CAPACITORS February, 2008

Capacitors Part I

A simple Capacitor  Remove the battery  Charge Remains on the plates.  The battery did WORK to charge the plates  That work can be recovered in the form of electrical energy – Potential Difference WIRES TWO PLATES Battery WIRES

INSIDE THE DEVICE

Two Charged Plates (Neglect Fringing Fields) d Air or Vacuum Area A - Q +Q E V=Potential Difference Symbol ADDED CHARGE

Where is the charge? d Air or Vacuum Area A - Q +Q E V=Potential Difference AREA=A  =Q/A

One Way to Charge:  Start with two isolated uncharged plates.  Take electrons and move them from the + to the – plate through the region between.  As the charge builds up, an electric field forms between the plates.  You therefore have to do work against the field as you continue to move charge from one plate to another.

Capacitor

More on Capacitors d Air or Vacuum Area A - Q +Q E V=Potential Difference Gaussian Surface Same result from other plate!

DEFINITION DEFINITION - Capacity  The Potential Difference is APPLIED by a battery or a circuit.  The charge q on the capacitor is found to be proportional to the applied voltage.  The proportionality constant is C and is referred to as the CAPACITANCE of the device.

UNITS  A capacitor which acquires a charge of 1 coulomb on each plate with the application of one volt is defined to have a capacitance of 1 FARAD  One Farad is one Coulomb/Volt

The two metal objects in the figure have net charges of +79 pC and -79 pC, which result in a 10 V potential difference between them. (a) What is the capacitance of the system? [7.9] pF (b) If the charges are changed to +222 pC and -222 pC, what does the capacitance become? [7.9] pF (c) What does the potential difference become? [28.1] V

NOTE  Work to move a charge from one side of a capacitor to the other is qEd.  Work to move a charge from one side of a capacitor to the other is qV  Thus qV=qEd  E=V/d As before ( we omitted the pesky negative sign but you know it is there, right?)

Continuing…  The capacitance of a parallel plate capacitor depends only on the Area and separation between the plates.  C is dependent only on the geometry of the device!

Units of  0 pico

Simple Capacitor Circuits  Batteries Apply potential differences  Capacitors  Wires Wires are METALS. Continuous strands of wire are all at the same potential. Separate strands of wire connected to circuit elements may be at DIFFERENT potentials.

Size Matters!  A Random Access Memory stores information on small capacitors which are either charged (bit=1) or uncharged (bit=0).  Voltage across one of these capacitors ie either zero or the power source voltage (5.3 volts in this example).  Typical capacitance is 55 fF (femto= )  Question: How many electrons are stored on one of these capacitors in the +1 state?

Small is better in the IC world!

TWO Types of Connections SERIES PARALLEL

Parallel Connection V C Equivalent =C E

Series Connection V C 1 C 2 q -q The charge on each capacitor is the same !

Series Connection Continued V C 1 C 2 q -q

More General

Example C 1 C 2 V C3C3 C1=12.0  f C2= 5.3  f C3= 4.5  d (12+5.3)pf series (12+5.3)pf

More on the Big C  We move a charge dq from the (-) plate to the (+) one.  The (-) plate becomes more (-)  The (+) plate becomes more (+).  dW=Fd=dq x E x d +q -q E=  0 A/d +dq

So….

Not All Capacitors are Created Equal  Parallel Plate  Cylindric al  Spherica l

Spherical Capacitor

Calculate Potential Difference V (-) sign because E and ds are in OPPOSITE directions.

Continuing… Lost (-) sign due to switch of limits.

Capacitor Circuits

A Thunker If a drop of liquid has capacitance 1.00 pF, what is its radius? STEPS Assume a charge on the drop. Calculate the potential See what happens

Anudder Thunker Find the equivalent capacitance between points a and b in the combination of capacitors shown in the figure. V(a  b) same across each

Thunk some more … C 1 C 2 V C3C3 C1=12.0  f C2= 5.3  f C3= 4.5  d (12+5.3)pf

More on the Big C  We move a charge dq from the (-) plate to the (+) one.  The (-) plate becomes more (-)  The (+) plate becomes more (+).  dW=Fd=dq x E x d +q -q E=  0 A/d +dq

So…. Sorta like (1/2)mv 2

DIELECTRIC

Polar Materials (Water)

Apply an Electric Field Some LOCAL ordering Larger Scale Ordering

Adding things up Net effect REDUCES the field

Non-Polar Material

Effective Charge is REDUCED

We can measure the C of a capacitor (later) C 0 = Vacuum or air Value C = With dielectric in place C=  C 0 (we show this later)

How to Check This Charge to V 0 and then disconnect from The battery. C0C0 V0V0 Connect the two together V C 0 will lose some charge to the capacitor with the dielectric. We can measure V with a voltmeter (later).

Checking the idea.. V Note: When two Capacitors are the same (No dielectric), then V=V 0 /2.

Messing with Capacitors + V - + V The battery means that the potential difference across the capacitor remains constant. For this case, we insert the dielectric but hold the voltage constant, q=CV since C   C 0 q    C 0 V THE EXTRA CHARGE COMES FROM THE BATTERY! Remember – We hold V constant with the battery.

Another Case  We charge the capacitor to a voltage V 0.  We disconnect the battery.  We slip a dielectric in between the two plates.  We look at the voltage across the capacitor to see what happens.

No Battery q0q0 qq q 0 =C 0 V o When the dielectric is inserted, no charge is added so the charge must be the same. V0VV0V

Another Way to Think About This  There is an original charge q on the capacitor.  If you slide the dielectric into the capacitor, you are adding no additional STORED charge. Just moving some charge around in the dielectric material.  If you short the capacitors with your fingers, only the original charge on the capacitor can burn your fingers to a crisp!  The charge in q=CV must therefore be the free charge on the metal plates of the capacitor.

A Closer Look at this stuff.. Consider this virgin capacitor. No dielectric experience. Applied Voltage via a battery. C0C V0V0 q -q

Remove the Battery V0V0 q -q The Voltage across the capacitor remains V 0 q remains the same as well. The capacitor is fat (charged), dumb and happy.

Slip in a Dielectric Almost, but not quite, filling the space V0V0 q -q q’ +q’ E0E0 E E’ from induced charges Gaussian Surface

A little sheet from the past q -q -q’ +q’ 0 2xE sheet 0

Some more sheet…

A Few slides back No Battery q0q0 qq q=C 0 V o When the dielectric is inserted, no charge is added so the charge must be the same. V0VV0V

From this last equation

Another look +-+- VoVo

Add Dielectric to Capacitor  Original Structure  Disconnect Battery  Slip in Dielectric +-+- VoVo V0V0 Note: Charge on plate does not change!

What happens? +-+-  i   i  oooo Potential Difference is REDUCED by insertion of dielectric. Charge on plate is Unchanged! Capacitance increases by a factor of  as we showed previously

SUMMARY OF RESULTS

APPLICATION OF GAUSS’ LAW

New Gauss for Dielectrics