Ch. 20 Electric Potential and Electric

Potential Energy

Electric Potential Energy

• Electrical potential energy is the energy contained in a configuration of charges. Like all potential energies, when it goes up the configuration is less stable; when it goes down, the configuration is more stable.

• The unit is the Joule.

Equation

• U = - W = q0Ed

• q0 – test charge

• E – Electric field• d - distance

Electrical potential energy increaseswhen charges are brought into less

favorable configurations

Electrical potential energy decreases when charges are brought into more favorable

configurations.

Work must be done on the charge to increase the electric potential energy

• For a positive test charge to be moved upward a distance d, the electric force does negative work.

• The electric potential energy has increased and U is positive (U2 > U1)

• If a negative charge is moved upward a distance d, the electric force does positive work.

• The change in the electric potential energy U is negative (U2 < U1)

Electric Potential (V)

Electric potential is hard to understand, but easy to measure.

• We commonly call it “voltage”, and its unit is the Volt.

• 1 V = 1 J/C• Electric potential is easily related to both the

electric potential energy, and to the electric field.

• The change in potential energy is directly related to the change in voltage.

U = qVV = U/q

• U: change in electrical potential energy (J)• q: charge moved (C)• V: potential difference (V)• All charges will spontaneously go to lower potential

energies if they are allowed to move.

• Since all charges try to decrease UE, and UE = qV, this means that spontaneous movement of charges result in negative U.

• V = U / q• Positive charges like to DECREASE their potential

(V < 0)• Negative charges like to INCREASE their potential.

(V > 0)

Sample Problem: A 3.0 μC charge is moved through a potential difference of 640 V. What is its potential energy change?

Sample Problem: A 3.0 μC charge is moved through a potential difference of 640 V. What is its potential energy change?

Electrical Potential in Uniform Electric Fields

The electric potential is related in a simple way to a uniform electric field.

V = -Ed• V: change in electrical potential (V)• E: Constant electric field strength (N/C or

V/m)• d: distance moved (m)

• Sample Problem: An electric field is parallel to the x-axis. What is its magnitude and direction if the potential difference between x =1.0 m and x = 2.5 m is found to be +900 V?

• Sample Problem: An electric field is parallel to the x-axis. What is its magnitude and direction if the potential difference between x =1.0 m and x = 2.5 m is found to be +900 V?

Sample Problem

What is the voltmeter reading between A and B? Between A and C? Assume that the electric field has a magnitude of 400 N/m.

1.0 2.0 x(m)

y(m)

1.0

A B

C

Sample Problem

How much work would be done BY THE ELECTRIC FIELD in moving a 2 mC charge from A to C? From A to B? from B to C?. How much work would be done my an external force in each case?

1.0 2.0 x(m)

y(m)

1.0

A B

C

Sample Problem: If a proton is accelerated through a potential difference of 2.000 V, what is its change in potential energy?

How fast will this proton be moving if it started at

rest?

Sample Problem: A proton at rest is released in a uniform electric field. What potential difference must it move through in order to acquire a speed of 0.20 c?

Electric Potential Energy forSpherical Charges

• Electric potential energy is a scalar, like all forms of energy.

U = kq1q2/r• U: electrical potential energy (J)• k: 8.99 × 109 N m2 / C2

• q1, q2 : charges (C)• r: distance between centers (m)This formula only works for spherical charges or

point charges.

Sample ProblemWhat is the potential energy of the configuration shown below?

y (m)

1.0

2.0

2 C 4 C

x (m)2.01.0

Sample ProblemHow much work was done in assembling the charge configuration

shown below?

y (m)

1.0

2.0

2 C

-3 C

4 C

x (m)2.01.0

Sample ProblemHow much work was done in assembling the charge configuration

shown below?

y (m)

1.0

2.0

2 C

-3 C

4 C

x (m)2.01.0

Absolute Electric Potential(spherical)

• For a spherical or point charge, the electric potential can be calculated by the following

Formula: V = kq/r• V: potential (V)• k: 8.99 x 109 N m2/C2

• q: charge (C)• r: distance from the charge (m)• Remember, k = 1/(4o)

Sample Problem

What is the electric potential at (2,2)?

y (m)

1.0

2.0

2 C

-3 C

4 C

x (m)2.01.0

Electric Field and Electric Potential

E = - V / dTwo things about E and V: • The electric field points in the direction of

decreasing electric potential.• The electric field is always perpendicular to the

equipotential surface.

20-4 Equipotential Surfaces and the Electric Field

For two point charges:

Equipotential Surfaces and the Electric Field

An ideal conductor is an equipotential surface. Therefore, if two conductors are at the same potential, the one that is more curved will have a larger electric field around it. This is also true for different parts of the same conductor.

Equipotential Surfaces and the Electric Field

There are electric fields inside the human body; the body is not a perfect conductor, so there are also potential differences.

An electrocardiograph plots the heart’s electrical activity.

Equipotential Surfaces and the Electric Field

An electroencephalograph measures the electrical activity of the brain:

Sample ProblemDraw field lines for the charge configuration below. The field is 600

V/m, and the plates are 2 m apart. Label each plate with its proper potential, and draw and label 3 equipotential surfaces between the plates. You may ignore edge effects.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

Fill in the following table for spherical charges

PotentialField

Potential EnergyForce

Capacitor

• Named for the capacity to store electric charge and energy.

• A capacitor is two conducting plates separated by a finite distance:

The capacitance relates the charge to the potential difference:

Sample Problem: A 0.75 F capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor?

Sample Problem: A 0.75 F capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor?

V = 16 V, C = 0.75 F = 0.75 x 10-6 F Q =?C = Q/V or Q = CVQ = (0.75 x 10-6)(16)Q = 1.2 x 10-5 C

A simple type of capacitor is the parallel-plate capacitor. It consists of two plates of area A separated by a distance d.

By calculating the electric field created by the charges ±Q, we find that the capacitance of a parallel-plate capacitor is:

The general properties of a parallel-plate capacitor – that the capacitance increases as the plates become larger and decreases as the separation increases – are common to all capacitors.

Capacitor GeometryThe capacitance of a

capacitor depends on HOW you make it.

2

212

o

o

1085.8

constantty permittivi vacuum

alityproportion ofconstant

platesbeteween distance d

plate of aread

1

Nm

Cx

d

AC

A

CAC

o

d

AC o

• Sample Problem: What is the AREA of a 1 F capacitor that has a plate separation of 1 mm?

C = 1 F, d = 1 mm = 0.001 m, = 8.85 x 10-12 C2/(Nm2)

Sides

A

Ax

D

AC o

001.01085.81 12

1.13 x 108 m2

10629 m

Is this a practical capacitor to build?

NO! – How can you build this then?

The answer lies in REDUCING the AREA. But you must have a CAPACITANCE of 1 F. How can you keep the capacitance at 1 F and reduce the Area at the same time?

Add a DIELECTRIC!!!

A dielectric is an insulator; when placed between the plates of a capacitor it gives a lower potential difference with the same charge, due to the polarization of the material. This increases the capacitance.

DielectricRemember, the dielectric is an insulating material placed between the

conductors to help store the charge. In the previous example we assumed there was NO dielectric and thus a vacuum between the plates.

Dielectric

kd

AkC o

All insulating materials have a dielectric constant associated with it. Here now you can reduce the AREA and use a LARGE dielectric to establish the capacitance at 1 F.

Sample Problem: A parallel plate capacitor is constructed with plate of an area of 0.028 m2 and a separation of 0.55 mm. Find the magnitude of the charge of this capacitor when the potential difference between the plate is 20.1 V.

A = 0.028 m2 d = 0.55 mm = 0.00055 m, V = 20.1 VQ = ?Q = CV

C = A/d C=(8.85 x 10 -12)(0.028) /(0.00055) C = 4.51 x 10-10 F

Q = (4.51 x 10-10 )(20.1) Q = 9.06 x 10-9 C

The polarization of the dielectric results in a lower electric field within it; the new field is given by dividing the original field by the dielectric constant κ:

Therefore, the capacitance becomes:

The dielectric constant is a

property of the material; here

are some examples:

Using MORE than 1 capacitorLet’s say you decide that 1

capacitor will not be enough to build what you need to build. You may need to use more than 1. There are 2 basic ways to assemble them together

• Series – One after another

• Parallel – between a set of junctions and parallel to each other.

Capacitors in SeriesCapacitors in series each charge each other by INDUCTION. So they each

have the SAME charge. The electric potential on the other hand is divided up amongst them. In other words, the sum of the individual voltages will equal the total voltage of the battery or power source.

Capacitors in ParallelIn a parallel configuration, the voltage is the same because

ALL THREE capacitors touch BOTH ends of the battery. As a result, they split up the charge amongst them.

Capacitors “STORE” energyAnytime you have a situation where energy is “STORED” it is called

POTENTIAL. In this case we have capacitor potential energy, Uc

Suppose we plot a V vs. Q graph. If we wanted to find the AREA we would MULTIPLY the 2 variables according to the equation for Area.

A = bh

When we do this we get Area = VQ

Let’s do a unit check!

Voltage = Joules/CoulombCharge = CoulombsArea =

ENERGY

Potential Energy of a CapacitorSince the AREA under the line is a triangle, the ENERGY(area) =1/2VQ

C

QQ

C

QU

CVVCVU

V

QCVQU

C

C

C

2)(2

1

21)(2

1

21

2

2

This energy or area is referred as the potential energy stored inside a capacitor.

Note: The slope of the line is the inverse of the capacitance.

most common form

Sample Problem: In a typical defibrillator, a 175 F, is charged until the potential difference between the plates is 2240 V. A.) What is the charge on each plate?

V = 2240 V, C = 175 F = 175 x 10-6 FQ = ?Q = CVQ = (175 x 10-6)(2240)Q = 0.392 C

B.) Find the energy stored in the charged up defibrillator.

U = ? Since you now know Q, C, & V. You

may use any of the 3 equations to find U.

U = ½ CV2 U = ½ QV U = Q2/(2C)U = ½ (0.392)(2240)

U = 439 J

• The energy stored in a capacitor can be put to a number of uses: a camera flash; a cardiac defibrillator; and others. In addition, capacitors form an essential part of most electrical devices used today.

• If we divide the stored energy by the volume of the capacitor, we find the energy per unit volume; this result is valid for any electric field:

If the electric field in a dielectric becomes too large, it can tear the electrons off the atoms, thereby enabling the material to conduct.

This is called dielectric

breakdown; the field at which this happens is called

the dielectric strength.