Unit 2 Capacitors (Student Copy) y

UNIT 2: Capacitor And Dielectrics

UNIT 2: Capacitor And Dielectrics

2.1 Capacitance And Capacitor In Series And Parallel

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Learning Objectives

Define and use capacitance,

Derive and determine the effective capacitance of capacitors in series and parallel.

Derive and use energy stored in a capacitor.

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2.1 CapacitanceA capacitor is a device that is capable of storing electric charges or electric potential energy.

It is consist of two conducting plates separated by a small air gap or a thin insulator - called a dielectric. Dielectric prevent charges from flowing across the capacitor. The dielectric in figure below is air.

The electrical symbol for a capacitor is

+Q-QV1/12/2013SF0274

2.1.1 CapacitanceThe capacitance of a capacitor is defined as the ratio of the charge on either plate to the potential difference between them.

Mathematically,

The unit of capacitance is the farad (F)

whereQ: Charge on one of the plateV: potential difference across two plates1/12/2013SF0275

2.1.1 Capacitance1 farad is defined as the charge of 1 coulomb stored on each of the conducting plates as a result of a potential difference of 1 volt between the two plates.1 F = 1 C V-1

Since the capacitance of a capacitor is constant, by rearranging the equation,

The charge Q in a capacitor is directly proportional to the potential difference V across the capacitor.

,

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Farad is a very large unit, therefore the capacitance are usually stored in pico, nano or milli - farad6

2.1.2 Capacitors in Series+Q-Q+Q-Q-Q+QC1C2C3V1V2V3V-Q+QVCEquivalent capacitor1/12/2013SF0277

2.1.2 Capacitors in SeriesWhen the circuit is complete, electrons are transferred onto the plates such that the magnitude of the charge Q on each plate is the same.

Thus the total charge (Q) on the equivalent capacitor is

The potential difference across each capacitor C1 , C2 and C3 are V1 , V2 and V3 respectively. Hence

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2.1.2 Capacitor is SeriesIf C is the equivalent capacitor, and , then

Therefore the equivalent (effective) capacitance Ceq for n capacitors connected in series is given by

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2.1.2 Capacitors in Parallel

+Q1-Q1+Q2-Q2+Q3-Q3C2C2C3VXY

VC-Q+QEquivalent capacitor1/12/2013SF02710

2.1.2 Capacitors in ParallelThe potential difference across each capacitor is the same as the supply voltage (V). Thus the total potential difference (V) on the equivalent capacitor is

The charges stored by each capacitor C1,C2 and C3 are Q1,Q2 and Q3 respectively. Hence

Since the total charge Q on the equivalent capacitor is given by

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2.1.2 Capacitors in ParallelSince the total charge Q on the equivalent capacitor is given by

Therefore the equivalent (effective) capacitance Ceq for n capacitors connected in parallel is given by

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Example 1Determine the equivalent capacitance of the configuration shown in figure below. All the capacitors are identical and each has capacitance of 1 F.

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Example 1 - SolutionLabel all the capacitors in the circuit.

To find the equivalent capacitance for circuit above, it is easier to solve it from the end of the circuit (left) to the terminal (right) shown by an arrow in figure above.

Capacitors C1, C2 and C3 connected in series, then

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Capacitors CX and C4 connected in parallel:

Capacitors CY, C5 and C6 connected in series, then

Capacitors Cz and C7 connected in parallel, then

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Example 2In the circuit shown in figure above, C1= 2.00 F, C2 = 4.00 F and C3 = 9.00 F. The applied potential difference between points a and b is Vab = 61.5 V. Calculatea. the charge on each capacitor.b. the potential difference across each capacitor.c. the potential difference between points a and d.similar to (Young & Freedman,pg.936.no.24.14)

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Example 2 - Solution1/12/2013SF02717

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Exercise 1Four capacitors are connected as shown in figure below.

Calculatea. the equivalent capacitance between points a and b.b. the charge on each capacitor if Given Vab=15.0 V. (Serway & Jewett,pg.823,no.21)

Answer : 5.96 F, 89.5 C on 20 F, 63.2 C on 6 F, 26.3 C on 15 F and on 3 F

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Exercise 2Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in figure below.

Take C1 = 5.00 F, C2 = 10.0 F and C3 = 2.00 F. (Serway & Jewett,pg.824,no.27)

Answer : 6.04 F1/12/2013SF02721

2.1.3 Energy Stored in A CapacitorWhen the switch is closed in figure 2.6a, charges begin accumulate on the plates.

The total work W required to increase the accumulated charge from zero to Q is given by

A small amount of work (dW ) is done in bringing a small amount of charge (dQ) from the battery to the capacitor. This is given by and

ORORNote : No charges will accumulate on each plate if the capacitor is not charged.1/12/2013SF02722

Example 3Two capacitors, C1= 3.00 F and C2 = 6.00 F are connected in series and charged with a 4.00 V battery as shown in figure below.

Calculatea. the total capacitance for the circuit above.b. the charge on each capacitor.c. the potential difference across each capacitor.d. the energy stored in each capacitor.e. the area of the each plate in capacitor C1 if the distance between two plates is 0.01 mm and the region between plates is vacuum.(Given permittivity of free space, 0 = 8.85 x 10-12 F m-1)

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Example 4Consider the circuit shown in figure below, where C1= 6.00 F, C2 = 3.00 F and V = 20.0 V.

Capacitor C1 is first charged by the closing of switch S1. Switch S1 is then opened, and the charged capacitor is connected to the uncharged capacitor by the closing of S2. Calculate the initial charge acquired by C1 and the final charge on each capacitor.(Serway & Jewett,pg.824,no.23)

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2.2 Charging And Discharging of Capacitors

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Learning ObjectivesDefine and use time constant, = RC

Sketch and explain the characteristic of Q-t and I-t graph for charging and discharging of a capacitor.

Use for discharging

for charging

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Charging a capacitors through a resistorWhen the switch S is closed, current I0 immediately begins to flow through the circuit and accumulates in the capacitor.Electrons will flow out from the negative terminal of the battery, through the resistor R and accumulate on the plate B of the capacitor.Then electrons will flow into the positive terminal of the battery, leaving a positive charge on the plate A. As charge accumulates on the capacitor, the potential difference across it increases and the current is reduced until eventually the maximum voltage across the capacitor equals the voltage supplied by the battery, V0.At this time, no further current flows (I = 0) through the resistor R and the charge Q on the capacitor thus increases gradually and reaches a maximum value Q0.AB

V0SRCIQVRVc1/12/2013SF02731

The number of charges in the capacitor increases exponentially until fully charged where Q0 is the maximum number of charges.As the number of charges increases, the voltage across the capacitor increases at the same rate.As the capacitor store more and more charges, the rate of charge flow (current) decreases and finally when the capacitor is full, there will be no more charges flow and the current in the circuit falls to zero.

and

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Discharging a capacitors through a resistorWhen a capacitor is already charged to a voltage V0 and it is allowed to discharge through the resistor R as shown in figure above.When the switch S is closed, electrons from plate B begin to flow through the resistor R and neutralizes positive charges at plate A.Initially, the potential difference (voltage) across the capacitor is maximum, V0 and then a maximum current I0 flows through the resistor R.When part of the positive charges on plate A is neutralized by the electrons, the voltage across the capacitor is reduced.The process continues until the current through the resistor is zero.At this moment, all the charges at plate A is fully neutralized and the voltage across the capacitor becomes zero.

QV0SIRABe1/12/2013SF02733

The number of charges in the capacitor decreases exponentially until fully discharged when the charges are reduced to zero.As the number of charges decreases, the voltage across the capacitor decreases at the same rate.As it discharges more charges, the rate of charge flow (current) increases until all the charges are used up by the resistor.

and

Note : For calculation of current in discharging process, ignore the negative sign in the formula.

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The negative sign indicates that as the capacitor discharges, the current direction opposite its direction when the capacitor was being charged34

Time constant, The quantity RC that appears in the exponent for all equation is called time constant or relaxation time of the circuit or mathematically

Its dimension is the dimension of time, then the unit is second (s).It is a measure of how quickly the capacitor charges or discharges.

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For charging process:The time constant is defined as the time required for the capacitor to reach (1-e-1)=0.63 or 63% of its maximum charge/voltage.ORThe time constant is defined as the time required for the current to drop to 1/e = 0.37 or 37% of its initial value(I0).

For discharging process:The time constant is defined as the time required for the charge on the capacitor/voltage across it/current in the resistor decrease to 1/e = 0.37 or 37% of its initial value.1/12/2013SF02736

Example 5In the RC circuit shown in figure below, the battery has fully charged the capacitor.

Then at t = 0 s the switch S is thrown from position a to b. The battery voltage is 20.0 V and the capacitance C = 1.02 F. The current I is observed to decrease to 0.50 of its initial value in 40 s. Determinea. the value of R.b. the time constant, b. the value of the charge, Q on the capacitor at t = 0.c. the value of Q at t = 60 s

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2.3 Capacitors with dielectrics1/12/2013SF02740

2.3.1 Parallel Plate CapacitorConsider two parallel metallic plate capacitor of equal area A are separated by a distance d and the space between plates is vacuum or air as shown in figure below.

One plate carries a charge +Q and the other carries a charge Q then the potential difference between this two parallel plates is V.Because d is small compared to the dimensions of each plate so that the electric field strength E is uniform between them.

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The magnitude of the electric field strength is given by and

Since Q=CV, equation (1) can be written as

Because the field between the plates is uniform, the potential difference between the plates is

Substituting this relation into eq. (2), thus the capacitance of a parallel-plate capacitor is

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(2)(1)

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Parallel-plate capacitor separated by a vacuum

Parallel-plate capacitor separated by a dielectric material

The capacitance of a parallel-plate capacitor is proportional to the area of its plates and inversely proportional to the plate separation where0 : permittivity of free space

: permittivity of dielectric materialA : area of the plated : distance between the two plates

Example 6The plates of a parallel-plate capacitor are 8.0 mm apart and each has an area of 4.0 cm2. The plates are in vacuum. If the potential difference across the plates is 2.0 kV, determinea. the capacitance of the capacitor.b. the amount of charge on each plate.c. the electric field strength was produced.d. the surface charge density on each plate.(Given permittivity of free space, 0 = 8.85 x 10-12 C2 N-1 m-2)

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Example 7A circular parallel-plate capacitor with radius of 10 cm is connected to a 15 V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. If the separation distance between plates is 35 mm and the medium between plates is air.a. Find the amount of charge on each plate.If their separation is increases to 50 mm after the battery is disconnected, determineb. the amount of charge on each plate.c. the potential difference between plates.d. the capacitance of the capacitor.(Given permittivity of free space, 0 = 8.85 x 10-12 C2 N-1 m-2)

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Example 7 - Solution 1/12/2013SF02748

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Exercise 3A parallel-plate, air-filled capacitor has circular plates separated by 1.80 mm. The charge per unit area on each plate has magnitude 5.60 pC m-2. Find the potential difference between the plates of the capacitor. (Young & Freedman,pg.934.no.24.4) Answer:1.14 mV

An electric field of 2.80x105 V m-1 is desired between two parallel plates each of area 21.0 cm2 and separated by 0.250 cm of air. Find the charge on each plate. (Giancoli,pg.628.no. 14) Answer:5.20x10-9 C

(Given permittivity of free space, 0 = 8.85 x 10-12 C2 N-1 m-2)1/12/2013SF02750

Exercise 4A 10.0 F parallel-plate capacitor with circular plates is connected to a 12.0 V battery. Calculatea. the charge on each plate.b. the charge on each plate if their separation were twice while the capacitor remained connected to the battery.c. the charge on each plate if the capacitor were connected to the 12.0 V battery after the radius of each plate was twice without changing their separation (Young & Freedman,pg.934.no.24.5) Answer: 120 C, 60 C, 480 C

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2.3.2 DielectricDefinition is defined as the non-conducting (insulating) material placed between the plates of a capacitor.When a dielectric (such as rubber, glass or waxed paper) is inserted between the plates of a capacitor, the capacitance increases.This capacitance increases by a factor or r which is called the dielectric constant (relative permittivity) of the material.The advantages of inserting the dielectric between the plates of the capacitor areIncrease in capacitanceIncrease in maximum operating voltage.Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C.

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Dielectric Constant, (r)Definition is defined as the ratio between the capacitance of given capacitor with space between plates filled with dielectric, C with the capacitance of same capacitor with plates in a vacuum, C0.Mathematically,

It is dimensionless constant (no unit).For parallel-plates capacitor:andthen

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From the definition of the capacitance,

From the relationship between E and V for uniform electric field,

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andQ is constant

and

The dielectric strength is defined as the electric field strength at which dielectric breakdown occurs and the material becomes a conductor.

Since V=Ed for a uniform electric field, the dielectric strength determines the maximum potential difference that can be applied across a capacitor per meter of plate spacing.

Summary :

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Example 8A parallel-plate capacitor has plates of area A = 2x10-10 m2 and separation d = 1 cm. The capacitor is charged to a potential difference V0 = 3000 V. Then the battery is disconnected and a dielectric sheet of the same area A is placed between the plates as shown in figure below.

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dielectric

In the presence of the dielectric, the potential difference across the plates is reduced to 1000 V. Determine

a. the initial capacitance of the air-filled capacitor.b. the charge on each plate before the dielectric is inserted.c. the capacitance after the dielectric is in place.d. the relative permittivity.e. the permittivity of dielectric sheet.f. the initial electric field.g. the electric field after the dielectric is inserted.(Given permittivity of free space, 0 = 8.85 x 10-12 F m-1)


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2.3.3 Dielectric Effect On The Parallel Plate Capacitors(a) Polar dielectricsThe molecules of some dielectrics like water have permanent electric dipole moments where the concentration of positive and negative charges are separated. When no electric fields is present the polar molecules are oriented randomly as shown in figure (a). The electric dipoles tend to line up when the external electric field is applied to them as in figure (b).

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The alignment of the electric dipoles produces an electric field that is directed opposite the applied field and smaller in magnitude.

(a)(b)

(b) Non-polar dielectricsNon-polar molecules such as glass or paraffin oil have their positive and negative charge centres at the same point in the absence of an external electric field as shown in figure (c).

When the non-polar molecules are placed in an external electric field, these centres become separated slightly and the molecules acquire induced dipole moments. These induced dipole moments tend to align with the electric field and the dielectric is polarized as shown in figure (d).

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(c)(d)

(c) Dielectric in a parallel-plate capacitorConsider a capacitor whose plates are separated by a dielectric material (either polar or non-polar). This capacitor has a charge +Q on one plate and Q on the other, so that the electric field E0 is produced between the plates. Because of the electric field, all the dielectric molecules tend to become oriented as shown in figure (e).

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(e)

The net effect in either case polar or non-polar is as if there were a net negative charge on the outer edge of the dielectric facing the positive plates and a net positive charge on the opposite side as shown in figure (f).

The electric field lines do not pass through the dielectric but instead end of charges induced on the surface of the dielectric as shown in figure 2.8f. Therefore the electric field within the dielectric is less than in air.

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(f)

According to figure (g), the electric field within the dielectric E is given by

Since , then

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OR

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Unit 2 Capacitors (Student Copy) y