1. Capacitor Charging and Discharging 18-Jan-08 Mr.NGAN HON SHING

2. Structure of capacitor Capacitor =conductor (metal plate) + insulator (dielectric) + conductor (metal plate)

3. Type of capacitor Non-polar capacitors Polarized capacitors

4. Charging up a capacitor After connected to the battery, the capacitor is charging up. More and more charges are on the plates of the capacitor. ++ + - - - ++ +++ - - ---

5. Current Flow (I) when charging up Current flows from the positive terminal of the supply to the negative terminal. ++ + - - - ++ +++ - - --- Current I is decreasing and then comes to be zero.

6. Voltage of capacitor (V c ) when charging up A voltage V c is built up when charging. ++ + - - - ++ +++ - - --- e.m.f. E.m.f. of the battery is constant while voltage across capacitor (V c ) is increasing until it is equals to e.m.f. of the battery. V c =0 Vc ↑Vc ↑ V c =e.m.f. i.e.: I=0 The capacitor is charged up.

7. Discharging If the p.d. across the capacitor is equal to that of the battery, discharging doesn’t occur. If a charged capacitor is connected to a resistor, current flows from its positive terminal. ++ +++ - - --- ++ + - - - Current I is decreasing and flows at the direction which is opposite as before. Also, V c is decreasing (expotentially)

8. Capacitance Which one is at a higher potential? A B + + +

9. Capacitance (C) The potential of a conductor is proportional to the charge stored on it. V  Q ∵ V = kQ/R i.e.: Q/V = constant = C (capacitance)

10. + + + + + + + + + V A = 1000V V B =1000V Q A = 6mCQ B = 3mC C = 6  F C = 3  F Capacitance If V A =V B, which one has higher capacitance? A B ANSWER

11. Unit of capacitance C = Q/V Unit of capacitance = 1C V -1 Or Farad (F) 1F = 1C V -1

12. Capacitance of a conducting sphere A conducting sphere can store charges. Voltage at the surface of a conducting sphere = k Q / R V = k Q / R C = Q / V = R/k = 4  o R So, the capacitance of a conducting sphere is proportional to the radius of the sphere.

13. Capacitance of a parallel plate capacitor Recall E=kQ/r 2 = Q/(4  o r 2 )=Q/(4  r 2  o )=  /  o  = Q / A E= -dV/dr = V/d ( ∵ E is uniform between plates) E=  /  o = Q /  o A = V / d C = Q / V =  o A / d

14. Capacitance I a) b)c) rr Area A d C 1 = C 2 = C 3 =

15. Capacitance II A = effective area = overlapping area

16. Function of dielectric Plates induced charges at dielectric Dielectric formed an opposite E-field which reduced both E-field and V of metal plates. By C=Q/V => C=Q/V’ where V’ < V => C  E’ = E/  Also V’ = V/ 