Relation between Electric Field and Potential in Physics

E = -V Relation between field and potential r 2 V(r) = - -(V2-V1) along any path ∫E.dl = ∫E.dl 1 ∞ • Set reference point 1 at infinity where V1 = 0 • Use linear path from infinity to desired point 2, coordinate r

Poisson’s equation Laplace’s equation 2V = 0 Free space • x E = 0 E = -V • . E = r/e0 2V = - r/e0 Equation satisfied by potential

    Continuous distribution of charges E = ________ R’ d3 R ’ V = ________ d3 R ’ R ‘ rv(R’) 4pe0|R’|3 rv(R’) 4pe0|R’|

V = ? Charged Sheet Charged Hollow Ball Charged Ball Various examples Charged Disc Charged line Calculate V directly, or from E obtained through Gauss’ Law What do equipotentials and field lines look like?

Equipotentials Connect pts. with same V E = -V runs perpendicular to it Familiar examples Equipotentials Point charge

How to draw equipotentials?

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

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

So dipoles annihilate each other, thereby countering the field that separated and created them in the first place. In other words, they conspire to produce a polarizing field.

http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.htmlhttp://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html

Conductors are equipotentials • Conductor  Static Field inside zero (perfect screening) • Since field is zero, potential is constant all over • E is perpendicular to the conducting surface

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

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

How much charge can we store on a metal? We can calculate the voltage on a metal for a given charge. Conversely, we can calculate the charge we need to store to create a given voltage on a metal. How would we quantify the charge that is needed to create 1 volt on a metal? The ‘Capacitance’

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

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

Dipoles Screen field - + E=(D-P)/e0 - + -P (opposing polarization Field) - + D (unscreened Field) - + - + - + - + - + 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

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

Relation between Electric Field and Potential in Physics