Charge that is spread out
A charged rod, a ring or a sheet is not a point. Its charge is spread over its length, its area or its volume. For a thin rod or wire we use the linear charge density \(\lambda\), the charge per unit length. For a sheet it is the surface charge density \(\sigma\), charge per unit area. For a solid body it is the volume charge density \(\rho\), charge per unit volume. Each is uniform when the charge is spread evenly.
A charge of \(10\,\mu\text{C}\) spread evenly over a rod \(50\,\text{cm}\) long has \(\lambda\) \(= 10\,\mu\text{C}/0.5\,\text{m}\) \(= 20\,\mu\text{C/m}\). A piece \(5\,\text{cm}\) long then holds \(20 \times 0.05 = 1\,\mu\text{C}\). A very short piece of length \(dx\) holds \(dq = \lambda\,dx\). In the same way a small patch of a sheet holds \(\sigma\,dA\), and a small bit of a solid holds \(\rho\,dV\).
A small element is so small that it acts as a point charge. Its field at a point P is \(k\,dq/r^2\), directed along the line from the element to P. The field of the whole body is the vector sum of these, an integral. This is the slicing you used for the centre of mass of a rod and for the gravitational field of a rod and a ring.
The method has four steps, and every lesson from here uses them. Cut the body into elements \(dq\). Write the field \(dE = k\,dq/r^2\) of one element. Use symmetry to see which components cancel. Then integrate the component that survives.
A thin ring of radius \(10\,\text{cm}\) carries \(3.14\,\mu\text{C}\) spread evenly round it. What is its linear charge density, in μC/m?
A rod seen from its middle
Take a rod \(60\,\text{cm}\) long with \(\lambda = 5\,\mu\text{C/m}\), and a point P \(40\,\text{cm}\) from its middle on the perpendicular bisector. Cut the rod into small elements. Each one sends a field to P, pointing from the element to P. For every element above the middle there is one below, at the same distance and the same tilt.
Watch the pair of tilted arrows in Fig 5.1. Their parts along the rod point opposite ways and cancel. Their parts away from the rod point the same way and add. So only the perpendicular component survives. The far pairs add less than the near ones, for two reasons. They are farther from P, and their fields are more tilted, so a smaller share of each points away from the rod.
The field of a rod at any point
Let P be a distance \(r\) from the line of the rod, and let O be the foot of the perpendicular from P. Measure each element by its angle \(\theta\) from the perpendicular PO. The two ends of the rod are seen at angles \(\alpha\) and \(\beta\) on either side, as in Fig 5.2. The derivation below uses \(\theta\) instead of the position \(x\), because then every element contributes a very simple amount.
\[ x = r\tan\theta, \qquad dx = r\sec^2\theta\,d\theta, \qquad \text{distance} = r\sec\theta \]
Why. an element at angle \(\theta\) from the perpendicular sits at \(x = r\tan\theta\) from O. Differentiating gives the length \(dx\) that a small angle \(d\theta\) covers.
\[ dE = \dfrac{k\lambda\,dx}{r^2\sec^2\theta} = \dfrac{k\lambda}{r}\,d\theta \]
Why. the element holds \(\lambda\,dx\) and acts as a point charge at distance \(r\sec\theta\). The two \(\sec^2\theta\) cancel, so each equal angle gives an equal field.
\[ E_{\perp} = \dfrac{k\lambda}{r}\int_{-\beta}^{\alpha}\cos\theta\,d\theta = \dfrac{k\lambda}{r}(\sin\alpha + \sin\beta) \]
Why. the part of \(dE\) perpendicular to the rod is \(dE\cos\theta\). The elements run from \(-\beta\) on one side to \(\alpha\) on the other.
\[ E_{\parallel} = \dfrac{k\lambda}{r}\int_{-\beta}^{\alpha}\sin\theta\,d\theta = \dfrac{k\lambda}{r}(\cos\beta - \cos\alpha) \]
Read it. the part along the rod is \(dE\sin\theta\), and it points away from each element. The longer side wins, so the net part points towards the shorter side.
On the perpendicular bisector the two angles are equal, \(\alpha = \beta\). Then \(E_{\parallel} = 0\), as the mirror pairs promised, and \(E_{\perp}\) \(= (2k\lambda/r)\sin\alpha\).
A rod \(60\,\text{cm}\) long carries \(\lambda = 5\,\mu\text{C/m}\). Find the field at P, \(40\,\text{cm}\) from its middle on the perpendicular bisector. Compare it with the field if the rod's whole charge sat at its middle.
(a) Each end is \(\sqrt{0.3^2 + 0.4^2}\) \(= 0.5\,\text{m}\) from P, so \(\sin\alpha = 0.3/0.5 = 0.6\).
(b) \(E\) \(= (2k\lambda/r)\sin\alpha\) \(= (2 \times 9 \times 10^9 \times 5 \times 10^{-6}/0.4) \times 0.6\) \(= 1.35 \times 10^5\,\text{N/C}\), away from the rod.
(c) The rod holds \(Q = \lambda L = 3\,\mu\text{C}\). At its middle this would give \(kQ/r^2\) \(= 9 \times 10^9 \times 3 \times 10^{-6}/0.16\) \(= 1.69 \times 10^5\,\text{N/C}\).
(d) The rod's field is smaller: its ends are farther than \(0.4\,\text{m}\), and their fields are tilted.
Values used, the \(k\) route: \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).
A uniformly charged rod has a total charge \(Q\). Compare the field at a point on its perpendicular bisector with the field of a point charge \(Q\) placed at the rod's middle.
An equilateral triangle is made of three identical thin rods, each with the same positive \(\lambda\). What is the field at its centre?
A long straight wire
Now make the rod very long compared with \(r\). Seen from P, both ends go off towards the horizon, so \(\alpha\) and \(\beta\) both approach \(90°\). Then \(\sin\alpha + \sin\beta\) approaches 2, and \(\cos\beta - \cos\alpha\) approaches 0. The field of a long straight line charge is \(2k\lambda/r\), straight away from the wire.
The field of a long wire falls as \(1/r\), more slowly than the \(1/r^2\) of a point charge. A wire with \(5\,\mu\text{C/m}\) gives \(2 \times 9 \times 10^9 \times 5 \times 10^{-6}/0.1\) \(= 9 \times 10^5\,\text{N/C}\) at \(10\,\text{cm}\). At \(20\,\text{cm}\) it gives half of that, not a quarter. Move away from a point charge and all of it recedes. Move away from a long wire and more of it comes into view.
In the figure above, a rod \(10\,\text{cm}\) long gives only \(4.0 \times 10^5\,\text{N/C}\). At \(40\,\text{cm}\) it already gives \(8.0 \times 10^5\), and at \(120\,\text{cm}\) it gives \(8.9 \times 10^5\), within 2 % of the long-wire value. A wire needs to be only a few times longer than the distance to P for \(2k\lambda/r\) to hold. The far parts are distant and tilted, so they add almost nothing.
A long straight wire carries \(3\,\mu\text{C/m}\). Find the field \(30\,\text{cm}\) from it, in units of \(10^5\,\text{N/C}\).
A wire that ends opposite the point
Take a very long straight wire that starts at O and runs off to the right. Let P be a distance \(r\) from O, on the perpendicular to the wire at O. From P, the near end is seen at \(\beta = 0\) and the far end at \(\alpha = 90°\). So \(E_{\perp}\) \(= (k\lambda/r)(1 + 0)\) \(= k\lambda/r\) and \(E_{\parallel}\) \(= (k\lambda/r)(1 - 0)\) \(= k\lambda/r\).
The two parts are equal, so the field is \(\sqrt2\,k\lambda/r\), at \(45°\) to the wire. It leans away from the wire's side, because all the charge lies on one side of P. For \(\lambda = 2\,\mu\text{C/m}\) and \(r = 10\,\text{cm}\), each part is \(1.8 \times 10^5\,\text{N/C}\), and the field is \(2.55 \times 10^5\,\text{N/C}\).
A very long wire with \(\lambda = 2\,\mu\text{C/m}\) starts at O and runs along \(+x\). Point P is \(10\,\text{cm}\) from O along \(+y\). What is the field at P?
A point on the line of the rod
Now let P lie on the line of the rod itself, a distance \(a\) beyond its near end. Every element pushes P straight along the rod, away from itself. All the fields point the same way, so nothing cancels. The sizes simply add.
\[ dE = \dfrac{k\lambda\,dx}{x^2} \]
Why. an element of length \(dx\) at distance \(x\) from P holds \(\lambda\,dx\) and acts as a point charge. Its field at P points along the rod.
\[ E = k\lambda\int_a^{a+L}\dfrac{dx}{x^2} = k\lambda\left[-\dfrac1x\right]_a^{a+L} \]
Why. every field points the same way, so the sizes add. The rod runs from \(x = a\) to \(x = a + L\).
\[ E = k\lambda\left(\dfrac1a - \dfrac{1}{a + L}\right) = \dfrac{k\lambda L}{a(a + L)} = \dfrac{kQ}{a(a + L)} \]
Read it. put in the limits and combine the fractions. The total charge is \(Q = \lambda L\).
A rod \(30\,\text{cm}\) long carries \(6\,\mu\text{C}\), and P is \(20\,\text{cm}\) beyond its near end. Then \(E\) \(= 9 \times 10^9 \times 6 \times 10^{-6}/(0.2 \times 0.5)\) \(= 5.4 \times 10^5\,\text{N/C}\). Far from the rod, \(a\) is much larger than \(L\), so \(a + L \approx a\) and \(E \approx kQ/a^2\). From far away, a rod looks like a point charge, as it must.
A rod \(20\,\text{cm}\) long carries \(4\,\mu\text{C}\) spread evenly. A point P lies on the line of the rod, \(10\,\text{cm}\) beyond its near end. Find the field at P, in units of \(10^5\,\text{N/C}\).
In \(kQ/(a(a + L))\), \(a\) is measured from the near end of the rod, not from its middle. Putting the middle distance into \(kQ/a^2\) gives a field that is too small, because the near half of the rod counts for more.
A Geiger counter, used to check for radioactive contamination, has a thin wire along the axis of a metal tube. The wire is kept at a high positive voltage. Its field grows as \(1/r\) towards the wire, so it becomes very strong in a thin layer around it. An electron freed there by radiation is accelerated so hard that it knocks more electrons free. The small avalanche of charge makes the click you hear.
- Adding the sizes of the element fields on the bisector.Only the perpendicular parts add; the parts along the rod cancel in mirror pairs.
- Using \(1/r^2\) for a long wire.A long wire gives \(2k\lambda/r\), which falls as \(1/r\).
- Measuring \(a\) from the middle of the rod in \(kQ/(a(a + L))\).\(a\) is the distance from P to the near end of the rod.
- Forgetting the part along the wire for a half-infinite wire.When the charge lies on one side of P, the field also has a part along the wire.