Electric Potential. Electric Potential Energy and Potential Difference Relation between Electric Potential and Electric Field Electric Potential Due to

Electric Potential

• Electric Potential Energy and Potential Difference

• Relation between Electric Potential and Electric Field

• Electric Potential Due to Point Charges

• Potential Due to Any Charge Distribution

• Equipotential Surfaces

• Electric Dipole Potential

• E Determined from V

• Electrostatic Potential Energy; the Electron Volt

• Cathode Ray Tube: TV and Computer Monitors, Oscilloscope

The electrostatic force is conservative – potential energy can be defined.

Change in electric potential energy is negative of work done by electric force:

Electrostatic Potential Energy and Potential Difference

Electric potential is defined as potential energy per unit charge:

Unit of electric potential: the volt (V):

1 V = 1 J/C.


Only changes in potential can be measured, allowing free assignment of V = 0:



A negative charge.

Suppose a negative charge, such as an electron, is placed near the negative plate at point b, as shown here. If the electron is free to move, will its electric potential energy increase or decrease? How will the electric potential change?

Analogy between gravitational and electrical potential energy:



Electrical sources such as batteries and generators supply a constant potential difference. Here are some typical potential differences, both natural and manufactured:


Electron in CRT.

Suppose an electron in a cathode ray tube is accelerated from rest through a potential difference Vb – Va = Vba = +5000 V. (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (m = 9.1 × 10-31 kg) as a result of this acceleration?

Relation between Electric Potential and Electric Field

The general relationship between a conservative force and potential energy:

Substituting the potential difference and the electric field:

Relation between Electric Potential and Electric FieldThe simplest case is a uniform field:


Electric field obtained from voltage.

Two parallel plates are charged to produce a potential difference of 50 V. If the separation between the plates is 0.050 m, calculate the magnitude of the electric field in the space between the plates.


Charged conducting sphere.

Determine the potential at a distance r from the center of a uniformly charged conducting sphere of radius r0 for (a) r > r0, (b) r = r0, (c) r < r0. The total charge on the sphere is Q.


The previous example gives the electric potential as a function of distance from the surface of a charged conducting sphere, which is plotted here, and compared with the electric field:


Breakdown voltage.

In many kinds of equipment, very high voltages are used. A problem with high voltage is that the air can become ionized due to the high electric fields: free electrons in the air (produced by cosmic rays, for example) can be accelerated by such high fields to speeds sufficient to ionize O2 and N2 molecules by collision, knocking out one or more of their electrons. The air then becomes conducting and the high voltage cannot be maintained as charge flows. The breakdown of air occurs for electric fields of about 3.0 × 106 V/m. (a) Show that the breakdown voltage for a spherical conductor in air is proportional to the radius of the sphere, and (b) estimate the breakdown voltage in air for a sphere of diameter 1.0 cm.

Point Charges

To find the electric potential due to a point charge, we integrate the field along a field line:

Point ChargesSetting the potential to zero at r = ∞ gives the general form of the potential due to a point charge:

Point Charges

Work required to bring two positive charges close together.

What minimum work must be done by an external force to bring a charge q = 3.00 μC from a great distance away (take r = ∞) to a point 0.500 m from a charge Q = 20.0 µC?

Point Charges

Potential above two charges.

Calculate the electric potential (a) at point A in the figure due to the two charges shown, and (b) at point B.

Charge Distribution

The potential due to an arbitrary charge distribution can be expressed as a sum or integral (if the distribution is continuous):

or

Charge Distribution

Potential due to a ring of charge.

A thin circular ring of radius R has a uniformly distributed charge Q. Determine the electric potential at a point P on the axis of the ring a distance x from its center.

Charge Distribution

Potential due to a charged disk.

A thin flat disk, of radius R0, has a uniformly distributed charge Q. Determine the potential at a point P on the axis of the disk, a distance x from its center.

An equipotential is a line or surface over which the potential is constant.

Electric field lines are perpendicular to equipotentials.

The surface of a conductor is an equipotential.

Equipotential Surfaces


Point charge equipotential surfaces.

For a single point charge with Q = 4.0 × 10-9 C, sketch the equipotential surfaces (or lines in a plane containing the charge) corresponding to V1 = 10 V, V2 = 20 V, and V3 = 30 V.

Equipotential SurfacesEquipotential surfaces are always perpendicular to field lines; they are always closed surfaces (unlike field lines, which begin and end on charges).


A gravitational analogy to equipotential surfaces is the topographical map – the lines connect points of equal gravitational potential (altitude).

The potential due to an electric dipole is just the sum of the potentials due to each charge, and can be calculated exactly. For distances large compared to the charge separation:

Electric Dipole Potential

E E Determined from V

If we know the field, we can determine the potential by integrating. Inverting this process, if we know the potential, we can find the field by differentiating:

This is a vector differential equation; here it is in component form:

E Determined from V

E for ring and disk.

Use electric potential to determine the electric field at point P on the axis of (a) a circular ring of charge and (b) a uniformly charged disk.

Electrostatic Potential Energy; the Electron Volt

The potential energy of a charge in an electric potential is U = qV. To find the electric potential energy of two charges, imagine bringing each in from infinitely far away. The first one takes no work, as there is no field. To bring in the second one, we must do work due to the field of the first one; this means the potential energy of the pair is:

One electron volt (eV) is the energy gained by an electron moving through a potential difference of one volt:

1 eV = 1.6 × 10-19 J.

The electron volt is often a much more convenient unit than the joule for measuring the energy of individual particles.



Disassembling a hydrogen atom.

Calculate the work needed to “disassemble” a hydrogen atom. Assume that the proton and electron are initially separated by a distance equal to the “average” radius of the hydrogen atom in its ground state, 0.529 × 10-10 m, and that they end up an infinite distance apart from each other.

Summary• Electric potential is potential energy per unit charge:

• Potential difference between two points:

• Potential of a point charge:

• Equipotential: line or surface along which potential is the same.

• Electric dipole potential is proportional to 1/r2.

• To find the field from the potential:

Summary

Capacitance and Dielectrics

• Capacitors

• Capacitance

• Capacitors in Series and Parallel

• Electric Energy Storage

• Dielectrics

• Molecular Description of Dielectrics*

•A capacitor is a device that stores charge and electrical energy.

•A capacitor consists of two conductors separated by an insulator.

Capacitors

Parallel-plate capacitor connected to battery. (b) is a circuit diagram.

Capacitors

When a capacitor is connected to a battery, the charge on its plates is proportional to the voltage:

The quantity C is called the capacitance.

Unit of capacitance: the farad (F):

1 F = 1 C/V.

Capacitance

Capacitance

For a parallel-plate capacitor as shown, the field between the plates is

E = Q/ε0A.

In a uniform field, V=Ed:

Vba = Qd/ε0A.

This gives the capacitance:

C=Q/V= ε0A/d

Capacitance

Capacitor calculations.

(a) Calculate the capacitance of a parallel-plate capacitor whose plates are 20 cm × 3.0 cm and are separated by a 1.0-mm air gap. (b) What is the charge on each plate if a 12-V battery is connected across the two plates? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 1 F, given the same air gap d.

CapacitanceCylindrical capacitor.

A cylindrical capacitor consists of a cylinder (or wire) of radius Rb surrounded by a coaxial cylindrical shell of inner radius Ra. Both cylinders have length l which we assume is much greater than the separation of the cylinders, so we can neglect end effects. The capacitor is charged (by connecting it to a battery) so that one cylinder has a charge +Q (say, the inner one) and the other one a charge –Q. Determine a formula for the capacitance.

CapacitanceSpherical capacitor.

A spherical capacitor consists of two thin concentric spherical conducting shells of radius ra and rb as shown. The inner shell carries a uniformly distributed charge Q on its surface, and the outer shell an equal but opposite charge –Q. Determine the capacitance of the two shells.

Determination of Capacitance

Capacitance of two long parallel wires.

Estimate the capacitance per unit length of two very long straight parallel wires, each of radius R, carrying uniform charges +Q and –Q, and separated by a distance d which is large compared to R (d >> R).

Capacitors in parallel have the same voltage across each one:

Capacitors in Parallel

321ab

eq

ab321321

333222111

321ab

)(

, , ,

,

CCCV

QC

VCCCQQQQ

VCQVCQVCQ

VVVV

Capacitors in series have the same charge:

Capacitors in Series

321eq

eq321

321ab

3

33

2

22

1

11

321

1111

,111

, , ,

,

CCCC

C

Q

CCCQ

VVVV

C

QV

C

QV

C

QV

QQQQ

Capacitors in Series and ParallelDetermine the capacitance of a single capacitor that will have the same effect as the combination shown.

Solution:

3223 CCC

In parallel,

In series,

321

321eq

231eq

)(

111

CCC

CCCC

CCC

Capacitors in Series and Parallel

Determine the charge on each capacitor and the voltage across each, assuming C = 3.0 μF and the battery voltage is V = 4.0 V.

Capacitors in Series and ParallelCapacitors reconnected.

Two capacitors, C1 = 2.2 μF and C2 = 1.2 μF, are connected in parallel to a 24-V source as shown. After they are charged they are disconnected from the source and from each other and then reconnected directly to each other, with plates of opposite sign connected together. Find the charge on each capacitor and the potential across each after equilibrium is established.

A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor. The work needed to transfer charge dq is dW=Vdq=(q/C)dq. The total work done is

This work is stored as electrical potential energy:

Electric Energy Storage

C

QdqC

qW

Q

2

2

0

22

E 2

1

2

1

2CVQV

C

QU


A camera flash unit stores energy in a 150-μF capacitor at 200 V. (a) How much electric energy can be stored? (b) What is the power output if nearly all this energy is released in 1.0 ms?

W3000sJ/100.3/P

J,0.3V200F101505.02

1

3-

262

tU

CVU


A parallel-plate capacitor carries charge Q and is then disconnected from a battery. The two plates are initially separated by a distance d. Suppose the plates are pulled apart until the separation is 2d. How has the energy stored in this capacitor changed?


The plates of a parallel-plate capacitor have area A, separation x, and are connected to a battery with voltage V. While connected to the battery, the plates are pulled apart until they are separated by 3x. (a) What are the initial and final energies stored in the capacitor? (b) How much work is required to pull the plates apart (assume constant speed)? (c) How much energy is exchanged with the battery?

The capacitance of a parallel-plate capacitor is C=ε0A/d, and the potential difference between its plates is V=Ed. The energy stored can be written as

Since the volume where the electric fieled exists is Ad, the energy density is defined as:


)(2

1)(

22

1 20

202E AdEEd

d

ACVU

202

1Eu

A dielectric is an insulator, introduced between the plates of a capacitor, and will increase the capacitance. It is characterized by a dielectric constant κ.

Capacitance of a parallel-plate capacitor filled with dielectric:

Dielectrics

d

AC 0

Using the dielectric constant, we define the permittivity:

0

Dielectric strength is the maximum field a dielectric can experience without breaking down.

Dielectrics

Dielectrics•The capacitor connected to a battery

•The voltage remaining constant

•A dielectric inserted

000

0 ,

QVCCVQ

CC

Dielectrics•The capacitor connected to a battery

•The capacitor then disconnected from the battery

•The charge remaining constant

•A dielectric inserted

000

0

0

0

,

,

E

d

V

d

VE

V

C

Q

C

QV

CC

DielectricsA parallel-plate capacitor, filled with a dielectric with K = 3.4, is connected to a 100-V battery. After the capacitor is fully charged, the battery is disconnected. The plates have area A = 4.0 m2 and are separated by d = 4.0 mm. (a) Find the capacitance, the charge on the capacitor, the electric field strength, and the energy stored in the capacitor. (b) The dielectric is carefully removed, without changing the plate separation nor does any charge leave the capacitor. Find the new values of capacitance, electric field strength, voltage between the plates, and the energy stored in the capacitor.

0

0

0

00

,

,

,

A

QE

d

VE

d

AC

VCQ

d

+Q

-Q

E

dt

+Q

-Q

E

E

Ed

A

QEE

0D

inderted, dielectric awith

)1/1(

,)()(

0

0D

td

A

V

QC

ttd

A

QtEtdEV

The molecules in a dielectric, when in an external electric field, tend to become oriented in a way that reduces the external field.

Molecular Description of Dielectrics

This means that the electric field within the dielectric is less than it would be in air, allowing more charge to be stored for the same potential. This reorientation of the molecules results in an induced charge – there is no net charge on the dielectric, but the charge is asymmetrically distributed.

The magnitude of the induced charge depends on the dielectric constant:

Molecular Description of Dielectrics

• Capacitor: nontouching (separated) conductors carrying equal and opposite charge.

• Capacitance:

• Capacitance of a parallel-plate capacitor:

Summary

Summary

• Capacitors in parallel:

• Capacitors in series:

• Energy density in electric field:

• A dielectric is an insulator.

• Dielectric constant gives ratio of total field to external field.

• For a parallel-plate capacitor:

Summary

Documents

Electric Potential. Electric Potential Energy and Potential Difference Relation between Electric Potential and Electric Field Electric Potential Due to