The ring on its axis
A thin ring of radius \(R\) carries a charge \(Q\) spread evenly round it. Find the field at P on its axis, a distance \(x\) from the centre. Every element of the ring is the same distance, \(\sqrt{R^2 + x^2}\), from P. So every element sends a field of the same size. For each element there is another directly opposite it on the ring. Their sideways parts cancel, and only their parts along the axis add.
\[ dE = \dfrac{k\,dq}{R^2 + x^2} \]
Why. each element \(dq\) acts as a point charge at distance \(\sqrt{R^2 + x^2}\) from P.
\[ dE_x = dE\cos\phi = \dfrac{k\,dq}{R^2 + x^2}\cdot\dfrac{x}{\sqrt{R^2 + x^2}} \]
Why. \(\phi\) is the angle between the element's field and the axis, and \(\cos\phi = x/\sqrt{R^2 + x^2}\). The sideways parts cancel in opposite pairs.
\[ E = \dfrac{kx}{(R^2 + x^2)^{3/2}}\int dq = \dfrac{kQx}{(R^2 + x^2)^{3/2}} \]
Read it. the factor in front is the same for every element, so it comes out of the integral, and the elements add up to \(Q\).
Three cases can be read straight from the formula. At the centre, \(x = 0\), so \(E = 0\): opposite elements cancel completely. Far away, \(x\) is much larger than \(R\), so \(R^2 + x^2 \approx x^2\) and \(E \approx kQ/x^2\); the ring looks like a point charge. In between, the field must rise from zero and then fall again, so it has a greatest value.
Drag P along the axis in the figure above. The field rises from zero, reaches its peak at \(x = 21\,\text{cm}\), then falls towards the dashed point-charge curve. Setting \(dE/dx = 0\) gives \(R^2 - 2x^2 = 0\), so the peak is at \(x = R/\sqrt2\). Put that back in, and \(E_{\max} = 2kQ/3\sqrt3\,R^2\).
A ring of radius \(30\,\text{cm}\) carries \(5\,\mu\text{C}\). Find the field on its axis \(40\,\text{cm}\) from the centre. Where on the axis is the field greatest, and how large is it there?
(a) \(R^2 + x^2\) \(= 0.09 + 0.16\) \(= 0.25\,\text{m}^2\), so \((R^2 + x^2)^{3/2}\) \(= 0.125\,\text{m}^3\).
(b) \(E\) \(= 9 \times 10^9 \times 5 \times 10^{-6} \times 0.4/0.125\) \(= 1.44 \times 10^5\,\text{N/C}\), along the axis, away from the ring.
(c) The peak is at \(x\) \(= R/\sqrt2\) \(= 0.3/1.414\) \(= 0.21\,\text{m}\).
(d) There \(E_{\max}\) \(= 2 \times 9 \times 10^9 \times 5 \times 10^{-6}/(3\sqrt3 \times 0.09)\) \(= 1.92 \times 10^5\,\text{N/C}\).
Values used, the \(k\) route: \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).
A ring of radius \(30\,\text{cm}\) carries \(4\,\mu\text{C}\). Find the field on its axis \(40\,\text{cm}\) from its centre, in units of \(10^5\,\text{N/C}\).
A charged ring gives zero field at its centre. Why?
The ring's axial field gives a neat force. Lay a very long thread with charge \(\lambda\) per metre along the axis, starting at the centre. The force on the thread is \(\int_0^\infty \lambda\,E\,dx\) \(= kQ\lambda\int_0^\infty x\,dx/(R^2 + x^2)^{3/2}\) \(= kQ\lambda/R\). The thread is infinitely long, yet the force is finite, because the ring's field dies away as \(1/x^2\).
The centre of an arc
Now keep only part of the ring: an arc of radius \(R\) subtending an angle \(\theta\) at its centre O. It carries charge \(\lambda\) per metre. Measure each element by its angle \(\phi\) from the arc's line of symmetry. An element of angle \(d\phi\) has length \(R\,d\phi\) and charge \(\lambda R\,d\phi\). It sits at distance \(R\) from O.
\[ dE = \dfrac{k\lambda R\,d\phi}{R^2} = \dfrac{k\lambda}{R}\,d\phi \]
Why. the element holds \(\lambda R\,d\phi\) and is a distance \(R\) from O, so each equal angle gives an equal field.
\[ dE_{\text{sym}} = \dfrac{k\lambda}{R}\cos\phi\,d\phi \]
Why. elements at \(+\phi\) and \(-\phi\) are mirror images, so their parts across the line of symmetry cancel. Only the part along it survives.
\[ E = \dfrac{k\lambda}{R}\int_{-\theta/2}^{\theta/2}\cos\phi\,d\phi = \dfrac{2k\lambda}{R}\sin\dfrac{\theta}{2} \]
Read it. the arc runs from \(-\theta/2\) to \(+\theta/2\) about its line of symmetry.
With \(\lambda = 2\,\mu\text{C/m}\) and \(R = 10\,\text{cm}\), \(2k\lambda/R\) \(= 3.6 \times 10^5\,\text{N/C}\). A quarter circle gives \(3.6 \times 10^5 \sin 45°\) \(= 2.55 \times 10^5\,\text{N/C}\). A semicircle gives the full \(3.6 \times 10^5\,\text{N/C}\), the largest of all. Opening the arc further adds elements on the far side, whose fields point back. At \(270°\) the field is \(2.55 \times 10^5\) again, and the full ring gives zero, as the figure above shows.
A semicircular wire of radius \(10\,\text{cm}\) carries a total charge of \(2\,\mu\text{C}\). Find the field at its centre, in units of \(10^6\,\text{N/C}\).
A ring of radius \(10\,\text{cm}\) has \(+2\,\mu\text{C/m}\) on its upper half and \(-2\,\mu\text{C/m}\) on its lower half. What is the field at the centre?
A disc built from rings
A flat disc of radius \(R\) carries charge \(\sigma\) per square metre. Cut it into thin rings. A ring of radius \(r\) and width \(dr\) has area \(2\pi r\,dr\) and charge \(dq = \sigma\,2\pi r\,dr\). On the axis, each ring gives its field along the axis, and every ring's field points the same way. So the rings' fields simply add.
\[ dE = \dfrac{k\,(\sigma\,2\pi r\,dr)\,x}{(r^2 + x^2)^{3/2}} = \dfrac{\sigma x}{2\varepsilon_0}\,\dfrac{r\,dr}{(r^2 + x^2)^{3/2}} \]
Why. this is the ring's axial field with \(Q\) replaced by the ring's charge. Then \(k \times 2\pi = 1/2\varepsilon_0\).
\[ \int_0^R \dfrac{r\,dr}{(r^2 + x^2)^{3/2}} = \left[-\dfrac{1}{\sqrt{r^2 + x^2}}\right]_0^R = \dfrac1x - \dfrac{1}{\sqrt{R^2 + x^2}} \]
Why. substitute \(u = r^2 + x^2\), so \(r\,dr = du/2\), and integrate \(u^{-3/2}\).
\[ E = \dfrac{\sigma}{2\varepsilon_0}\left(1 - \dfrac{x}{\sqrt{x^2 + R^2}}\right) \]
Read it. multiply by \(\sigma x/2\varepsilon_0\) and simplify.
Very close to the disc, \(x\) is tiny compared with \(R\), so the bracket is nearly 1 and \(E = \sigma/2\varepsilon_0\). Far away, the disc looks like a point charge \(\sigma\pi R^2\). Take \(\sigma\) \(= 1.77 \times 10^{-8}\,\text{C/m}^2\) and \(R = 30\,\text{cm}\). Then \(\sigma/2\varepsilon_0\) \(= 1.77 \times 10^{-8}/(2 \times 8.85 \times 10^{-12})\) \(= 1000\,\text{N/C}\) at the surface. At \(x = 40\,\text{cm}\) the bracket is \(1 - 0.4/0.5 = 0.2\), so \(E = 200\,\text{N/C}\).
A large sheet: the same field everywhere
Now let the disc grow until \(R\) is much larger than \(x\). The bracket becomes 1 at every distance, and the field is \(\sigma/2\varepsilon_0\). It does not depend on \(x\) at all. Near a large flat sheet of charge the field is uniform. It has the same size everywhere and points straight away from the sheet on each side. The field lines are straight, parallel and evenly spaced.
Why does the field not weaken as you move away? Moving away puts every patch of the sheet farther off, which weakens its field. But it also brings more of the sheet into view at a useful angle, the way a long wire did in the last lesson. For an infinite sheet the two effects balance exactly. A sheet with \(\sigma\) \(= 8.85 \times 10^{-9}\,\text{C/m}^2\) gives \(500\,\text{N/C}\) at \(1\,\text{cm}\) and at \(1\,\text{m}\) alike.
A large flat sheet carries \(\sigma\) \(= 1.77 \times 10^{-7}\,\text{C/m}^2\). What is the field \(5\,\text{cm}\) from it, in N/C?
Two sheets
Put two large parallel sheets side by side. Each makes \(\sigma/2\varepsilon_0\) on both of its sides, whatever the other does, so at every point the two fields add as vectors. A positive sheet's field points away from it; a negative sheet's field points towards it. Everything depends on whether the two parts point the same way or opposite ways.
With opposite charges, as in the first chip of Fig 6.1, both fields point from the positive sheet to the negative one between them. They add to \(\sigma/\varepsilon_0\) \(= 1000\,\text{N/C}\). Outside, the two fields point opposite ways and cancel. So all the field is trapped between the sheets, uniform there and zero outside. This is the arrangement inside a parallel-plate capacitor. With like charges the fields cancel between the sheets and add to \(1000\,\text{N/C}\) outside, pointing away on both sides.
A ball hanging beside a sheet
A small ball of mass \(2\,\text{g}\) carrying \(3\,\mu\text{C}\) hangs by a thread beside a large vertical sheet with \(\sigma\) \(= 8.85 \times 10^{-8}\,\text{C/m}^2\). The sheet's field is uniform and horizontal, so the push \(qE\) on the ball is the same wherever it swings. The ball settles where its weight, the tension and the push balance, just like the balls on threads in lesson 3.
For the ball and sheet just described, take \(g = 10\,\text{m/s}^2\). Find the field of the sheet, the electric force on the ball, the angle of the thread and the tension.
(a) \(E\) \(= \sigma/2\varepsilon_0\) \(= 8.85 \times 10^{-8}/(2 \times 8.85 \times 10^{-12})\) \(= 5000\,\text{N/C}\), horizontal and away from the sheet.
(b) \(F\) \(= qE\) \(= 3 \times 10^{-6} \times 5000\) \(= 0.015\,\text{N}\), pushing the ball away from the sheet.
(c) The weight is \(mg = 0.02\,\text{N}\), so \(\tan\theta\) \(= qE/mg\) \(= 0.015/0.02\) \(= 3/4\), and \(\theta = 37°\).
(d) \(T\) \(= \sqrt{(mg)^2 + (qE)^2}\) \(= \sqrt{0.02^2 + 0.015^2}\) \(= 0.025\,\text{N}\).
The \(\varepsilon_0\) route, as for every sheet: \(\varepsilon_0\) \(= 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)\).
Two large parallel sheets each carry \(+\sigma\). What is the field in the space between them?
A wire bent into a square of side \(a\) carries \(\lambda\) per metre. One side is removed. What is the field at the centre of the square?
Two large parallel plates with opposite charges make the one field that is the same everywhere between them. That is why they are used wherever charged particles must be pushed evenly. The electron beam of an oscilloscope, the charged ink drops of a date-coding printer and Millikan's oil drops all move between such plates. In the chapter on Capacitance and Dielectrics the same pair of plates stores charge and energy.
- Thinking a ring's field is largest at its centre.It is zero at the centre and largest at \(x = R/\sqrt2\).
- Using \(2k\lambda/R\) for every arc.That is the semicircle. A general arc gives \((2k\lambda/R)\sin(\theta/2)\).
- Making a sheet's field weaken with distance.A large sheet gives \(\sigma/2\varepsilon_0\) at every distance.
- Using \(\sigma/\varepsilon_0\) for a single sheet.One sheet gives \(\sigma/2\varepsilon_0\); \(\sigma/\varepsilon_0\) is between two opposite sheets.