Ch25 capacitance

Rogers: Lectures based on Halliday, Resnick and Walker’s Fundamentals of Physics, Copyright 2005 by Wiley and Sons

Material, including many figures, is used with permission of John Wiley and Sons, Inc.Material is not to be further distributed in any format and is subject to Copyright Protection.

1

Ch25-1/57

Chapter 25: Capacitance p656-671 (not sec 8)

Capacitance is related to the ability of two isolated conductors to hold a charge.

We talk of two `plates’ but capacitors have many different shapes.

To start, consider nothing between plates.

We speak of the charge q on the capacitor, but the plates have equal and opposite charges +q & -q

The plates are conductors => are equipotentialsurfaces => all points on each plate are at same

potential.

Ch25-2/57

Capacitance

A parallel plate capacitor. Charges on facing plates are equal but of opposite sign. Charges are on inside of plates (proven on Ch23-51 for conducting plates).

The field lines in the interior are constant, from +ve to-ve charges. There are fringing fields near the end.

The exterior fields are much weaker (would be 0 for infinite plates - see Ch23-51).



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Ch25-3/57

CapacitanceThe potential difference across the plates is denoted by V rather than ∆V (for historical reasons).

The charge q and potential difference V on the plates are proportional to each other, i.e.

Proportionality constant C is called the capacitance.

Capacitance depends ONLY on the geometry of the plates, NOT on q nor V.

Capacitance is a measure of charge needed on plates to attain a certain potential difference.

Greater capacitance => more charge required.

Alternatively, for a given voltage, more capacitance => more charge held.

Ch25-4/57

The farad, F

The SI unit for capacitance is the farad (F), after Englishman Michael Faraday (1791-1867)

The farad is a huge unit.

It is common to use the microfarad (1 µF =10-6 F)

or

the picofarad (1 pF = 10-12 F) (the “puff”)



3

Ch25-5/57

Charging a capacitorNote the notation/symbolsSwitch S: Battery B: with the +ve side longerCapacitance C: 2 equal parallel lines even if shape very different.Battery -maintains potential difference V across terminals -sets up electric field in wires which draws e-

+ve terminal has the higher potential.Circuit incomplete until switch closed, then it is completed.Same potential V developed across h and l as across battery => no field in wires => no more current. Capacitor is fully charged. We assume (a) no leakage between plates

(b) capacitor stores the full charge indefinitely.

-from plate h (becomes +ve) -to plate l (becomes negative).

Ch25-6/57

Question

Does the capacitance C of a capacitor increase, decrease or remain the same(a) when the charge q is doubled?

(b) when the potential difference is tripled?



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Ch25-7/57

Question

Does the capacitance C of a capacitor increase, decrease or remain the same(a) when the charge q is doubled?

(b) when the potential difference is tripled?

Ans: Remains the same since it is a property of the geometry, not the charge

Ans: Remains the same since it is a property of the geometry, not the voltage

Ch25-8/57

1) assume charge q on plates2) calculate electric field E between plates using

Gauss’ Law (arranging for E and dA to be parallel) so

3) calculate the potential difference using

always following a field line from -ve to +ve plate=> E and ds will be in opposite directions =>

and

4) Calculate C from

Calculating capacitance



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Ch25-9/57

Assume plates large enough we can ignore fringing fields => E is constant between plates.

Consider the Gaussian surface and Gauss’ Law with E=0 in plate gives

which leads to

substituting E = V/d and then C=q/V

Capacitance of a parallel plate capacitor

C increases with plate area and as plates closer together

Ch25-10/57

Using k instead of would make this eqn much uglier, and this eqn is used more often.

Capacitance of a parallel plate capacitor (cont)



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Ch25-11/57

Take ds radially inwards (-ve to +ve) but then ds = -dr so:

C = q/V =>

Cylindrical capacitor, inner radius a,outer, b and length L. Assume L>>b => ignore fringing at ends. Opposite charges q on each surface.

Gaussian surface is a coaxial cylinder shown in red. E is radial =>

where A is area of cylinder’s curved surface (no flux thru the ends).

Hence (field from a line).

Capacitance of a cylindrical capacitor

Ch25-12/57

Take ds radially inwards (-ve to +ve) but then ds = -dr so:

C = q/V =>

Spherical capacitor, inner radius a, outer, b. Charge q on each surface.

Gaussian surface is a concentric sphereshown in red. E is radial =>

where A is area of sphere.

Hence

(field from an isolated charge/sphere).

Capacitance of a spherical capacitor



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Ch25-13/57

Consider a single isolated sphere, radius R, as part of a spherical capacitor with the outer radius at infinity.

Rewrite the standard eqn, viz:

as

then let , substitute R for “a”

=>

Capacitance of an isolated sphere

Ch25-14/57

Summary of capacitances

All the formulae are for εo multiplied a quantity with dimensions of length (F/m x m)

parallel plates

co-axial cylinders

concentric spheres

isolated sphere

a is the inner radius, b the outer



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Ch25-15/57

Question

For capacitors charged by the same battery, does the charge stored by the capacitor:

increase, decrease, or stay the same when:

a) plate separation is increased in a parallel plate capacitor

b) radius of the inner cylinder is doubled in a cylindrical capacitor

c) outer radius of a spherical capacitor is increased

Ch25-16/57

Question

For capacitors charged by the same battery, does the charge stored by the capacitor:

increase, decrease, or stay the same when:

a) plate separation is increased in a parallel plate capacitor

b) radius of the inner cylinder is doubled in a cylindrical capacitor

c) outer radius of a spherical capacitor is increased

=> C decreases as d increases => q=CV decreases

=> C & q increase as “a” increases (plates closer)

=> C & q decrease as “b” increases (plates farther apart)



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Ch25-17/57

Sample Problem

A storage capacitor on a random access memory chip (RAM) has a capacitance of 55 fF. If charged to 5.3 V, how many excess electrons are on the -ve plate.

Key idea: number of electrons on plate is n = q/e. q = CV

=> n = CV/e

(or q= 2.9x10-13 C)

Ch25-18/57

Several capacitors Replace a number of capacitors in a circuit by a single equivalent capacitor, i.e. a capacitor with same capacitance as the others.

Consider combinations of capacitors in parallel or series.

In parallel-directly wired together at each

plate- same potential difference across all

capacitors- total charge stored is the sum of

charges on each component-equivalent capacitor has same total

charge q and same V as actual capacitors



10

Ch25-19/57

Capacitors in parallelWe know q = CV, hence for capacitors in parallel:

so

Same V across each capacitor and total charge stored is sum of individual charges

Ceq is dominated by the larger Cj

Ch25-20/57

Capacitors in seriesCapacitors wired in series, one after another. Potential difference V is applied across

two ends of series.

Each capacitor has same charge when in series (chain of events, charging of 1 causes charging of next because there is nowhere else for the charge to go to/come from. But potential differences may vary)

The potential differences across all capacitors add up to give applied potential difference.



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Ch25-21/57

Capacitors in series (cont) In series, there is only one route for charge

from one capacitor to the next in a series.

If there is a second route => not in series.

The equivalent capacitor has the same chargeq and the same total potential difference V as the actual series capacitors

Ch25-22/57

Capacitors in series (cont)Since we have the same q on each capacitor in series

but we know the total V is just the sum so:

Ceq is dominated by the smallest Cj



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Ch25-23/57

QuestionA battery of potential V stores a charge q on a combination of two identical capacitors. What are the potential differences across each capacitor and the charge stored on each capacitor (in terms of q and V) if the capacitors are (a) in parallel or (b) in series?

i) V/2, q ii) V/2, q/2 iii) V, q iv) V, q/2

Ch25-24/57

QuestionA battery of potential V stores a charge q on a combination of two identical capacitors. What are the potential differences across each capacitor and the charge stored on each capacitor (in terms of q and V) if the capacitors are (a) in parallel or (b) in series?

i) V/2, q ii) V/2, q/2 iii) V, q iv) V, q/2

Ans: parallel => V same for all and:

so (iv) is correct

Ans: series => q same for all and:

so (i) is correct.



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Ch25-25/57

Sample problem Find the equivalent capacitance of the 3 capacitors to the right.

Key idea: Replace any capacitors in series or parallel by their equivalent capacitor. Which ones are in parallel or series.

Are C1 and C3 in series?

Are C1 and C2 in parallel?

What is the equivalent capacitance C12?

Ch25-26/57

Sample problem Find the equivalent capacitance of the 3 capacitors to the right.

Key idea: Replace any capacitors in series or parallel by their equivalent capacitor. Which ones are in parallel or series.

Are C1 and C3 in series?

Are C1 and C2 in parallel?

What is the equivalent capacitance C12?

Ans: No. Charge from C3 can go to C1or C2 => C1 and C3 are not in series.

Ans: Yes since their top and bottom plates are directly wired together

Ans:



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Ch25-27/57

Sample problem(cont)

Are C12 and C3 in series or parallel?

What is the equivalent capacitance of C123?

Ch25-28/57


Are C12 and C3 in series or parallel?

What is the equivalent capacitance of C123?

C123 = 3.57 µF if you stuff in the values.

Ans: Series

Ans:



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Ch25-29/57


If potential difference is V =12.5 V, what is charge on C1?

Key idea 1: work backwards to C1 from the equivalent capacitor of the set of 3, viz C123.

12.5 V is applied across C123 =>

Key idea 2: charge on capacitors in series is same, so q12 = q123

=>

Ch25-30/57


Key idea: voltage across C12 is same as that across two capacitors in parallel => V1 = V12 = 2.58 V

Hence:

QED

General approach: divide and conquer



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Ch25-31/57

Another sample problem

Capacitor 1 with C1 =3.55 µF is charged to a potential difference of V0 = 6.3 V using a 6.3 V battery. Battery is removed and capacitor is placed in circuit shown with an uncharged capacitor C2 = 8.95 µF. When switch S is closed, charge flows between capacitors until they have same potential difference V. What is V?

Are these capacitors in parallel or series or neither?

Ans: neither since no applied voltage.

Ch25-32/57

Another sample problem(cont)

Key idea: after switch closed, the charge on C1 is shared between C1 and C2 to give equal voltages on each. Hence:

and using q = CV:

Key idea: the capacitors share their charge to give equal voltages.



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Ch25-33/57

Question

In above circuit, replace C2 with two capacitors in series, C3and C4.

(a) what is the relationship between between the initial charge qo and the charges q1 and q34 ?

(b) If C3 > C4, is q3 > q4, q3 = q4, q3 < q4?

Ch25-34/57

Question

In above circuit, replace C2 with two capacitors in series, C3and C4.

(a) what is the relationship between between the initial charge qo and the charges q1 and q34 ?

(b) If C3 > C4, is q3 > q4, q3 = q4, q3 < q4?Ans:

Ans: they are in series so they must be equal



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Ch25-35/57

QuestionAre C1 and C2 in parallel? What is the equivalent C12? C1

C2

C3

C4Are C3 and C4 in parallel? What is equivalent C34?

What is the equivalent C of all 4 capacitors?

Ch25-36/57

QuestionAre C1 and C2 in parallel? What is the equivalent C12?

Ans: yes => C12 = C1 + C2

Ans: yes => C34 = C3 + C4

C1

C2

C3

C4Are C3 and C4 in parallel? What is equivalent C34?

What is the equivalent C of all 4 capacitors?

Ans: C12 & C34 in series =>



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Ch25-37/57

Strategy for solving multi-capacitor problems

• If capacitors, or groups of capacitors are in parallel => voltage across them is same. One can apply q=CV

• if capacitors are in parallel

– goal is usually to find voltage across parallel capacitors => can find qi or Ci,depending on question since qi = CiVcommon or Ci=Vcommon/qi

• if capacitors are in series

– goal is usually to find charge on a series of capacitors => each capacitor in the series has same charge

– from this you determine C or V as required

Ch25-38/57

Energy stored in an electric fieldWork must be done by an external agent to charge a capacitor (the battery uses its stored chemical energy).

Remember: U = Wapp = q V was the general relationship.

So if q’ has been transferred to a capacitor C, then V’ = q’/C and as we add a new small amount of charge q’

This work is stored as potential energy U in the capacitor

U is stored in electric field between plates of capacitor.



20

Ch25-39/57

Units of stored energy

Ch25-40/57

Energy stored in an electric field(cont)

For parallel plate capacitors (Ch25-8)

=> field is same in both cases above since q is.

for 2d case: => U’ is twice as large in

larger capacitor (consistent with concept of storage in field).

V’ is double to get same q in larger capacitor

=> U’ twice as large.

d2d

+q +q

-q

-q

CC’ = C/2



21

Ch25-41/57

Sample problem

Consider 2 parallel plates which form a capacitor which becomes charged to a potential difference V1 when their area of overlap was A1. What is the potential difference between them when their overlap area is A2 < A1?

Key ideas: q = CV, q is constant and C varies as

So how would you prevent a spark in the hospital?

picture p665

hypobaric oxygen treatment

Ch25-42/57

What is the ratio of the stored potential energy before vs after the movement?

Key ideas: and

and we just showed so easier to use 1st formula

hence:

Where did the extra potential energy come from?


Ans: From the force moving the gurney.



22

Ch25-43/57

We have a 26 µF capacitor which we charge up to 3 kV. How much energy is stored?

=> U = 26x10-6 F (3x103 V)2 = 234 J

How much energy is that?

Exploding wire demonstration

Ch25-44/57

Define the energy density, u, of an electric field as the potential energy stored in an electric field per unit volume. This builds on concept of potential energy of capacitor being stored in electric field inside capacitor.

Consider a parallel plate capacitor, ignoring fringing fields. The total energy stored is . Volume of space between plates is Ad.

=>

We know =>

Also know =>

Energy density



23

Ch25-45/57

We derived this for a parallel plate capacitor, but it holds in general.

If an electric vector E exists at any point in space, that point is a site of stored electric potential energy.

The amount stored per unit volume is given above.

Energy density (cont)

Ch25-46/57

Do the units make sense?

Energy density (cont)



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Ch25-47/57

Capacitor with a DielectricFaraday:fill space between plates of a capacitor with a dielectric, (an insulating material) => capacitance increases by a factor κ, the dielectric constant. κ is unity for vacuum by definition

Above a certain maximum field strength the material will break down and form a conducting pathbetween the plates (sparking).

=> there is a maximum voltagewhich can be applied.

The maximum field strength is called the dielectric strength,

Material Dielectric DielectricConstant Strength

κ kV/mmair 1.00054 3polystyrene 2.6 24paper 3.5 16transformer oil 4.5 25silicon 12water 80titania

ceramic 130strontium

titanate 310 8

Ch25-48/57

Capacitor with a Dielectric (cont)For any capacitor we can write:

where has dimensions of length. If a dielectric completely fills space between plates, then

With the dielectric added the charge increases on the plates by a factor κ

For a constant charge, the voltage decreasesby a factor κ



25

Ch25-49/57

Capacitor with a Dielectric (cont)In a region completely filled by a dielectric material of dielectric constant κ, all electrostatic eqns containing the permittivity constant εo are modified, replacing εo by κεo.

For example, inside a dielectric, electric field produced by a point charge is given by:

and the electric field just outside any isolated conductorin a dielectric is:

For a fixed distribution of charges, a dielectric weakensthe electric field that would otherwise be there.

Ch25-50/57

Sample problemA parallel-plate capacitor with C= 13.5 pF is charged by a battery to 12.5 V. The battery is then disconnected and a porcelain slab (κ = 6.50) is placed between the plates.

What is the potential energy of the capacitor

(1) before and (2) after the slab is inserted?

Key idea:



26

Ch25-51/57

Sample problem (cont)A parallel-plate capacitor with C= 13.5 pF is charged by a battery to 12.5 V. The battery is then disconnected and a porcelain slab (κ = 6.50) is placed between the plates. What is the potential energy of the capacitor (2) after the lab is inserted?

Key idea 2: Since battery is disconnected, charge on capacitor cannot change. Potential changes. Capacitance becomes κC.

i.e. when slab is inserted, potential energy decreases by a factor κ.

The energy was used `pulling’ slab into capacitor.

Ch25-52/57

QuestionIn the previous problem, if the battery stayed connected do the following increase, decrease or stay the same:

a) potential difference across the plates?

b) the capacitance?

c) the charge on the capacitor?

d) the potential energy of the device?

e) the electric field between the plates, given that charge on the capacitor is not fixed?



27

Ch25-53/57

QuestionIn the previous problem, if the battery stayed connected do the following increase, decrease or stay the same:

a) potential difference across the plates?

b) the capacitance?

c) the charge on the capacitor?

d) the potential energy of the device?

e) the electric field between the plates, given that charge on the capacitor is not fixed?Ans: (a) constant because of battery(b) increased by factor κ.

(c) increased by κ because of fixed V and increased capacitance

(d) increased since V is constant and C has gone up(e) same because same potential difference and distance V=Ed

Ch25-54/57

Dielectrics: an atomic viewWhat happens when we put a dielectric in an electric field?

Polar dielectrics

Dipoles in an E get rotated and aligned in direction of electric field -thermal motion => not perfect alignment.

Each dipole generates an E in dielectric -net effect is complex -in direction of dipole moment on axis, - in other direction & stronger closer in (goes as 1/r3).

Net effect is to produce an Epolar which is in opposite direction to applied field.



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Ch25-55/57

Dielectrics: an atomic view (cont)

Nonpolar dielectrics

Dipoles are induced by external field. Net charge stays 0, but near plates there are net equal and opposite charges induced which offset charges on plates.

Net effect is to produce an E’ which is in opposite direction to applied field.

For both polar and nonpolar dielectrics, the net effect is to weaken any applied field, by a factor κ.

Why was dielectric “pulled” into capacitor?Ans: Opposite charges on plates/dielectric are attractive

Ch25-56/57

Problem solving techniquesCapacitor with dielectrics as parallel capacitors

A parallel-plate capacitor with two different dielectrics in the above formation can be treated as two capacitors in parallel since the upper and lower surfaces are at the same voltage and share the charge between them.

Must take into account relative areas and different dielectric constants for two parts of capacitor.

+q

-q



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Ch25-57/57

Capacitor with dielectrics as capacitors in series

A parallel-plate capacitor with two different dielectrics in the left formation above can be treated as two capacitors in series. On the right, the `virtual pair of capacitors’ has the same charge distribution as on left since the middle pair of plates exactly cancel each other’s charge. Calculate the effective capacitance using the standard rule for a series, viz

Take into account thickness and different dielectricconstants for the two parts of capacitor.

+q

-q

+q

-q

-q+q

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Ch25 capacitance