Topic 2: Electric and magnetic fields

Part 2: CAPACITORS

Prepared by: Pn Siti Fatimah binti Saipuddin INTEC

Topic 2: Electric and magnetic fields

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SPECIFICATIONSinvestigate and use the expression C = Q/V recognise and use the expression W = QV for the energy stored by a capacitor, derive the expression from the area under a graph of potential difference against charge stored, and derive and use related expressions, for example, W = CV2investigate and recall that the growth and decay curves for resistorcapacitor circuits are exponential, and know the significance of the time constant RCrecognise and use the expression Q = Q0 e-t/RC and derive and use related expressions, for exponential discharge in RC circuits, for example, I = Io e-t/RC2

FLOW OF CHARGE3Electric field is set up within the conducting materialCauses electron to feel force and move through wires and components in the circuitNeed complete circuit for electron to flowSwitch on power supplyElectric field created in conducting wires Electron flow towards +ve terminalElectrons accumulate at one plate -vely chargeElectrons at +ve terminal plate flow towards +ve terminal +vely chargePlates of opposite charges attract each other, creates electric field btwn platesPd increases until pd equals to power supply

In CircuitsIn Capacitors

CAPACITORCapacitor pair of plates with insulator in between them

Charge will build up on a capacitor until pd across the plates = pd of power supply (fully charged)

Store of charge

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Electric Field In Capacitor5

ELECTRIC FIELDS BTW THE PLATES OF A PARALLEL-PLATE CAPACITOR IS UNIFORM NEAR THE CENTER BUT NONUNIFORM NEAR THE EDGES

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CAPACITANCEAmount of charge a capacitor can store per unit volt applied across itUnit = Farads (F)Depends on:Size of platesSeparation btwn platesNature of the insulator between plates

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Charging begins with zero charge stored on itcharge slowly fills up as the pd increases until charge at V is given by QWhen increase voltage and pushing the charge onEnergy required8

Energy Stored On A Charged Capacitor

9Energy Stored On A Charged Capacitor

Charging CapacitorFully charged capacitorPd = power supply = 6V0.6 mC of charge10

Discharging CapacitorSlow down discharging process using ResistorLamp At first, electrons rush out at maximumCurrent at maximum = 0.06 AAs electrons flow, pd reduced Curent lessLight dimmer almost off if so little electrons flowFully discharged capacitorNo more electrons flowCurrent = 0 A

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12Charging CapacitorDischarging Capacitor

Time Constant, Time constant, T (tau)The rate of discharge of a capacitorT = RCUnit = secondsTells how many seconds it takes for the current to fall to 37% of starting valueFor charging capacitorT rate of charging a capacitorInfluenced byResistance (higher, longer time to charge up)Capacitance (larger, longer time to charge up)13

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Discharging MathCharge remaining on the capacitor, Q at a time, t:

Therefore: 15

Plate area, AElectrons flow into ve plate and out of +ve plateIf plates are large The accumulated charges are able to distribute themselves over the areaAmount of charge that can be stored on a plate for a given pd increases as plate area increasedCapacitance proportional to plate areaPlate separation, dIf battery has constant pd btw its terminalsElectric field increases as d decreasedCloser the plate separation, the charge on the capacitor increasesCapacitance inversely proportional to d

16Geometry Effects On Capacitance

Geometry Effects On CapacitanceSurface charge density on plate, = Q / AElectric field btw the plates, E = / o = Q / o A fields btw the plates are uniformThe magnitude of p.d:V = Ed = Qd / o AHence, capacitance:C = Q / V = o A / dThe capacitance of a parallel-plate capacitor is proportional the area of its plates & inversely proportional to the plate separation

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Capacitors In Parallel

The individual potential differences across capacitors connected in parallel are the same and equal to the potential difference applied across the combinationThe total charge on capacitors connected in parallel is the sum of the charges on the individual capacitorsThe equivalent capacitance of a parallel combination of capacitors is the sum of the individual capacitance and is greater than any individual capacitance in the combination18

19Total charge stored by the two capacitors:Since the voltage across the capacitors are the same:

Hence:

Capacitors In Parallel

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The individual charges on capacitors connected in series are the sameThe total potential differences across capacitors connected in series is the sum of the potential differences across the individual capacitorsThe inverse of the equivalent capacitance of a series combination of capacitors is the sum of the inverses of the individual capacitance and always less than any individual capacitance in the combination20

Capacitors In Series

Total potential difference btw the two capacitors:

Since the charges on the capacitors are the same:

Hence: 21

Capacitors In Series

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Exercise:Find the equivalent capacitors for the combinations of capacitors in figure. All capacitance are in microfarads22

Exercise:

A 5000 F capacitor is charged to 12.0 V and discharged through a 2000 resistor.

What is the time constant? What is the voltage after 13 s? What is the half-life of the decay? How long would it take the capacitor to discharge to 2.0 V

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Exercise:A 470 F capacitor, charged up to 12.0 V is connected to a 100 k resistor. (a) What is the time constant? (b) What is the voltage after 10 s? (c) How long does it take for the voltage to drop to 2.0 V?

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Exercise:

An uncharged capacitor and a resistor are connected in series to a battery. If = 12V , C = 5F, and R = 8 x 105 , find the time constant for the circuit and the charge and current as functions of time.

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Exercise:

Consider a capacitor of capacitance C that has been discharged through a resistor of resistance R.

After how many time constant is the charge on the capacitor one-fourth its initial value?The energy stored in the capacitor decreases with time as the capacitor discharges. After how many time constant is this stored energy one-fourth its initial value?26

27End of Capacitors