CAPACITORS February 11, 2008

Calendar of the Week Exams have been returned. If you did badly in the exam you need to have a plan to succeed. Let me know if you want any help with this. Regular problem solving sessions on Wednesday and Friday. Room 306. 7:39 AM. Quiz on Friday – Capacitors. WebAssign should appear shortly if it hasn’t done so already.

Capacitors Part I

A simple Capacitor TWO PLATES WIRES WIRES Battery 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 WIRES Battery

INSIDE THE DEVICE

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

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

The capacitor therefore stores energy! 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. The capacitor therefore stores energy!

Capacitor Demo

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

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

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 e0 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=10-15) 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 CEquivalent=CE

Series Connection The charge on each capacitor is the same ! q -q q -q V C1 C2 q -q q -q The charge on each capacitor is the same !

Series Connection Continued V C1 C2 q -q q -q

More General

Example C1 C2 series (12+5.3)pf (12+5.3)pf V C3 C1=12.0 uf C2= 5.3 uf C3= 4.5 ud C1 C2 series (12+5.3)pf (12+5.3)pf V C3

More on the Big C +q -q E=e0A/d +dq 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

So….

Not All Capacitors are Created Equal Parallel Plate Cylindrical Spherical

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 … C1 C2 (12+5.3)pf V C3 C1=12.0 uf C2= 5.3 uf C3= 4.5 ud C1 C2 (12+5.3)pf V C3

More on the Big C +q -q E=e0A/d +dq 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

So…. Sorta like (1/2)mv2

What's Happening? 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

Non-Polar Material Effective Charge is REDUCED

We can measure the C of a capacitor (later) C0 = Vacuum or air Value C = With dielectric in place C=kC0 (we show this later)

How to Check This Charge to V0 and then disconnect from The battery. Connect the two together V C0 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=V0/2.

Messing with Capacitors 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  kC0 qk kC0V THE EXTRA CHARGE COMES FROM THE BATTERY! + V - + - + V - Remember – We hold V constant with the battery.

Another Case We charge the capacitor to a voltage V0. 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 q0 q0 =C0Vo When the dielectric is inserted, no charge + - q0 =C0Vo When the dielectric is inserted, no charge is added so the charge must be the same. V0 V qk

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. ++++++++++++ ------------------ V0 q -q C0

Remove the Battery ------------------ The Voltage across the ++++++++++++ ------------------ V0 q -q The Voltage across the capacitor remains V0 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 Gaussian Surface ++++++++++++ ------------------ V0 q -q - - - - - - - - + + + + + + -q’ +q’ E E’ from induced charges E0

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

Some more sheet…

A Few slides back No Battery q0 q=C0Vo When the dielectric is inserted, no charge is added so the charge must be the same. + - V0 V qk

From this last equation

Another look + - Vo

Add Dielectric to Capacitor + - Vo Original Structure Disconnect Battery Slip in Dielectric + - V0 + - Note: Charge on plate does not change!

Potential Difference is REDUCED by insertion of dielectric. What happens? + - si + si - so Potential Difference is REDUCED by insertion of dielectric. Charge on plate is Unchanged! Capacitance increases by a factor of k as we showed previously

SUMMARY OF RESULTS

APPLICATION OF GAUSS’ LAW

New Gauss for Dielectrics