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Lecture 5 : Conductors and Dipoles • Conductors • Capacitors • Electric dipoles Recap • Gauss’s Law 𝐸. 𝑑 𝐴 = 𝑄𝑒𝑛𝑐 𝜀0 and Maxwell’s 1st equation 𝛻. 𝐸 = 𝜌 𝜀0 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 Conductors • We’ll now consider the behaviour of 𝐸 in materials, which we divide into conductors and insulators Conductors • In a conductor, charges can flow freely • In practice, this usually involves free electrons moving within an ionic lattice 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 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 Conductors • An application of this effect is electrostatic shielding 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 crosssectional area 𝐴 crossing the surface 𝐴 𝐸=0 • 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 𝑄𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 = 𝜎𝐴 • Applying Gauss’s Law: 𝑬= 𝝈 𝜺𝟎 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 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 Capacitors • Capacitors are useful for storing charge (or, potential energy) and then releasing it Capacitors • What is the capacitance of a parallel-plate capacitor, where the plates have area 𝐴 and separation 𝑑? Separation 𝑑 Area 𝐴 +𝑄 𝐸= 𝜎 𝜀0 −𝑄 • From Gauss’s Law (previous slides), electric field 𝐸 = • Capacitance 𝐶 = 𝑄 𝑉 = 𝜎×𝐴 𝐸×𝑑 = 𝜀0 𝐴 𝑑 𝜎 𝜀0 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/𝜀0 or 𝐸 = 𝜆/2𝜋𝜀0 𝑟 • Potential difference between the 𝑏 cylinders is 𝑉 = 𝑎 𝐸 𝑑𝑟 = 𝑏1 𝜆 2𝜋𝜀0 𝑎 𝑟 𝑑𝑟 = • Capacitance 𝐶 𝜆 𝑏 𝑙𝑜𝑔𝑒 2𝜋𝜀0 𝑎 = 𝜆 𝑉 = 2𝜋𝜀0 𝑏 𝑙𝑜𝑔𝑒 𝑎 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 Electric dipoles What is the electric field strength along the axis? 𝑃 𝐸 𝑟 • Electric potential: 𝑉𝑃 = • Electric field 𝐸𝑥 = − • 𝐸∝ 1 𝑟2 𝑑𝑉 𝑑𝑟 𝑞 𝑑 4𝜋𝜀0 (𝑟−2 ) = + −𝑞 𝑑 4𝜋𝜀0 (𝑟+2 ) ≈ 𝑝 4𝜋𝜀0 𝑟 2 2𝑝 4𝜋𝜀0 𝑟 3 for a point charge, and 𝐸 ∝ 1 𝑟3 for a dipole 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 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 𝑄 potential difference is 𝑉, then 𝐶 = 𝑉 • Two charges ±𝑞 separated by distance 𝑑 form an electric dipole 𝑝 = 𝑞 𝑑, which produces an electric field 𝐸 ∝ 1/𝑟 3