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## CAPACITORS

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**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!**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**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.