2. 1 CAPACITORS

3. 2 BASIC CONSTRUCTION INSULATOR CONDUCTOR + - TWO OPPOSITELY CHARGED CONDUCTORS SEPARATED BY AN INSULATOR - WHICH MAY BE AIR The Parallel Plate Capacitor

4. 3 CAPACITORS STORE CHARGE & ENERGY USES 1. Storing energy as in flash photography 2. Time delays in electronic circuits 3. As filters in electronic circuits 4. In tuning circuits

5. © 4 Charge stored [Q] depends on p.d. [Volts] applied [V] Q V Gradient = C = The capacitance, C, is defined as the charge required to raise the potential by one volt. Hence Q = C V C is measured in FARADS, [F], - more often :  F, nF or pF

6. 5 Q Q Q C ------ ++++++ +Q -Q Q = CV Charging a Capacitor

7. 6 WORK DONE BY THE CELL W = Q V As VOLTS = Joules per Coulomb The pd across the capacitor builds up as more charge is added Voltage Charge 0 Q V Work done in charging = average volts x charge

8. 7 Volts V Charge Q qq v The work done in adding such an infinitesimally small amount of charge,  q, that V remains constant is given by:  w = V.  q This is the area of the strip. Hence the total work done = the energy stored in the capacitor is the area under the graph Or if we combine with Q = CV, or

9. 8 What is the charge and the energy stored when a 50  F capacitor is charged to (a) 100 V, (b) 200 V? (a) Q = CV = 50 x 10 -6 x 100 = 0.005 C E = 0.5CV 2 = 0.5 x 50 x 10 -6 x 100 2 = 0.25 J (b) Q = CV = 50 x 10 -6 x 200 = 0.01 C E = 0.5CV 2 = 0.5 x 50 x 10 -6 x 200 2 = 1.00 J Why does doubling the p.d. in (b) quadruple the energy stored??

10. 9 At all times after the switch is closed V = V R + VCVC Initially Vc Vc = 0, so V = Vr Vr and so initial current = ? Finally I =0 when capacitor is charged and V C = V time Volts VCVC I0I0 time 0 VCVC C R V VRVR Charging a Capacitor

11. 10 CAPACITOR DISCHARGE R + - I V Initially V = V0V0 Hence initial current, I 0 = ? Final Current = ? Volts, V Time, t 0 V0V0 i.e. the decay is exponential What if R and / or C is larger?

12. 11 CAPACITOR DISCHARGE The current also falls exponentially and is given by : Current I Time, t I0I0 Note the area under this graph is the initial charge stored.

13. 12 CAPACITOR DISCHARGE R times C has units of seconds and is called the time constant of the circuit If we put t=RC into the equation above, it becomes V=V 0 e -1, which works out to: i.e. when time is equal to R x C the p.d. across the capacitor has dropped to 37% of its original p.d. The capacitor is almost discharged in 5 time constants. Try putting t = 5 x R x C V=0.37V 0

14. 13 CAPACITOR DISCHARGE Taking logs to the base “e” y = c + m x In the form Hence Plot ln V t Gradient is negative and equal to

15. 14 The Parallel Plate Capacitor Area of Plate overlap = A d d = plate separation Medium relative permittivity =  r 0 0 = the permittivity of free space = 8.86. X 10 -12 F m -1 For air or a vacuum,  r = 1

16. 15 Question 1A 1000  F capacitor is charged to 100 V and then discharged through a 2000  resistor. (a)What will be the initial current? (b)What will be the current after 4 s? (a) (b)

17. 16 Question 2A capacitor is charged to 100 V and then discharged through a resistor whose value is gradually reduced in order to maintain a constant current of 200 mA though it. The capacitor becomes discharged after 10 s. (a)What is the initial value of the resistor? (b)What is the capacitance of the capacitor? (a) Initially the p.d. is 100 V, so (b) Q = I t = 0.200 x 10 = 2.0 C Q = C V, so 2.0 = C x 100 Hence C = 2 x 10 -2 F or 20 000 FF The capacitor was initially charged to 200 V with 2.0 C of charge