switch goes shut. battery establishes e-field in wires, charge builds up on the capacitor

Switch goes shut.

Battery establishes E-field in wires, charge builds up on the capacitor.

Chemical energy in battery goes down and electrical potential energy stored in capacitor goes up.

Electrical potential energy results from charge separation across capacitor’s plates

Ch 26.4 – Energy Stored in a Capacitor Ch 26.4 – Energy Stored in a Capacitor – charging a capacitor– charging a capacitor

To charge the capacitor, an external agent (battery) must do work to separate the charges.

Step by step, the battery “grabs” a small amount of charge dq off one capacitor plate and move it to the other.

At first, this requires no work, because the uncharged capacitor has no electric field to resist the movement of charge.

But, once dq has been transferred, the capacitor starts to develop a potential difference.

Now, to move more charge from one plate to the other, the battery must do some amount of work “dW,” to overcome the rising potential difference across the plates.

As more and more charge is transferred, the work required to transfer the same amount of charge, dq, goes up.

Ch 26.4 – Energy Stored in a CapacitorCh 26.4 – Energy Stored in a Capacitor – charging a capacitor – charging a capacitor

Suppose q is the amount of charge on the capacitor at some instant during the charging process.

At the same instant, the potential difference across the capacitor is ΔV = q/C

The amount of work necessary to transfer dq amount of charge across this potential difference is

dW = ΔVdq = (q/C)dq

Ch 26.4 – Energy Stored in a CapacitorCh 26.4 – Energy Stored in a Capacitor – charging a capacitor – charging a capacitor

Thus, the total amount of work required to charge the capacitor from q = 0 to a final charge of q = Q is

Q

Q

C

Qqdq

Cdq

C

qW

00

2

2

1

But, in an isolated system with no non-conservative forces, total mechanical energy must be conserved.

Therefore, the work done to charge the capacitor must equal the change in the system’s potential energy.

Ch 26.4 – Energy Stored in a CapacitorCh 26.4 – Energy Stored in a Capacitor

22

2

1

2

1

2VCVQ

C

QU

Energy stored in a charged capacitor


EdV

d

AC 0

-It’s not obvious, but the potential energy stored in the capacitor actually resides in its electric field.

-This implies we should be able to solve the density of the energy stored in the field (J/m3).

-For a parallel plate capacitor, we already know:

-and, its capacitance is just:

-Substituting these into the purple equation,

20

220

2

1

2

1EAddE

d

AU

202

1EuE

-Dividing by the volume in between the plates of the capacitor (V=Ad), we get

Energy per unit volume in a capacitor (J/m3)

Ch 26.4 – Energy Stored in a CapacitorCh 26.4 – Energy Stored in a Capacitor 2

2

2

1

2

1

2VCVQ

C

QU

-We don’t attempt it here, but it can be shown that this result is valid for any electric field!

202

1EuE Energy per unit volume in

an electric field.

-In a very real sense, electric fields “carry” energy.


Two capacitors, C1 and C2 (C1 > C2), are charged to the same initial potential difference, ΔVi. The charged capacitors are removed from the battery, and their plates are connected with opposite polarity, as shown. The switches, S1 and S2, are then closed.

(a) Find the final potential difference ΔVf between a and b after the switches are closed.(b) Find the total energy stored in the capacitors before and after the switches are closed and

determine the ratio of the final energy to the initial energy.

EG 26.4 – Rewiring two Charged CapacitorsEG 26.4 – Rewiring two Charged Capacitors

A dielectric is something you stick in between the plates of a capacitor to change (increase) it’s capacitance.

The term comes from the fact that, at the atomic level, such materials can be polarized into arrays of dipoles.

Ch 26.5 – Capacitors with dielectricsCh 26.5 – Capacitors with dielectrics

Common dielectric materials are: wax, paper, oil, polymers, fluid (electrolyte), etc.

Dielectrics are insulators.

Since you stick the dielectric material into the region of the capacitor’s E-field, it changes how “good” the capacitor is at establishing the E-field

ΔV↓ for the same amount of Q on the capacitor plate.


What happens?

Consider parallel-plate capacitor where ΔV0 = Q0/C0

Assume no battery is connected Q can’t change

When you stick a dielectric in between the plates


0V

V

Consider parallel-plate capacitor where ΔV0 = Q0/C0

Assume no battery is connected Q can’t change

When you stick a dielectric in between the plates

-where κ is a dimensionless constant called the “dielectric constant”


0

0

0

0

00

CC

V

QVQ

V

QC

-Q on the capacitor does not change

-Therefore:

-the capacitance is changed by a factor of κ.

-as κ goes up, C goes up.


d

AC

d

AC

0

00

-For a parallel plate capacitor

To make capacitance ↑-decrease d-increase A-increase κ

- Only limited by “dielectric strength” of the dielectric


Example values of dielectric constant

“Dielectric strength” is themaximum field in the dielectric before breakdown.(a spark or flow of charge)

max maxE V / d

A parallel-plate capacitor is charged with a battery to a charge of Q0. The battery is then removed, and a slab of material that has a dielectric constant κ is inserted between the plates. Identify the system as the capacitor and the dielectric.

Find the energy stored in the system before and after the dielectric is inserted.

EG 26.5 – Energy stored before and afterEG 26.5 – Energy stored before and after

0

20

0 2C

QU

C

QU

2

20

0

A parallel-plate capacitor is charged with a battery to a charge of Q0. The battery is then removed, and a slab of material that has a dielectric constant κ is inserted between the plates. Identify the system as the capacitor and the dielectric.

Find the energy stored in the system before and after the dielectric is inserted.

0

0

20

0 2

U

C

QU

Before:

After:

Where did the energy go?

EG 26.4 – Rewiring two Charged CapacitorsEG 26.4 – Rewiring two Charged Capacitors

The combination of two equal charges of opposite sign, +q and –q, separated by a distance 2a

Every dipole can be characterized by it’s “dipole moment.”- vector which points from –q to +q-magnitude p = 2aq

1p

2p

1 2p p p

Ch 26.6 – Electric Dipole in an Electric FieldCh 26.6 – Electric Dipole in an Electric Field

What happens when we pop this baby in an external E-field?




-external field exerts F=qE on each charge

-net torque about the dipole’s center

-dipole rotates to “align” with the field



-external field exerts F=qE on each charge

-net torque about the dipole’s center

-dipole rotates to “align” with the field

sin

sin

Fa

Fa

sin2Fanet

qEF aqp 2

sinsin2 pEaqE

but,

and

Thus:


sin2Fanet

qEF aqp 2

sinsin2 pEaqE

Ep

but,

and

Thus:


The dipole and the external field are a system

-electric force is an internal conservative force

we can describe its work using a potential energy

In other words, different configurations of the dipole-field system have different potential energies.


As the dipole aligns with the field, the system’s potential energy goes down.


-Work must be done to “un-align” the dipole from the field.

-in an isolated system, the work input must correspond to an increase in potential energy.


-Work must be done to “un-align” the dipole from the field.

-in an isolated system, the work input must correspond to an increase in potential energy.

W = ΔK + ΔU


ddW

sinpE

-To rotate the dipole through some small angle dθ, an amount dW of work must be done.

but,


ddW

sinpE

-To rotate the dipole through some small angle dθ, an amount dW of work must be done.

f

i

f

i

f

i

dpEdpEdUU if

sinsin

)cos(cos]cos[ fipEpE f

i

but,

-so, to rotate the dipole from θi to θf, the change in potential energy is:


0iU 90i

Let’s define the zero potential energy as being when the dipole is at θ = 90,

when


0iU 90i


when

We’ll use this reference energy as an anchor point.

At any time, we can write the system’s instantaneous potential energy, U, with respect to the zero-point potential energy.

090 ffif UUUUUU


0iU 90i


when

cospEU



090 ffif UUUUUU

ffi pEpEpE f

i

cos)cos(cos]cos[

But, we already know


0iU 90i


EpU

when

cospEU



090 ffif UUUUUU

ffi pEpEpE f

i

cos)cos(cos]cos[

But, we already know


A water molecule has an electric dipole moment of 6.3x10-30 Cm. A sample contains 1021 water molecules. All of the dipoles are oriented in the direction of an external E-field, which has a magnitude of 2.5x105 N/C.

How much work is required to rotate all the dipoles from this orientation (θ = 0) to one in all the dipoles are perpendicular to the external field (θ = 90)?

EG 26.6 – The Water MoleculeEG 26.6 – The Water Molecule

A water molecule has an electric dipole moment of 6.3x10-30 Cm. A sample contains 1021 water molecules. All of the dipoles are oriented in the direction of an external E-field, which has a magnitude of 2.5x105 N/C.

How much work is required to rotate all the dipoles from this orientation (θ = 0) to one in all the dipoles are perpendicular to the external field (θ = 90)?

EG 26.6 – The Water MoleculeEG 26.6 – The Water Molecule

WU )0cos()90cos(

090NpENpEUUW

)/105.2)(103.6)(10( 53021 CNmCNpE

J3106.1

Example P26.9

When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the spacing between the plates?

0 AQ Vd

12 2 2

09 2 4 2 2

8.85 10 C N m 150 V4.42 m

30.0 10 C cm 1.00 10 cm m

Vd

Example P26.21

Four capacitors are connected as shown in Figure P26.21.

(a) Find the equivalent capacitance between points a and b.

(b) Calculate the charge on each capacitor if ΔVab = 15.0 V.

1

2.50 F

2.50 6.00 8.50 F

1 15.96 F

8.50 F 20.0 F

s

p

eq

C

C

C

1 1 115.0 3.00sC

5.96 F 15.0 V 89.5 CQ C V

89.5 C4.47 V

20.0 F15.0 4.47 10.53 V

6.00 F 10.53 V 63.2 C on 6.00 F

QV

C

Q C V

89.5 63.2 26.3 C

Example P26.27

Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C1 = 5.00 μF, C2 = 10.0 μF, and C3 = 2.00 μF.

1

1

2

1

1 13.33 F

5.00 10.0

2 3.33 2.00 8.66 F

2 10.0 20.0 F

1 16.04 F

8.66 20.0

s

p

p

eq

C

C

C

C

Example P26.35

A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled?

2 12d d 2 112

C C stored energy doubles,

. Therefore, the

Example P26.43Determine (a) the capacitance and (b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm2 and plate separation of 0.040 0 mm.

12 4 2110

5

2.10 8.85 10 F m 1.75 10 m8.13 10 F 81.3 pF

4.00 10 m

AC

d

6 5max max 60.0 10 V m 4.00 10 m 2.40 kVV E d

Example P26.59A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00 × 108 V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4 000 V. Find the minimum area of the capacitor plates.

8 maxmax 2.00 10 V m

VE

d

60 0.250 10 FA

Cd

62max

12 80 0 max

0.250 10 40000.188 m

3.00 8.85 10 2.00 10

C VCdA

E

3.00

switch goes shut. battery establishes e-field in wires, charge builds up on the capacitor

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