CAPACITORS September 29, 2008
How did you do? • Great • OK • Poor • Really bad • I absolutely flunked!
Calendar of the Day • Exams will be returned within a week. • 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. • Quiz on Friday – Potential or Capacitance. • WebAssign will appear shortly if it hasn’t done so already. • There is a WA on board for potential. • Quizzes are in the bin on the third floor through the double doors.
Two +q charges are separated by a distance d. What is the potential at a point midway between the charges on the line connecting them • Zero • Kq/d • Kq/d • 2Kq/d • 4kq/d
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!
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
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!
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) • Probably less these days! • Question: How many electrons are stored on one of these capacitors in the +1 state?
Cap-II October 1, 2008
Note: • I do not have the grades yet. Probably by Friday. • Quiz on Friday … Potential or Capacitors. • Watch WebAssign for new stuff.
Last Time • We defined capacitance: • C=q/V • Q=CV • We showed that • C=e0A/d • And • E=V/d
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
In the drawing below, find the equivalent capacitance of the combination. Assume that C1 = 8 µF, C2 = 4 µF, and C3 = 3 µF. 5.67µF
In the diagram, the battery has a potential difference of 10 V and the five capacitors each have a capacitance of 20 µF. What is the charge on ( a) capacitor C1 and (b) capacitor C2?
In the figure, capacitors C1 = 0.8 µF and C2 = 2.8 µF are each charged to a potential difference of V = 104 V, but with opposite polarity as shown. Switches S1 and S2 are then closed. (a) What is the new potential difference between points a and b? 57.8 VWhat are the new charges on each capacitor?(b)46.2µC (on C1)(c)162µC (on C2)
AnudderThunker Find the equivalent capacitance between points a and b in the combination of capacitors shown in the figure. V(ab) same across each
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.