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

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

Type of capacitor Non-polar capacitors Polarized capacitors

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

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.

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.

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)

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

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)

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

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

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.

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

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

Capacitance II A = effective area = overlapping area

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/ 