilon_0}$.

Check yourself

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

1. A ring has a charge $Q$. What is the field at its centre? Explain in one sentence.

2. A thin wire $0.80$ m long carries $4.0$ nC spread evenly. What is its linear charge density?

3. The field of a very long charged wire is $E_1$ at a distance of $0.10$ m. What is it at $0.40$ m?

4. A ring of radius $R$ has a charge $Q$. At what distance from its centre on the axis is the field greatest, and what is the value there?