ELECTRIC POTENTIAL ENERGYAND ELECTRIC POTENTIAL

• POTENTIAL ENERGY • ELECTRIC POTENTIAL • WORK-ENERGY THEOREM • CAPACITANCE • COMBINATIONS OF CAPACITORS • STORED ENERGY

Written by Dr. John K. Dayton

cosA B

A B

A B A B

A B

W F x

F qE

W qE x

U W

U qE x

POTENTIAL ENERGY IN A UNIFORM ELECTRIC FIELD:

In this example a charged particle is moved from point A to point B in a uniform field by the electrostatic force. The work done by the force and the change in potential energy of the particle can be calculated in the usual way. Remember, energy is a scalar quantity.

ELECTRIC POTENTIAL:

Electric potential is defined as the electric potential energy per unit charge.

U UV V

q q

In a uniform electric field, where U = -qE x, the change in electric potential will be V = -Ex.

The SI unit for electric potential is the volt, V. V=J/C.

To find a particle’s change in potential energy, use:

U q V

Electric potential is a scalar quantity.

EXAMPLE: What is the change in a particle’s potential energy if it moves from a position of 100 volts to a position of 150 volts? Let the particle have an electric charge of -3C.

Click For Answer

3 150 10

150

0

U q V

U C

U J

V V

V vs r for a positive charge

V

r

1 1

A B B A

A BB A

A BB A

qV k

rV V V

q qV k k

r r

V kqr r

r

V

V vs r for a negative charge

If q is a point charge (positive or negative), then the electric potential a distance r from q is given by the top equation. Note, V will be negative if q is negative.

The difference between the potentials at two points is also shown.

THE ELECTRIC POTENTIAL OF A POINT CHARGE:

THE SUPERPOSITION OF ELECTRIC POTENTIALS OFSEVERAL POINT CHARGES:

Let P be a point in space near several point charges such that it is adistance r1 from q1, r2 from q2, and r3 from q3. Then the net electricpotential at P is:

V V V V

V kqr

kqr

kqr

P P P P

P

, , ,1 2 3

1

1

2

2

3

3

+

-

q1 = -4C

10cm

q2 = +3C

15cm

P

r2 = .1803m

r1 = .15m

2 2

2

,1 ,2

.1 .15 0.1803

P P P

r m m m

V V V

This diagram is the first step in a well planned solution.

Solution continues on next slide.

EXAMPLE: Calculate the electric potential on the x axis at 15.0 cm produced by two point charges; q1 = -4.0 C on the origin and q2 = +3.0 C at 10.0cm on the y axis.

Click For Answer

2 2

2 2

1 2

1 2

9 6 6

4

98.99 10 4 10 8.99 10 3 10

.15 .180

9 03 10

3

.

P

Nm NmC

P

CP

kq kqV

r r

C CV

m

V V

m

Superposition of point charge potentials

Electric potential is a scalar so is much easier to work with than the electric field. No directions are involved. Don’t forget to use the sign of the charge when calculating an electric potential.

continued from previous slide…

A particle of charge q and mass m moves under the influence of an electric field from point A to point B.

In General:

THE WORK-ENERGY THEOREM FOR ELECTROSTATICS:

For a single point charge:

For a uniform field E:

For an infinite plane of charge, :2 o

E

ò

For a charged, infinite conducting sheet,

o

E

ò

2 21 12 2

0

0B A A B

K U

mv mv q V

:

1 1

2

A B B A

A B oB A

A B B A

A B B Ao

A B B Ao

V V V

V kqr r

V E x b

V x x

V x x

ò

ò

EXAMPLE: A proton is released from rest 5.0 cm from the surface of a charged sphere of radius 10.0 cm and charge Q = 4.0 C. What will the proton’s speed be when it has moved 1 meter?

2

2

212

19 9 6

27

6

0

0 0

2 1 1

2 1.6 10 8.99 10 4 10 1 1

1.67 10 .

7.21 10

15 1.15

f i f i

ff i

fi f

ms

mC

f

f

N

K K q V V

kQ kQmv e

r r

ekQv

m r r

C Cv

kg

v

m m

Work-Energy Theorem for Electrostatics

Working Equation with Potentials of a spherical charge

Solved for final velocity

Final Answer

The proton will move between ri = .15m and rf = 1.15m.

Click For Answer

CAPACITANCE AND THE CAPACITOR:

A capacitor is comprised of two charged, conducting bodies maintained at a potential difference. The charge in the capacitor is actually a charge separation.

Capacitance is defined as the ratio charge to potential difference within a capacitor.

QC

V

C = capacitance in SI units of coulombs/ volts. This combination is called the farad, F.

Q = charge on capacitor. One conductor has +Q while the other has -Q.

V = voltage difference between conductors.

A capacitor’s capacitance depends on its size and shape, not on the charge separation and not on the voltage difference.

THE PARALLEL PLATE CAPACITOR:

The parallel plate capacitor is comprised of two metal plates that face each other with each plate connected to a battery terminal. The inside area of each plate is A and they are separated by a distance d.

The plate connected to the positive battery terminal will have a charge of +Q on its inside surface. The plate connected to the negative terminal will have -Q on its inside surface.

The voltage difference between the plates will be the battery voltage, V. The electric field between the plate will be uniform given by pointing from the positive plate to the negative plate.

+Q

dA

E

-QV

o

o

o

Q A E A

V Ed

Q E AC

V EdA

Cd

ò

ò

ò

/ oE ò

EXAMPLE: Calculate the capacitance of a parallel plate capacitor made of two circular disks of radius 6.0 cm and separated by 0.5mm. How large a radius should they have if the capacitor is to be 1.0 F?

2

2

2

2

1

2

212

0

3

2

12

3

3

8.85 10 .06

.5 10

.5 10 1

8.85 10

2.00 10 200

4.24 10

o o

CNm

o o

o

CNm

C F pF

r

A rC

d d

mC

m

A r dCC r

d d

m F

m

r

ò ò

ò òò

(a)

(b)

Click For Answer

CAPACITORS CONNECTED IN SERIES:

Beginning with a group of capacitors connected in series, find the single, equivalent capacitor.

In series each capacitor has the same charge on it: Q1 = Q2 = Qeq = Q.

In series the voltages across each capacitor add to the battery voltage:V1 + V2 = Veq = V.

1 21 2

1 2

1 2

1 2

1 1 1

eq

eq

eq

Q Q QV V V

C C C

V V V

Q Q Q

C C C

C C C

C2C1

+

-V

Ceq

+

-V

EXAMPLE: C1 = 4.0 F and C2 = 6.0 F are connected in series to a 24V battery. What is the stored charge in C1?

C2C1

+

-V

Ceq

+

-V

1

1 2

2

1 1 1 1 1

4 6

2.4 24 57.6

2.4

57.6

eq

eq

eq

eq

C F

C C C F F

Q C V F

Q

V

Q C

C

Use series equation

Capacitors in series each have the same charge.

Charge stored in Ceq.

Click For Answer

CAPACITORS CONNECTED IN PARALLEL:

Beginning with a group of capacitors connected in parallel, find the single, equivalent capacitor.

In parallel each capacitor has the same voltage across it: V1 = V2 = Veq = V.

The individual charges add to the charge on the equivalent capacitor: Q1 + Q2 = Qeq.

Q V C Q V C

Q V C

Q Q Q

V C V C V C

C C C

eq eq

eq

eq

eq

1 1 2 2

1 2

1 2

1 2

C2

C1

+

-V

Ceq

+

-V

EXAMPLE: C1 = 4.0 F and C2 = 6.0 F are connected in parallel to a 24.0 V battery. What is the stored charge in C1?

C2

C1

+

-V

Ceq

+

-V

1 2

1 2

1 1 1

4 6

10 24 240

24

4 24

10

96

eq

eq eq

eq

C C C F F

Q C V F V C

V V V V

Q C V F V

F

C

Equation for parallel

Charge in Equivalent Capacitor

Capacitors in parallel have the same voltage

Click For Answer

CAPACITORS CONNECTED IN GENERAL:

C C

Ceq C

eq i

i

1 1

In Parallel:

In Series:

C2C1

+

-V

C3

In the diagram C1 and C2 are not in series; they cannot be directly combined. C2 and C3 are in parallel and can be combined. Once C2,3 is known it can be combined with C1

in series.

ENERGY STORED IN A CAPACITOR:

Capacitors store energy within their electric fields. Assume the average voltage during the charging process is one-half the final voltage, V/2. Thus the charge that separates crosses this voltage and the net change in potential energy is QV/2. This is the energy stored in the capacitor:

221 1 1

2 2 2

QU QV CV

C

Calculate the energy density within a parallel plate capacitor:

o AC Vol Ad V Ed

d ò

The final expression is good for any capacitor.

212

2

12

212

o

o

U CVu

Vol VolA

Eddu

Ad

u E

ò

ò

EXAMPLE: Calculate the equivalent capacitance of the circuit and the energy stored in each of the original capacitors.

+

-V

C1 = 2F

C3 = 2F

C2 = 3F

A series of reduced circuits. C2 and C3 can be combined first because they are in parallel.

C2,3C1

+

-V

Ceq

+

-V

Solution Continues on Next Slide

Click For Answer

continued from previous slide…

2,3 2 3

1 2.3

3 2

1 1 1 1 1

2 5

5

1.4286

eq

eq

C C C F F

C C C

F

C F

F F

1 2,3

11

1

2.32,3

2.3

1.4286 24 34.2857

34.2857

34.285717.1429

2

34.28576.8571

5

eq eq BatQ C V F V C

Q Q C

Q CV V

C F

Q CV V

C F

Compute equivalent capacitor

Distribute charge and voltage up one circuit level

(Because they are in series)

continued from previous slide…

2 3 2.3

2 2 2

3 3 3

221 11 1 12 2

221 12 2 22 2

221 13 3 32 2

12

293.87

6.8571

3 6.8571 20.5713

2 6.8571 13.7142

2 17.1429

3 6.8571

2 6.8571

41

9

70.5297

47.0198

1.429

eq e

V V V V

Q C V F V C

Q C V F V C

U C V F V

U C V F V

U C V F V

to

J

J

J

U

J

tal

C

22 12 1.4286 24 411.429q BatV F V J

(Because they are in parallel)

The total of the individual stored energies must equal the energy stored in the equivalent capacitor.

ANSWERS

End Of Presentation