Electric Potential and Electric EnergyChapter 17

Potential Energy

Let's go back to junior physics for a second :)

What is gravitational potential energy? Energy that depends on an object's mass

and its position relative to some point i.e. To calculate someone's potential

energy relative to the surface of the Earth you'd need mass, g and height above the surface

Electric Potential Energy

The idea of electric potential energy is similar to that of gravitational potential energy

Electric potential energy for a charge is calculated based on the magnitude of the charge and its position relative to some point

Gravitational vs Electric Potential Energy (p. 504)

Caption: (a) Two rocks are at the same height. The larger rock has more potential energy. (b) Two charges have the same electric potential. The 2Q charge has more potential energy.

Recall point charges in electric fields (ch 16)

Let's say you have an electric field of magnitude 4500 N/C pointing toward the right

If you place a proton in that field, what is the magnitude and direction of the force acting on that proton?

F=qE=7.2 x 10-16 N Right

+E

The proton in the electric field

So since the force acting on the proton is toward the right, it will accelerate toward the right

What will happen to the proton's kinetic energy and electric potential energy?

Kinetic Energy will increase EPE will decrease (conservation of

energy)

Two charged plates (capacitor) Let's say we've got two

charged plates that are separated by a small distance (this is a capacitor)

The E-field points from left to right

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Two charged plates (capacitor) A proton between these

two plates would move towards the negative plate (right)

An electron between these two plates would move towards the positive plate (left)

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E-field betweenTwo charged plates

Two charged plates (capacitor)p. 503

A proton has the highest potential energy when it's near the positive plate

An electron has the highest potential energy when it's near the negative plate

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E-field betweenTwo charged plates

Potential for pos/neg charges

By convention, the positive plate is at a higher potential than the negative plate

Positively charged objects move from higher potential to lower potential (i.e. towards negative plate)

Negatively charged objects move from lower potential to higher potential (i.e. towards positive plate)

Potential for positive/negative charges

Electric Potential (V)

Electric Potential, V, is the potential energy per unit charge

Unit is Volts (1 V= 1J/1 C)

If a point charge, q, has an electric potential energy at some point a, then the electric potential is

V= PE/q

¿q> ¿

¿PEa>¿¿

V=¿

Electric potential and Potential Energy

The change in potential energy of a charge, q, when moved between two points a and b

ΔPE = PEb-PE

a=qV

ba

Sample Problem p. 505

An electron in a television set is accelerated from rest through a potential difference V

ba=+5000 V What is the change in PE of the electron? What is the speed of the electron as a

result of the acceleration? Repeat for a proton that accelerates

through a potential difference of -5000 V

Change in PE of electron

ΔPE = Peb-PE

a=qV

ba

ΔPE = qVba

=(-1.6 x 10-19 C)(5000 V)

ΔPE = -8 x 10-16 J Potential Energy was lost!

What is the speed of the electron as a result of the acceleration?

Conservation of Energy! The amount of PE lost, must be equal to

the amount of KE gained!

KE= 8 x 10-16 J=0.5mv2

V=4.2 x 107 m/s

For the proton

ΔPE = qVba

=(1.6 x 10-19 C)(-5000 V)

ΔPE = -8 x 10-16 J (Same as electron)

Velocity is less because speed is greater V=9.8 x 105 m/s

Potential Difference Since potential energy is always measured

relative to some other point, only differences in potential energy are measurable

Potential Difference is also known as voltage

Potential Difference

In order to move a charge between two points a and b, the electric force must do work on the charge

Vab

=Va-V

b= -W

ba/q

The potential difference between two points a and b is equal to the negative of the work done by the electric force to move the charge from point b to point a, divided by the charge

Sample Problem p. 522 #2

How much work is needed to move a proton from a point with a potential of +100 V to a point where it is -50 V?

Break it down We're moving the proton from +100 V to -50

V Therefore point A is +100 V, point B is -50

V We're looking for the work done by the field -W

ba= qV

ab=q(V

a-V

b)

-Wba

= (1.6 x 10-19 C)(100V -(-50V))

Wba

= -2.4 x 10-17 J

Back to the parallel plates!

For two parallel plates, the relationship between electric field and electric potential is below

E=Vba

/d d is the distance

between the plates

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E-field betweenTwo charged plates

The electron volt

The electron volt is another unit for energy 1 ev= 1.6 x 10-19 J Problem: A proton has 2 MeV of kinetic

energy, how fast is it moving? 2 x106 eV= 3.2 x 10-13 J= 0.5mv2

V= 1.96 x 107 m/s

Section 17-Equipotential Lines

Equipotential lines are used to represent electric potential

Equipotential lines are always perpendicular to electric field lines

Equipotential Lines (p. 507)

Equipotential lines (green) are perpendicular to the electric field lines (red)

17-5 Electric Potential due to Point Charges (p. 509)

The electric potential at a distance r from a single point charge q is : V=kQ/r

Potential is zero at infinity The potential near a positive charge is large

and decreases toward zero at large distances

Electric potential p. 509

The potential near a negative charge is negative and increases toward zero at large distances

Bringing charges togetherEx 17-3 p. 509

What minimum work is required by an external force to bring a charge q = 3.00 microC from a great distance away to a point 0.500 m from a charge Q= 20.0 microC?

Analyze the problem

Basically, we're taking the charge q from a place of zero potential, to a place of nonzero potential

Use our trusty equation:Vab

=Va-V

b= -W

ba/q

Figure out the work done

The charge is coming from infinity, so Va=0

What is Vb?

Vb=KQ/r=(9x109 Nm2/C2)(20x10-6C)/0.500m

Vb= 360,000 V

Wba

= -q(Va-V

b)=-(3.00x10-6C)(0-360000V)

W= 1.08 J

Electric potential of multiple charges

Electric fields are vectors, but electric potential is a scalar!

When determining the electric potential at a point you can just add the electric potential from each charge, just be sure to include the correct sign of the charge when calculating potential

Example Calculate the electric field at a point midway

between a -0.5 microC charge and a -0.8 microC charge that are separated by 0.50 m.

For the -0.5 microC charge, E= 72000 N/C left

For the -0.8 microC charge, E= 115,200 N/C right

Therefore E is 43200 N/C right

Electric Potential

Calculate the electric potential at a point midway between a -0.5 microC charge and a -0.8 microC charge that are separated by 0.50 m.

For the -0.5 microC charge, V=kQ/r= (9x109 Nm2/C2)(-0.5 x 10-6 C)/0.25m

V= -18000 N/C

Electric Potential

For the -0.8 microC charge, V=kQ/r= (9x109 Nm2/C2)(-0.8 x 10-6 C)/0.25m

V= -28800 V

Total V= -46800 V This is much easier! No directions...just

make sure you include the sign!

Section 17-7- Capacitance

A capacitor stores electric charge and consists of two conducting objects that are placed next to each other but not touching

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E-field betweenTwo charged plates

Capacitance p. 513• If a voltage is applied to a

capacitor (i.e. connected to a battery), then it becomes charged

• Amount of charge for each plate:

• C= Capacitance of the capacitor (different for each capacitor)

• Unit for C is farad (F)

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CVQ

Capacitance of the Capacitor

d

AC 0

•A= Area of plates

•If A increases, C increases

•d= distance between the plates

•If d increases, C decreases

•ε0 = 8.85 x 10-12 C2/Nm2

(This is the permitivity of free space)

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d

Storage of Electrical Energy

A charged capacitor stores electric energy

C

QCVQVU

22

2

1

2

1

2

1Energy Stored

Sample Problem p. 524 #41

A 7.7 µF capacitor is charged by a 125 V battery and then is disconnected from the battery. When this capacitor (C1) is connected to a second, uncharged capacitor (C2), the voltage on the first drops to 15 V. What is the value of C2? (Charge is conserved)

Solve the Problem

• For the first capacitor:

• When the capacitors are connected, the voltage on the first one is 15 V. That means the new charge on C1 is:

CxVFxCVQ 46 10625.9)125)(107.7(

CxVFxCVQ 46 10155.1)15)(107.7(

Solving the problem

• What happens to the rest of the charge?

• It must be on capacitor 2 because charge is conserved

• Since the two capacitors are connected, the voltage for the second one must also be 15 V

CxCxCxQ 4442 1047.810155.110625.9

FxCx

V

QC 5

42 106.5

15V

1047.8

Connected Capacitors• Capacitors can be

connected in series or parallel

• When capacitors are connected in parallel, the equivalent capacitance is the sum

• The voltage across each capacitor is the same

Capacitors in Parallel

.....321 CCCCeq

321 VVV

...321 QQQQ

Capacitors in Series

If the capacitors are connected in series, the equivalent capacitance is given by the following expression

...1111

321

CCCC

Capacitors in Series

For capacitors in series, the total voltage must equal the sum of the voltages across each capacitor

The charge on each capacitor is the same as the charge on the equivalent capacitor for capacitors in series

321 VVVVtotal

321 QQQ

Sample Problem •What is the equivalent capacitance for this combination of capacitors?

•C2 and C3 are connected in parallel

•Combine them into one capacitor

•C23=C2 + C3 = 35 µFC1 = 12 µFC2 = 25 µFC3 = 10 µF

Simplify the Combination

C23 and C1 are connected in series

231123

111

CCC

C1 = 12 µFC23 = 35 µF

µF35

1

µF12

11

123

C

FC 94.8123

Sample Problem Continued•How much charge is stored on each capacitor?

•Q=CV

•V1= 50 V (this is the voltage across C1)

CxVFxVCQ 46111 100.6501012

Sample Problem Continued

C1 and C23 are connected in series, therefore the charge on C23 is the same as the charge on C1

CxQQ 4231 100.6

Sample Problem Continued

C2 and C3 are connected in parallel, therefore:

232323 VCQ

V 14.171035

100.66

4

23

2323

Fx

Cx

C

QV

3223 VVV

Sample Problem Finished!

The charge on C2 is:

The charge on C3 is:

CxVFxVCQ 46222 1029.414.171025

CxVFxVCQ 46333 1071.114.171010