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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. TWO PLATES. WIRES. WIRES. Battery.
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CAPACITORS February, 2008
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 TWO PLATES WIRES WIRES Battery
d Air or Vacuum E - Q +Q Symbol Area A V=Potential Difference Two Charged Plates(Neglect Fringing Fields) ADDED CHARGE
Where is the charge? +++++ + - - - - - - d AREA=A s=Q/A Air or Vacuum E - Q +Q Area A V=Potential Difference
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
d Air or Vacuum E - Q +Q Gaussian Surface Area A V=Potential Difference More on Capacitors 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 ( 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 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?
TWO Types of Connections SERIES PARALLEL
V CEquivalent=CE Parallel Connection
q -q q -q V C1 C2 Series Connection The charge on each capacitor is the same !
q -q q -q V C1 C2 Series Connection Continued
Example 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
E=e0A/d +dq +q -q 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
Parallel Plate Cylindrical Spherical Not All Capacitors are Created Equal
Calculate Potential Difference V (-) sign because E and ds are in OPPOSITE directions.
Continuing… Lost (-) sign due to switch of limits.
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(ab) same across each
Thunk some more … C1=12.0 uf C2= 5.3 uf C3= 4.5 ud C1 C2 (12+5.3)pf V C3
E=e0A/d +dq +q -q 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
So…. Sorta like (1/2)mv2
What's Happening? DIELECTRIC
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) 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. C0 V0 Connect the two together V C0 will lose some charge to the capacitor with the dielectric. We can measure V with a voltmeter (later).
V Checking the idea.. 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 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.