Electrostatic pressure, the force between two halves of a charged sphere and the charged soap bubble go beyond the Class 12 syllabus. They are JEE Advanced material. The breakdown of air, in the last part of the lesson, is for every reader.
A push on every patch
The charges on a conductor repel one another, so every part of the surface is pushed outward. To find the push, take a small patch of the surface, with charge density \(\sigma\). The field just outside is \(\sigma/\varepsilon_0\), and just inside it is zero. Which of these acts on the patch's own charge? Neither, because a charge feels no force from its own field.
Split the field into two parts: the field of the patch itself and the field of the rest of the conductor. Seen from very close, the patch is a small flat sheet. It gives \(\sigma/2\varepsilon_0\) outward just outside and \(\sigma/2\varepsilon_0\) inward just inside. The rest of the conductor is far from both points, so it gives one field, the same at both.
\[ E_\text{rest} + \dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma}{\varepsilon_0} \]
Why. just outside, the total field of a conductor is \(\sigma/\varepsilon_0\); the patch gives \(\sigma/2\varepsilon_0\) outward there.
\[ E_\text{rest} = \dfrac{\sigma}{2\varepsilon_0} \]
Why. the same value makes the field inside zero, since there the patch gives \(\sigma/2\varepsilon_0\) inward.
\[ dF = (\sigma\,dA)\,\dfrac{\sigma}{2\varepsilon_0} \]
Why. the force on the patch's charge \(\sigma\,dA\) comes only from the field of the rest.
\[ P = \dfrac{dF}{dA} = \dfrac{\sigma^2}{2\varepsilon_0} = \tfrac12\varepsilon_0E^2 \]
Read it. divide by the area. With \(\sigma = \varepsilon_0E\), where \(E\) is the field just outside, the second form follows.
The pressure depends on \(\sigma^2\), so it pushes outward whatever the sign of the charge. A negatively charged surface has its field pointing inward, and the force on negative charge is then outward. This outward push is electrostatic pressure. The same expression, \(\tfrac12\varepsilon_0E^2\), returns in the chapter Capacitance and Dielectrics as the energy stored in each cubic metre of field.
The field just outside a charged conductor is \(\sigma/\varepsilon_0\). Why is the force per unit area on its surface \(\sigma^2/2\varepsilon_0\) and not \(\sigma \times \sigma/\varepsilon_0\)?
How large is the push?
Take a strong field of \(10^6\,\text{N/C}\) just outside a conductor. The charge density is \(\sigma\) \(= \varepsilon_0E\) \(= 8.85 \times 10^{-6}\,\text{C/m}^2\). The pressure is \(\tfrac12\varepsilon_0E^2\) \(= 0.5 \times 8.85 \times 10^{-12} \times 10^{12}\) \(\approx 4.4\,\text{N/m}^2\). The air presses on the same surface with about \(10^5\,\text{N/m}^2\). Electrostatic pressure is small, but it acts on every patch.
Now cut a charged sphere in two halves. Each patch of the right half is pushed along its own normal. The sideways parts of these pushes cancel in pairs. What remains is the part along the axis. It adds up to the pressure times the projected area of the half, a disc \(\pi R^2\), as for a liquid pressing on a curved wall.
The field just outside a charged metal sphere of radius \(10\,\text{cm}\) is \(10^6\,\text{N/C}\). Find the charge density, the pressure on its surface and the force pushing its two halves apart. Take the \(\varepsilon_0\) route.
\(\sigma\) \(= \varepsilon_0E\) \(= 8.85 \times 10^{-12} \times 10^6\) \(= 8.85 \times 10^{-6}\,\text{C/m}^2\).
\(P\) \(= \sigma^2/2\varepsilon_0\) \(= (8.85 \times 10^{-6})^2/(2 \times 8.85 \times 10^{-12})\) \(\approx 4.4\,\text{N/m}^2\).
\(F\) \(= P\pi R^2\) \(= 4.4 \times \pi \times 0.01\) \(\approx 0.14\,\text{N}\). If the sphere were made of two loose halves, they would need this force to hold them together.
Values used: \(\varepsilon_0\) \(= 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)\).
The field just outside a charged conductor is \(2 \times 10^6\,\text{N/C}\). What is the electrostatic pressure on its surface, in N/m²? Take \(\varepsilon_0\) \(= 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)\).
A bubble held open by charge
A soap bubble has two surfaces, and its film pulls inward. In the chapter on Surface Tension, the air inside a bubble must be at an excess pressure \(4T/r\) to hold it open. Here \(T\) is the surface tension and \(r\) the radius. Now charge the bubble. The charge spreads over it and pushes the film outward with \(\sigma^2/2\varepsilon_0\). Charged just enough, the bubble needs no extra air pressure at all.
A soap bubble of radius \(1\,\text{cm}\) has surface tension \(T = 0.03\,\text{N/m}\). How much charge must it carry for the electrostatic pressure to balance the surface-tension pressure exactly? Take the \(\varepsilon_0\) route.
Balance: \(\sigma^2/2\varepsilon_0 = 4T/r\).
\(4T/r\) \(= 4 \times 0.03/0.01\) \(= 12\,\text{N/m}^2\), so \(\sigma\) \(= \sqrt{2\varepsilon_0 \times 12}\).
\(\sigma\) \(= \sqrt{2 \times 8.85 \times 10^{-12} \times 12}\) \(\approx 1.46 \times 10^{-5}\,\text{C/m}^2\). Then \(Q\) \(= \sigma \times 4\pi r^2\) \(\approx 1.46 \times 10^{-5} \times 1.26 \times 10^{-3}\) \(\approx 1.8 \times 10^{-8}\,\text{C}\).
The bubble needs about \(18\,\text{nC}\). The field just outside is then \(\sigma/\varepsilon_0\) \(\approx 1.6 \times 10^6\,\text{V/m}\), below the breakdown of air, so the air holds.
A soap bubble of radius \(2\,\text{cm}\) has \(T = 0.03\,\text{N/m}\). What charge makes its electrostatic pressure balance its surface-tension pressure, in nC? Take \(\varepsilon_0\) \(= 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)\).
The limit set by the air
Dry air is a good insulator up to a point. When the field reaches about \(3 \times 10^6\,\text{V/m}\), stray electrons gain enough energy between collisions to knock more electrons out of molecules. The air then conducts, and a spark or a glow appears. This largest field an insulator can stand is its dielectric strength. For air it is about \(3 \times 10^6\,\text{V/m}\).
A charged sphere has its strongest field just outside its surface, \(kQ/R^2\). So charge can be added only until \(kQ/R^2\) reaches \(E_b\), the breakdown field. For a sphere of radius \(10\,\text{cm}\), \(Q_\text{max}\) \(= E_bR^2/k\) \(= 3 \times 10^6 \times 0.01/(9 \times 10^9)\) \(\approx 3.3\,\mu\text{C}\). Its potential is then \(V_\text{max}\) \(= kQ/R = E_bR\) \(= 3 \times 10^5\,\text{V}\).
\(V_\text{max} = E_bR\) grows with the radius. So a big smooth sphere can reach a high potential, and a small one cannot. A sphere of radius \(0.5\,\text{m}\) could reach \(1.5 \times 10^6\,\text{V}\) before the air broke down. A sharp point of radius \(1\,\text{mm}\) reaches the breakdown field at only \(3000\,\text{V}\). This is why a point leaks, as the lightning conductor of Lesson 2 does on purpose.
A metal sphere of radius \(20\,\text{cm}\) is charged slowly in dry air, which breaks down at \(3 \times 10^6\,\text{V/m}\). What is the largest charge it can hold, in μC? Take \(k\) \(= 9 \times 10^9\,\text{N m}^2/\text{C}^2\).
Why is the dome of a machine built to reach a very high potential large and smooth, with no sharp edges?
A school Van de Graaff generator has a dome of radius about \(15\,\text{cm}\). Estimate the highest potential it could reach in dry air, in volts.
The Van de Graaff generator at a school science fair carries charge up a moving rubber belt. The belt hands it to the inside of a metal dome. Charge given to the inside of a conductor moves at once to its outer surface. So the belt can keep adding charge, however high the dome's potential already is. A student who stands on an insulating stool and touches the dome finds her hair standing on end. Each hair carries charge of one sign and is repelled by the others.
- Using \(\sigma \times \sigma/\varepsilon_0\) for the force per unit area on a conductor.Half of the field outside is the patch's own. Only the rest, \(\sigma/2\varepsilon_0\), pushes on it, so \(P = \sigma^2/2\varepsilon_0\).
- Using the full curved area for the force on half a sphere.The sideways pushes cancel. The net force is the pressure times the projected area \(\pi R^2\).
- Expecting a smaller sphere to reach a higher potential because its field is stronger.The field limit is fixed by the air. A sphere reaches at most \(E_bR\), so a smaller sphere reaches a lower potential.