EE 3321 Electromagnetic Field Theory Potential Energy Energy

EE 3321 Electromagnetic Field Theory Potential Energy, Energy Density Capacitance, Polarization Boundary Conditions

Potential Energy �The electric field and potential energy are directly related: �As a test charge +q moves in the direction that the field opposed it, its potential energy increases. �The electrostatic potential energy is the energy of an electrically charged particle (at rest) in an electric field. �The energy difference between two potentials is given by U = q(V 2 – Vref) Joules (VAs)

Example Potential Energy U = +q. V Charge Q R V(∞→R) =Vref = 0 Test Charge +q Potential V = k. Q R

Observations �The potential at infinity is zero �A positive test charge +q gains potential as it gets closer to the charge +Q �A negative test charge –q loses potential as it gets closer to the charge +Q.

Cathode Ray Tube �The CRT is a vacuum tube containing an electron gun (a source of electrons) and a fluorescent screen used to create images in the form of light emitted from the fluorescent screen. �The image may represent electrical waveforms (oscilloscope), pictures (television, computer monitor), radar targets and others.

CRT Simplified Set Up �Once the electrons leave the cathode, they accelerate toward the grid. �Electrons entering the deflecting plate region and change directions depending on the voltage between the plates.

Exercise �An electron moves at a constant velocity v = vo ax. Assume that the electron enters in a field E = - 1 a z (V/m) at x =0. �Compute the potential energy U the electron loses as it moves from A (at z = 0 cm) to B (at z = 0. 5 cm). Recall e = 1. 602 x 10 – 19 As.

Energy Stored in an E Field �Using Gauss’ Law in differential form and the Divergence theorem and it can be shown that the energy density or energy per unit volume (J/m 3) of the electric field is: u = ½ є |E|2 (Joules/m 3) �The total energy stored in the electrostatic field is U = ∭ u d. V (Joules) where d. V is the volume differential.

Example �Let E = 9 V/mm in between the plates. Suppose that the area A = 1 cm 2 and the dielectric thickness is d = 1 mm. Find the energy stored by the capacitor for a relative permittivity of 2. 8. Neglect (field) fringing effects. �Notice that the field is constant �Calculate the energy density �Calculate the volume between the capacitor plates

Exercise �Calculate the energy stored in the field produced by a metal sphere of radius a holding a charge Q. �Determine the electric field E �Find the energy density u �Set up the integral for U �Integrate over space a<R< ∞

Capacitance �As shown above a capacitor consists of two conductors separated by a non-conductive region. �The non-conductive substance is called the dielectric medium. �The conductors contain equal and opposite charges on their facing surfaces, and the dielectric contains an electric field. �A capacitor is assumed to be self-contained and isolated, with no net electric charge and no influence from an external electric field.

Capacitance �An ideal capacitor is wholly characterized by its capacitance C (in Farads), defined as the ratio of charge ±Q on each conductor to the voltage V between them C = Q/V �More generally, the capacitance is defined in terms of incremental changes C = dq/dv

Parallel Plate Capacitor �From Gauss’ Law the charge and the electric field between the plates is related by �Likewise, the line integral relating the potential and the electric field simplifies to �Thus the capacitance is given by

Exercise �Consider a parallel plate capacitor. Derive an expression for the stored energy U in terms of the capacitance C and the potential V.

Polarization �Suppose that a capacitor is charged up by connecting it to a voltage source V which is then removed. � A fixed charge Q is placed on its upper plate and –Q on the lower plate. �Suppose the capacitor is air filled. In this case, �The capacitance is

Polarization �Next assume that the capacitor is filled with dielectric material as illustrated here. �Since the charge does not change, the electric flux D is the same as before. �However, the electric field E changes to

Polarization �The decrease of E is said to be due to the polarization P of the dielectric molecules which opposes E: �But the capacitance increases to

Exercise �The relative permittivity of air is 1. 0 and that of quartz is about 4. 5. Calculate the difference in capacitance for two capacitors with identical geometry using these two dielectric materials.

Electric Boundary Conditions �On a perfect conductor �The component of E parallel to the conducting surface is zero �The component of D normal to the conducting surface is numerically equal to the charge density �On a perfect dielectric material �The component of E parallel to the interface is continuous �The component of D normal to the interface is continuous

Perfect Dielectric Medium �Tangential components of E are continuous E 1 t = E 2 t Medium 1 (air) Medium 2 WATER DROPPLET

Perfect Dielectric Medium �Normal components of E are discontinuous ε 1 E 1 n = ε 2 E 2 n Medium 1 (air) no free charges Medium 2 WATER DROPPLET

Perfect Conductor Medium �Tangential components of E are zero E 1 t = E 2 t = 0 Medium 1 (air) X Short circuit Medium 2 WATER DROPPLET

Perfect Conductor Medium �Normal components of E are discontinuous E 1 n≠ 0 Medium 1 (air) E 2 n= 0 Medium 2 WATER DROPPLET

Examples of Field Lines

Exercise �Consider a dielectric interface at z = constant. �Let єr 1 = 2, єr 2 = 5, and E 1 = 2 ax + 3 ay + 5 az �Find E 2 єr 1 = 2 єr 2 = 5

Homework �Read textbook sections 4 -8, 4 -9, 4 -10, 4 -11 �Solve problems 4. 43, 4. 45, 4. 50, 4. 51, 4. 52, 4. 54