Capacity to store charge Capacitance Capacity to store charge C = Q/V L a b Q = Ll E = l/2pe0r V = -(l/2pe0)ln(r/a) C = 2pe0L/ln(b/a) Dimension e0 x L (F/m) x m

Capacity to store charge Capacitance Capacity to store charge C = Q/V Q = Ars E = rs/e0 V = Ed C = e0A/d A d (F/m) x m Increasing area increases Q and decreases C Increasing separation increases V and decreases Q

Capacitance Capacitor microphone – sound vibrations move a diaphragm relative to a fixed plate and change C Tuning  rotate two cylinders and vary degree of overlap with dielectric  change C Changing C changes resonant frequency of RL circuit Increasing area increases Q and decreases C Increasing separation increases V and decreases Q

Increasing C with a dielectric - + bartleby.com e/e0 = er C  erC To understand this, we need to see how dipoles operate They tend to reduce voltage for a given Q

Effect on Maxwell equations: Reduction of E Point charge in free space in a medium .E = rv/e0er .E = rv/e0

Point Dipole R >> d p. R 4pe0R2 V = _________ Note 1/R2 !

R >> d Point Dipole p. R 4pe0R2 V = _________ Note 1/R2 !  E = -V = -RV/R – (q/R)V/q p(2Rcosq + qsinq) 4pe0R3 _____________

How do fields create dipoles? + - J + - E J Let’s review what happens in a metal conductivity A field creates a current density, J=sE, which moves charges to opposite ends, creating an inverse field that completely screens out E

How do fields create dipoles? - + Charges are not free to move in a dielectric! But electrons can be driven by E a bit away from the Nucleus without completely leaving it, creating an excess Charge on one side and a deficit on the other, .... …. in other words, generating a dipole

+ - Rotating a Dipole F+ = qE F- = -qE T = (d/2 x F+) - (d/2 x F+) = qd x E = p x E

That’s how micro- wave ovens work! p = 6.2 x 10-30 Cm 2.5 GHz Radio wave source Absorbed by water, sugar, fats Aligns dipoles built in water molecule and excites atoms “Friction” during rotation in opp. directions causes heating

Polarization and Dielectrics - + + - P - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + - + Instead of creating new dipoles, E could align existing atomic dipoles (say on H20) creating a net polarization P  E

Metal conductor vs Dielectric Insulator Either way, the end result is excess charge on one end and a deficit on the other, like a metal… BUT… There are differences!! In a metal, E forms a current of freely moving charge, and the applied E gets cancelled completely In a dielectric, E creates a polarization of bound but aligned, distorted but immobile charges, and the applied E gets reduced partially (by the dielectric constant).

Metal conductor vs Dielectric Insulator A metal is characterized by a conductivity s which determines its resistance R to current flow, J=sE A dielectric is characterized by a susceptibility c (and thus a dielectric constant) which determines its capacitance C to store charge, P = Np = e0cE

Dipoles Screen field + E=(D-P)/e0 D = e0E + P - P = e0cE D = eE Displacement vector + - -P (opposing polarization Field) E=(D-P)/e0 D = e0E + P D (unscreened Field) P = e0cE D = eE e = e0(1+c) Relative Permittivity er Thus the unscreened external field D gets reduced to a screened E=D/e by the polarizing charges For every free charge creating the D field from a distance, a fraction (1-1/er) bound charges screen D to E=D/e

Free vs Bound Charges + E=(D-P)/e0 - -P (opposing polarization D Field) E=(D-P)/e0 D (unscreened Field) e0.E = rtotal = .D - .P = rfree + rbound

Effect on Capacitance + - + - e/e0 = K Since the same charge on the which is why capacitors employ dielectrics (bartleby.com) + Free charges on metal + - Polarizing charges in dielectric - e/e0 = K Since the same charge on the plates is now supported by a smaller, screened potential, the capacitance (charge stored by applying unit volt) has actually increased by placement of a dielectric inside! Due to screening, only few of the field lines originating on free charges on the metal plates survive in the Dielectric inside the capacitor

Effect on Maxwell equations: Reduction of E .D = rv  x E = 0 Differential eqns (Gauss’ law) Fields diverge, but don’t curl Defines Scalar potential E = -U Integral eqns  D.dS = q  E.dl = 0 D = eE Constitutive Relation (Thus, E gets reduced by er) er = 1 (vac), 4 (SiO2), 12 (Si), 80 (H20) ~2 (paper), 3 (soil, amber), 6 (mica),

Point charge in free space in a medium .E = rv/e0er .E = rv/e0

Electrostatic Boundary values Maxwell equations for E Supplement with constitutive relation D=eE .D = r  x E = 0 D1n  D.dS=q  E.dl = 0 rs D2n Use Gauss’ law for a short cylinder Only caps matter (edges are short!) D1n-D2n = rs Perpendicular D discontinuous

Electrostatic Boundary values Maxwell equations for E Supplement with constitutive relation D=eE .D = r  x E = 0  D.dS=q  E.dl = 0 E1t E2t No net circulation on small loop Only long edges matter (heights are short!) E1t-E2t = 0 Parallel E continuous

Electrostatic Boundary values Perpendicular D discontinuous Parallel E continuous Can use this to figure out bending of E at an interface (like light bending in a prism)

Example: Bending Parallel E continuous Perpendicular D continuous if no free charge rs at interface e1 e2 q1 q2 E1t=E2t cosq1=cosq2 tanq1/tanq2 = e2/e1 e1E1n=e2E2n D1n=D2n e1sinq1=e2sinq2 e2 > e1 means q2 < q1

Conductors are equipotentials Conductor  Static Field inside zero (perfect screening) Since field is zero, potential is constant all over Et continuity equation at surface implies no field component parallel to surface Only Dn, given by rs.

Field lines near a conductor + - Equipotentials bunch up here  Dense field lines Principle of operation of a lightning conductor Plot potential, field lines

Images So can model as Charge above Compare with Ground plane field of a Dipole! So can model as Charge + Image Charge above Ground plane (fields perp. to surface) Equipotential on metal enforced by the image

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