1. Book Reference : Pages 96-97 1.To understand that when a capacitor is charged it stores energy 2.To be able to calculate the amount of energy stored 3.To be able to solve problems involving the energy stored in a capacitor

2. We have seen from last lesson that when capacitor is charged, the pd across the plates increases in proportion to charge. Charge on plates / C Plate pd / V V Q v q q +  q Consider one step in the charging process, from q to q +  q. Work must be done to force the extra charge onto the plates & this can be given by  E = v  q Where v is the average voltage during this step

3. The work done in this small step from charging from q to q +  q is shown by the green vertical strip in the graph Charge on plates / C Plate pd / V V Q v q q +  q If we consider all the steps from a pd of 0 to a the full pd of V then the energy stored is given by the area under the graph. This can be calculated from ½ base x height Energy stored = E= ½QV

4. The energy equation : E= ½QV Can be re-written in alternative forms... Since Q = CV Substitute for Q : E= ½CV 2 Or Substitute for V : E= ½Q 2 / C

5. Doubling the charge also doubles the pd (voltage) & vice versa. If we look at the E= ½CV 2 Form of the equation we can see that this will quadruple the energy stored

6. While charging the power supply (battery) transfers energy equal to QV. Half of this energy is stored in the capacitor (E= ½QV) the other half is wasted through the resistance in the circuit and dissipated to the surroundings as heat

7. Calculate the charge and the energy stored if a 10  F capacitor is charged to 3.0V and 6.0V [30  C & 45  J, 60  C & 180  J]

8. In the following circuit a charged capacitor with value C is allowed to discharge through a light bulb. V Switch ChargeDischarge Joulemeter In Out Readings : 1.Charge the capacitor, read the pd and the initial joulemeter reading 2.Discharge the capacitor & take a further joulemeter reading C Results : Compare the difference between the two joulemeter readings with E= ½CV 2

9. ++++ ---- During a storm the cloud and the Earth below act like a pair of charged plates & a strong electric field exists If the cloud and ground are separated by d, then then potential difference V is given by : V = Ed Where E is the electric field strength

10. For a cloud carrying a charge of Q then the energy is given by ½QV which can be expanded to ½QEd If the wind moves the cloud to a new height d’ then the new energy stored will be ½QEd’ Note : The electric field strength E remains unchanged since it depends on charge per unit area The increase in energy is given by ½QEd’ - ½QEd

11. We can rewrite this as ½QE  d where  d = d’ – d This increase in stored energy has come from the work done by the wind overcoming the electrical attraction between the cloud and Earth which have opposite charges The insulating properties of air break down when the field strength reaches more than about 300kV/m

12. A 50,000  F capacitor is charged from a 9V battery and then discharged through a light bulb in a flash which lasts 0.2s. Calculate : The charge and energy stored before discharge The average power supplied to the light bulb [0.45C & 2J, 10W]