{Capacitance

Chapter 25 Halliday-Resnick 9th Ed.

a measure of how much charge must be puton the plates to produce a certain potentialdifference between them:

The greater the capacitance, the more chargeis required.

The SI unit of capacitance : coulomb per volt(a.k.a. farad [F])

1 farad = 1 F = 1 coulomb per volt = 1 e/v.

Capacitance =

Calculating the CapacitanceElectric Field= ∙ ==

Potential Difference− = − ∙= + ∙

Parallel platecapacitor:so large, so close togetherthe fringing of the electricfield can be neglected

Electric Field= ∙ ==Potential Difference− = − ∙= + ∙ =

= → = =Geometry dependent

Cylindrical Electric Field= ==Potential Difference= + ∙= −= ln= = ln /

Spherical Electric Field= ==Potential Difference= + ∙= − 4= 4 1 − 1

= = − = 1 − /

Capacitors in Parallel• = = =• The total charge =

sum of the chargesstored on all thecapacitors= + += + +== + +=

1 2 3

Capacitors in Series• = = =• The sum of the potential

differences across all thecapacitors is equal to theapplied potential difference= + += 1 + 1 + 1

== 11/ + 1/ + 1/1 = 1

At given instant:- A charge has been transferred- Potential difference between plates

is = /If extra increment charge is added

- = =- = ∫ = ∫ =

Energy Stored in an Electric Field

Stored as potentialenergy== 12

Energy density= / = 2 /= 12

Dielectric: an insulating material (mineral oil or plastic)what happens?- the capacitance increased by a numerical factor , (Faraday exp.)- =the dielectric constant of the insulating material

Capacitor with a Dielectric

>The effect of adielectric is to

weaken the electricfield

=

The effect of both

polar and nonpolar

dielectrics:

to weaken any

applied field within

them, as between

the plates of a

capacitor.

Effect of a Dielectric

Dielectric and Gauss’ Law(a) Electric Field∙ = == /(b) Electric Field∙ = = −= − /Dielectric: weaken the Electric Field:

= = −− =

1. The flux integral now involves , not just . (Thevector is sometimes called the electricdisplacement → ∮ ∙ =

2. The charge enclosed by the Gaussian surface isnow taken to be the free charge only.

3. Our original statement of Gauss' law is differentto the one above that is replaced by . Wekeep inside the integral to allow for cases inwhich is not constant over the entire Gaussiansurface.

∙ = ( )

(a) What is the capacitance before the dielectric slab is inserted?

(b) What free charge appears on the plates?

(c) What is the potential difference between the plates afterthe slab has been introduced?