1. Lecture 5 : Conductors and Dipoles
Capacitors
Electric dipoles

2. Recap
Gauss’s Law 𝐸 .𝑑 𝐴 = 𝑄 𝑒𝑛𝑐 𝜀 0 and Maxwell’s 1st equation 𝛻 . 𝐸 = 𝜌 𝜀 0 are equivalent integral and differential formulations for the electric field 𝐸 produced by charge density 𝜌
The electric field 𝐸 must also satisfy Maxwell’s 2nd equation 𝛻 × 𝐸 = 0 . This implies that the electric field can be generated by the gradient of an electrostatic potential 𝑉, 𝐸 =− 𝛻 𝑉

3. Conductors
We’ll now consider the behaviour of 𝐸 in materials, which we divide into conductors and insulators

4. Conductors In a conductor, charges can flow freely
In practice, this usually involves free electrons moving within an ionic lattice

5. Place charge 𝑄 on a conducting sphere
Conductors
What can we say about the electric field in and around a charge-carrying conductor in equilibrium?
Place charge 𝑄 on a conducting sphere

6. Conductors
What can we say about the electric field in and around a charge-carrying conductor in equilibrium?
First, all charge must be located on the surface (otherwise it would move under the effect of forces from other charges)
Hence from Gauss’s Law, 𝑬 = 𝟎 inside a conductor
Hence, all points of the conductor are at constant electrostatic potential

7. Conductors
An application of this effect is electrostatic shielding

8. Conductors What about the electric field just outside the conductor? 𝐸
There can be no component of 𝐸 parallel to the surface, otherwise charges would move
Consider a Gaussian cylinder of cross- sectional area 𝐴 crossing the surface
𝐸 is perpendicular to the surface and zero inside, such that 𝐸 .𝑑 𝐴 =𝐸×𝐴
Let the charge per unit area at the surface be 𝜎, then 𝑄 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 =𝜎𝐴
Applying Gauss’s Law: 𝑬= 𝝈 𝜺 𝟎 𝐸 𝐴
𝐸 = 0

9. Electric field around conductor
Conductors
The electric field just outside a conductor is perpendicular to the surface and proportional to the charge density
Electric field around conductor
Conductor in applied field

10. Capacitors
A capacitor is a very useful circuit component formed by two parallel conductors separated by an insulator (or “dielectric”)
When connected to a battery at potential 𝑉, charge ±𝑄 flows onto the plates. The capacitance is 𝐶=𝑄/𝑉 [unit: Farads, F] +𝑄 𝑉 -𝑄

11. Capacitors
Capacitors are useful for storing charge (or, potential energy) and then releasing it

12. Capacitors
What is the capacitance of a parallel-plate capacitor, where the plates have area 𝐴 and separation 𝑑?
From Gauss’s Law (previous slides), electric field 𝐸= 𝜎 𝜀 0
Capacitance 𝐶= 𝑄 𝑉 = 𝜎×𝐴 𝐸×𝑑 = 𝜀 0 𝐴 𝑑
Area 𝐴 +𝑄
𝐸= 𝜎 𝜀 0
Separation 𝑑 −𝑄

13. Capacitors
What is the capacitance of a pair of concentric cylinders of radii 𝑎 and 𝑏>𝑎?
Suppose the charge per unit length on the cylinders is ±𝜆
Applying Gauss’s Law to a cylinder of radius 𝑟 and length 𝐿, we find 𝐸× 2𝜋𝑟𝐿=𝜆𝐿/ 𝜀 0 or 𝐸=𝜆/2𝜋 𝜀 0 𝑟
Potential difference between the cylinders is 𝑉= 𝑎 𝑏 𝐸 𝑑𝑟 = 𝜆 2𝜋 𝜀 0 𝑎 𝑏 1 𝑟 𝑑𝑟 = 𝜆 2𝜋 𝜀 0 𝑙𝑜𝑔 𝑒 𝑏 𝑎
Capacitance 𝐶= 𝜆 𝑉 = 2𝜋 𝜀 0 𝑙𝑜𝑔 𝑒 𝑏 𝑎

14. Electric dipoles
An electric dipole consists of two charges +𝑞 and −𝑞 separated by distance 𝑑. It is neutral but produces an 𝐸 -field
A dipole is a good model for many molecules!
Dipole moment

15. What is the electric field strength along the axis?
Electric dipoles
What is the electric field strength along the axis? 𝑃 𝐸 𝑟
Electric potential: 𝑉 𝑃 = 𝑞 4𝜋 𝜀 0 (𝑟− 𝑑 2 ) + −𝑞 4𝜋 𝜀 0 (𝑟+ 𝑑 2 ) ≈ 𝑝 4𝜋 𝜀 0 𝑟 2
Electric field 𝐸 𝑥 =− 𝑑𝑉 𝑑𝑟 = 2𝑝 4𝜋 𝜀 0 𝑟 3
𝐸∝ 1 𝑟 2 for a point charge, and 𝐸∝ 1 𝑟 3 for a dipole

16. Electric dipoles
When a dipole 𝑝 is placed in an electric field 𝐸 , it feels no net force (since it is neutral) but it feels a net torque 𝜏 = 𝑝 × 𝐸 𝐸

17. Summary
Conductors are materials in which charges can flow freely. All charge will reside on the surface, and 𝐸 = 0 inside
Two separated conductors storing charge ±𝑄 form a capacitor 𝐶. If the potential difference is 𝑉, then 𝐶= 𝑄 𝑉
Two charges ±𝑞 separated by distance 𝑑 form an electric dipole 𝑝 =𝑞 𝑑 , which produces an electric field 𝐸∝ 1/ 𝑟 3