Lecture The use of dielectrics 1. Capacitance and E-field energy

Learning Objectives To calculate the energy stored in a capacitor To derive an expression for the energy per unit volume To obtain an expression for the force between the two plates of a parallel plate capacitor To introduce the use of dielectric materials

Learning Objectives To calculate the energy stored in a capacitor To derive an expression for the energy per unit volume To obtain an expression for the force between the two plates of a parallel plate capacitor To introduce the use of dielectric materials

Energy Stored in a Capacitor Transfer dq between the two conductors whilst at a potential difference of V q

Stored in the electric field between the “plates” of a capacitor

Consider a parallel plate capacitor d

Energy Density (energy stored per unit volume) Note this is generally valid

What is the Force Experienced by a Capacitor Plate? Cannot use Coulomb’s law directly(integration very complicated.) Easiest approach is to use potential energy Use (STMR)

Why is it negative? An attractive force between the plates

Summary on electrostatics Point charge: E,V, U Electric dipole: E and V Infinite plane of charge: E Charged cylinder, Charged sphere (solid as well as shell) Capacitors in planar geometry, cylindrical geometry, and spherical geometry The second half of the lecture module: magnetism

24-5 Dielectrics

Review and Summary Electrostatic potential energy U of a charged capacitor, is given by U represents the work required to charge the capacitor Stored energy density with an electric field

Dielectrics  E-Field in a Dielectric  Induced Surface Charge  Polarisation inside a Dielectric Note that this material in non-examinable, but it will be needed for your second year course

What is a dielectric? A dielectric is an insulator – e.g. air, glass, paper, teflon Where is it used? It is placed between conducting surfaces What are the functions of a dielectric? To keep the surfaces physically separated To raise the C of a capacitor To reduce the chance of electric breakdown. Larger V possible (more charge and energy stored)

_ _ _ _ _ _ _ Consider a parallel plate capacitor Perfect voltmeter V0V0 C0C0

_ _ _ _ _ _ _ Perfect voltmeter V0V0 C0C0 Consider a parallel plate capacitor

_ _ _ _ _ _ _ Perfect voltmeter V d < V 0 CdCd Charge is the same,  C d > C 0 Define the dielectric constant as Consider a parallel plate capacitor

The electric permittivity  is defined by the equation  = K  0 K is often referred to as the relative permittivity

Dielectrics are chosen to have the following properties: K independent of V or E - linearity K independent of the orientation of the dielectric - isotropy K has the same value for every part of the dielectric - homogeneity Linear, isotropic, homogeneous (LIH) dielectrics

What is the effect of K on electrostatic formulae? For a LIH dielectric replace  0 with , i.e.

24-6 Molecular View of a Dielectric

Molecular Explanation of Dielectric Behaviour Schematic representation of a molecule

Molecular Explanation of Dielectric Behaviour The molecule is polarised E ind

Polarisation of a dielectric gives rise to thin layers of bound charges on the surfaces.

E0E0 E = E 0 /K

(Gauss’s law in a dielectric)

Note: conductors have free charge carriers - they are completely polarised in that inside they make E = zero

Review and Summary Electrostatic potential energy U of a charged capacitor, is given by U represents the work required to charge the capacitor Stored energy density with an electric field

Review and Summary Capacitance with a Dielectric Capacitance increases by K (dielectric constant) Gauss’s Law with a Dielectric

Next An introduction to Magnetic Fields and their Properties

Cosmic rays bombard the Earth from outer space and experiments show that they are mostly positively charged protons arriving uniformly over the surface at a rate of 1 proton m-2 s-1 with an average energy of MeV. By treating the Earth as a conducting sphere of radius 6371 km calculate the rate of change of potential as a result of this bombardment. At this rate how long would it take for the Earth to charge sufficiently to repel all the cosmic rays? In view of the facts that the age of the Earth is approximately 4.5  10 9 years and that cosmic rays are arriving today, comment on your result.

Number of protons hitting the Earth per second = Therefore time to reach volts necessary to deflect protons =.