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Electric Flux and Gauss's Law · Lesson 5 of 5

The Solid Sphere and Concentric Shells

15 min read4 ideas4 checks1 figure to playBy Asheshanand Acharya
What you will learn 0 / 4
Field inside a uniformly charged sphere
Maths you need

Inside a solid charged sphere

Now spread the charge \(Q\) evenly through the volume of a solid ball of radius \(R\), made of an insulator. The charge on each cubic metre is the volume charge density \(\rho = Q/\tfrac43\pi R^3\). The field still points along the radius and depends only on \(r\), so a concentric Gaussian sphere works again. Outside the ball it encloses all of \(Q\), and \(E = kQ/r^2\), as for the shell.

Inside the ball, a Gaussian sphere of radius \(r\) encloses only part of the charge. Charge is spread evenly by volume, and volume grows as \(r^3\). So the enclosed charge is \(Q\,r^3/R^3\). Half-way to the surface, a sphere encloses only \((1/2)^3 = 1/8\) of the charge.

Deriving \(E = \dfrac{kQr}{R^3} = \dfrac{\rho r}{3\varepsilon_0}\) step 1 of 3
1

\[ q_\text{in} = Q\,\dfrac{r^3}{R^3} = \rho\cdot\tfrac43\pi r^3 \]

Why. the charge is spread evenly through the volume, and a sphere of radius \(r\) holds the fraction \(r^3/R^3\) of the volume.

RememberField of a uniformly charged solid sphere
\[ \begin{gathered} E = \dfrac{kQr}{R^3} = \dfrac{\rho r}{3\varepsilon_0} \quad (r \le R) \\ E = \dfrac{kQ}{r^2} \quad (r \ge R) \end{gathered} \]
Holds whenan insulating sphere of radius \(R\) with charge \(Q\) spread evenly through its volume; \(r\) is measured from its centre.
Symbols\(E\) field, \(Q\) total charge, \(R\) radius, \(r\) distance from the centre, \(\rho\) volume charge density, \(k = 1/4\pi\varepsilon_0\)
Units\(E\) in N/C, \(\rho\) in C/m³

The field inside grows in proportion to \(r\). At the centre it is zero, because the ball pulls equally from every side. It is largest at the surface, \(kQ/R^2\), where both forms agree. This is the rule for \(g\) below the Earth's surface in the chapter on Gravitational Force and Field, with charge in place of mass.

ExampleOne sphere, two routes

An insulating sphere of radius \(10\,\text{cm}\) carries \(2\,\mu\text{C}\) spread evenly through its volume. Find the field at \(5\,\text{cm}\) from the centre by the \(k\) route and by the \(\rho\) route. Then find the field at the surface and at \(20\,\text{cm}\).

Try it on paper first. The solution is one tap away when you are ready.
A Gaussian sphere growing through charged bodiesA dashed Gaussian sphere S, centred on a charged body, grows as a slider raises its radius r from 1 to 30 centimetres. Chips choose the body: a thin shell of 2 microcoulombs and radius 10 centimetres, a solid sphere of the same charge and radius with the charge spread through it, or that shell with a second shell of minus 2 microcoulombs at 20 centimetres. The readouts give the charge inside S and the field on it, and a dot runs along the E against r graph. Inside the thin shell E is 0 and it jumps to 18 times ten to the five newtons per coulomb at the surface. Inside the solid sphere E rises in a straight line. Outside every charged body, E falls as one over r squared, as for a point charge. Beyond the two shells E is 0. S 1020305101520Or (cm)E (10⁵ N/C) r = 15 cm, qin = 2 μC E = 8.0 × 10⁵ N/C, outward
SimThe Gaussian sphere of Lesson 4 again. Choose the chip for the solid sphere and move the slider. Inside the ball the enclosed charge grows as \(r^3\), and the field rises in a straight line to the surface.

The same argument works inside a long solid cylinder of radius \(R\) with charge spread through it. A coaxial Gaussian cylinder of radius \(r < R\) and length \(L\) encloses \(\rho\pi r^2L\). So \(E \times 2\pi rL\) \(= \rho\pi r^2L/\varepsilon_0\), and \(E = \rho r/2\varepsilon_0\). The Gaussian surface must lie inside the body, \(r < R\), to find the field inside.

CheckpointUnanswered

A uniformly charged solid sphere has radius \(R\). What is the ratio of the field at \(r = R/2\) to the field at \(r = 2R\)?

Checkpoint · numericUnanswered

The sphere of the example carries \(2\,\mu\text{C}\) through a radius of \(10\,\text{cm}\). How much charge lies within \(5\,\text{cm}\) of its centre, in μC?

μC± 2 %
The e r graph of a charged sphere
The E against r graph of a charged sphereThe field E, in units of ten to the five newtons per coulomb, against the distance r from the centre of a uniformly charged solid sphere. E rises in a straight line from 0 at the centre to its peak of 18 at the surface, r equal to 10 centimetres, and then falls as one over r squared: 9 at 5 centimetres, 4.5 at 20 centimetres. For a thin shell of the same charge and radius, E is 0 inside and jumps to the same peak at the surface, the dashed line; outside the two graphs are one.5101520253036912151821Or (cm)E (10⁵ N/C) (5, 9) (10, 18) (20, 4.5) E ∝ r E ∝ 1/r² dashed: thin shell
Fig 5.1The field of a solid sphere against the distance from its centre, with \(E\) in units of \(10^5\,\text{N/C}\). The dashed line is a thin shell of the same charge and radius.

Fig 5.1 is the graph of the example's sphere. It rises in a straight line from the origin, peaks, and then falls as \(1/r^2\). The peak is at the surface, so its position gives the radius, \(R = 10\,\text{cm}\). Its height gives the charge: \(Q\) \(= ER^2/k\) \(= 1.8 \times 10^6 \times 0.01/(9 \times 10^9)\) \(= 2\,\mu\text{C}\). Half-way up the straight line, at \(5\,\text{cm}\), \(E = 9 \times 10^5\,\text{N/C}\). At twice the radius it is a quarter of the peak, \(4.5 \times 10^5\,\text{N/C}\).

A thin shell of the same charge and radius has the dashed graph. It is zero all the way to the surface and then jumps to the peak. Outside, the two graphs are one line, because outside both bodies act as the same point charge. The solid sphere reaches half its peak twice: at \(R/2\) inside, and at \(\sqrt2 R \approx 14.1\,\text{cm}\) outside.

Checkpoint · numericUnanswered

The E-r graph of a uniformly charged solid sphere peaks at \(r = 10\,\text{cm}\) with \(E = 1.8 \times 10^6\,\text{N/C}\). Beyond the surface, at what distance from the centre is the field half the peak, in cm?

cm± 2 %
Two concentric shells

Shells inside shells

Two thin shells share a centre: one of radius \(10\,\text{cm}\) with \(+2\,\mu\text{C}\), and one of radius \(20\,\text{cm}\) with \(-2\,\mu\text{C}\). To find the field at any point, draw a Gaussian sphere through it and count the charge inside. A shell outside the point adds nothing, just as a single shell gives no field inside itself.

At \(P\), \(5\,\text{cm}\) from the centre, nothing is enclosed, so \(E = 0\). At \(Q\), \(15\,\text{cm}\) out, only the inner shell counts. So \(E\) \(= 9 \times 10^9 \times 2 \times 10^{-6}/(0.15)^2\) \(= 8 \times 10^5\,\text{N/C}\), outward. At \(S\), \(30\,\text{cm}\) out, the enclosed charge is \(+2 - 2 = 0\), so \(E = 0\) again. Choose the chip for two shells in the figure of Lesson 4 and move the slider to see all three.

A Gaussian sphere growing through charged bodiesA dashed Gaussian sphere S, centred on a charged body, grows as a slider raises its radius r from 1 to 30 centimetres. Chips choose the body: a thin shell of 2 microcoulombs and radius 10 centimetres, a solid sphere of the same charge and radius with the charge spread through it, or that shell with a second shell of minus 2 microcoulombs at 20 centimetres. The readouts give the charge inside S and the field on it, and a dot runs along the E against r graph. Inside the thin shell E is 0 and it jumps to 18 times ten to the five newtons per coulomb at the surface. Inside the solid sphere E rises in a straight line. Outside every charged body, E falls as one over r squared, as for a point charge. Beyond the two shells E is 0. S 1020305101520Or (cm)E (10⁵ N/C) r = 15 cm, qin = 2 μC E = 8.0 × 10⁵ N/C, outward
SimChoose the chip for two shells. Between them the field is the inner shell's alone. Beyond the outer shell the enclosed charge is zero, and so is the field.
Checkpoint · numericUnanswered

Thin concentric shells of radii \(10\,\text{cm}\) and \(20\,\text{cm}\) carry \(+2\,\mu\text{C}\) and \(+4\,\mu\text{C}\). What is the field \(30\,\text{cm}\) from the centre, in N/C? Take \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).

N/C± 2 %
When gauss cannot give e
Common mistake

Gauss's law is always true, but it does not always give \(E\). Take a finite charged rod, a charged cube or a disc. No closed surface around it has a field of one size, along the normal, wherever it crosses. The law still gives the flux, but the flux is an integral of a field that varies, and one number cannot tell how. For these bodies, go back to adding the fields of small pieces, as in the chapter on Coulomb's law.

Think of a finite rod and a Gaussian cylinder around it. Near the ends the field bends outward and crosses the flat faces. The curved surface then has a field that changes along its length. Gauss's law gives the total flux, \(q_\text{in}/\varepsilon_0\), but not the field at any one point. Symmetry is what turns the law into a way to find a field.

Deeper

Two cases for JEE Advanced use the solid sphere with a twist. First, cut a spherical cavity out of a charged sphere, with the cavity's centre displaced by \(\vec a\). Treat it as the full sphere plus a small sphere of charge \(-\rho\). The field in the cavity is \(\rho\vec a/3\varepsilon_0\), the same at every point. Second, a ball whose density falls as \(\rho_0(1 - r/R)\) has \(E\) \(= (\rho_0 r/3\varepsilon_0)(1 - 3r/4R)\) inside. This field is largest at \(r = 2R/3\), not at the surface.

Story · J. J. Thomson, 1904
An atom made of a charged sphere

In 1904 J. J. Thomson, who had found the electron, proposed a model of the atom. He pictured a ball of positive charge spread evenly through its volume, with electrons sitting inside it. Inside such a ball the field is \(\rho r/3\varepsilon_0\), so an electron moved off centre is pulled back with a force in proportion to its distance. Thomson hoped such oscillating electrons would explain the lines of atomic spectra. Rutherford's experiments soon showed the positive charge sits in a tiny nucleus instead.

Exam traps
  1. Using the whole charge \(Q\) inside a solid sphere.
    A Gaussian sphere of radius \(r < R\) encloses only \(Qr^3/R^3\), so \(E = kQr/R^3\).
  2. Treating a solid sphere and a shell alike inside.
    A shell gives zero inside; a solid sphere gives a field that grows with \(r\). Outside they are the same.
  3. Counting an outer shell's charge at a point inside it.
    Draw the Gaussian sphere through the point. Charges outside it add no flux, and by symmetry no field.