1. Capacitors A device storing electrical energy

2. Capacitor A potential across connected plates causes charge migration until equilibrium VV – + –q+q Charge stored q = C  V C = capacitance Unit = C/V = henry = H

3. Parallel Plate Capacitance Plate area A, separation d d A Capacitance = A  0 /d  0 = 8.85  10 –13 C2C2 N m 2

4. Circuit Element Symbols Potential Source + – VV Conductor Capacitoror Resistor

5. At Equilibrium VV C + – Capacitor charges to potential  V Capacitor charge Q = C  V + – VV

6. Energy in a Capacitor C = Q/  V so  V = Q/C VV Q Work to push charge  Q W =  V  Q = (Q/C)  Q slope = 1/C QQ area = W

7. Energy in a Capacitor Work to charge to Q is area of triangle W = 1/2 Q(Q/C) = 1/2 Q 2 /C VV Q Q/CQ/C CVCV Work to charge to  V W = 1/2  V (C  V) = 1/2 C(  V) 2

8. Combining Capacitors and Parallel Series

9. Parallel Components All have the same potential difference Capacitances add (conceptually add A’s)

10. Series Capacitors All have the same charge separation Reciprocals are additive (conceptually add d’s)

11. Gauss’s Law Electric flux through a closed shell is proportional to the charge it encloses.  E = Q in /  0  0 = 8.85  10 –13 C2C2 N m 2

12. Field around a Point Charge Shell Area = 4  r2 +q R  E = q/  0 = EA E = q 0A0A q 04r204r2 = q 4  0 r 2 = kq r2r2 =if k = 1 4  0

13. Field Around Infinite Plate With uniform charge density  = Q/A  E = AA 00  00 1 2, so E = = E(2A)

14. Infinite ||-Plate capacitor Individually –q 1/2  /  0 +q 1/2  /  0 –q+q /0/0 00 Together

15. Parallel Plate Capacitance Plate area A, plate separation d Field E =  00 = Q A0A0 Potential  V = Ed = Qd A0A0 Capacitance Q/  V = Q A  0 Qd A0A0 d =