AP-C Electric Potential

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1.a.b.

c.d.e.f.

g.

h.2.

a.b.

3.a.

i.

ii.

iii.b.c.

i.ii.

4.a.

i.ii.iii.

b.i.

ii.iii.iv.v.

vi.

c.d.

5.a.

b.

AP-C Objectives (from College Board Learning Objectives for AP Physics)Electric potential due to point charges

Determine the electric potential in the vicinity of one or more point charges.Calculate the electrical work done on a charge or use conservation of energy to determine the speed of a charge that moves through a specified potential difference.Determine the direction and approximate magnitude of the electric field at various positions given a sketch of equipotentials.Calculate the potential difference between two points in a uniform electric field, and state which point is at the higher potential.Calculate how much work is required to move a test charge from one location to another in the field of fixed point charges.Calculate the electrostatic potential energy of a system of two or more point charges, and calculate how much work is require to establish the charge system.Use integration to determine the electric potential difference between two points on a line, given electric field strength as a function of position on that line.State the relationship between field and potential, and define and apply the concept of a conservative electric field.

Electric potential due to other charge distributionsCalculate the electric potential on the axis of a uniformly charged disk.Derive expressions for electric potential as a function of position for uniformly charged wires, parallel charged plates, coaxial cylinders, and concentric spheres.

ConductorsUnderstand the nature of electric fields and electric potential in and around conductors.

Explain the mechanics responsible for the absence of electric field inside a conductor, and know that all excess charge must reside on the surface of the conductor.Explain why a conductor must be an equipotential, and apply this principle in analyzing what happens when conductors are connected by wires.Show that the field outside a conductor must be perpendicular to the surface.

Graph the electric field and electric potential inside and outside a charged conducting sphere.Understand induced charge and electrostatic shielding.

Explain why there can be no electric field in a charge-free region completely surrounded by a single conductor.Explain why the electric field outside a closed conducting surface cannot depend on the precise location of charge in the space enclosed by the conductor.

CapacitorsUnderstand the definition and function of capacitance.

Relate stored charge and voltage for a capacitor.Relate voltage, charge, and stored energy for a capacitor.Recognize situations in which energy stored in a capacitor is converted to other forms.

Understand the physics of a parallel-plate capacitor.Describe the electric field inside the capacitor and relate the strength of the field to the potential difference and separation between the plates.Relate the electric field to the charge density on the plates.Derive an expression for the capacitance of a parallel-plate capacitor.Determine how changes in the geometry of the capacitor will affect its capacitance.Derive and apply expressions for the energy stored in a parallel-plate capacitor as well as the energy density in the field between the plates.Analyze situations in which capacitor plates are moved apart or closer together, or in which a conducting slab is inserted between capacitor plates.

Describe the electric field inside cylindrical and spherical capacitors.Derive an expression for the capacitance of cylindrical and spherical capacitors.

DielectricsDescribe how insertion of a dielectric between the plates of a charged parallel-plate capacitor affects its capacitance and the field strength and voltage between the plates.Analyze situations in which a dielectric slab is inserted between the plates of a capacitor.

Electric Potential due to Point Charges

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F = − dUdl

= − ddr

14πε0

q1q2r

⎛

⎝⎜⎞

⎠⎟= 14πε0

q1q2r 2

Electric Force from Electric Potential Energy

1.a.b.c.d.e.f.g.h.

AP-C Objectives (from College Board Learning Objectives for AP Physics)Electric potential due to point charges

Determine the electric potential in the vicinity of one or more point charges.Calculate the electrical work done on a charge or use conservation of energy to determine the speed of a charge that moves through a specified potential difference.Determine the direction and approximate magnitude of the electric field at various positions given a sketch of equipotentials.Calculate the potential difference between two points in a uniform electric field, and state which point is at the higher potential.Calculate how much work is required to move a test charge from one location to another in the field of fixed point charges.Calculate the electrostatic potential energy of a system of two or more point charges, and calculate how much work is require to establish the charge system.Use integration to determine the electric potential difference between two points on a line, given electric field strength as a function of position on that line.State the relationship between field and potential, and define and apply the concept of a conservative electric field.

Electric potential (voltage) is the work per unit charge required to bring a charge from infinity to some point R in an electric field.

If there are multiple charges, just add up the electric potentials due to each of the charges.

Electric Potential due to a Point Charge

V =Wq= 14πε0q

q1q2R

= 14πε0

qR

V = 14πε0

qirii

∑ = 14πε0

qirii

∑

Finding Electric Potential from Electric FieldFinding Electric Field from Electric Potential

V = 14πε0

qr→ dVdr

= ddr

14πε0

qr

⎛

⎝⎜⎞

⎠⎟= q4πε0

ddr

1r

⎛⎝⎜

⎞⎠⎟= −q4πε0

ddrr −1( )→

dVdr

= −q4πε0r

2 →dVdrr = −q4πε0r

2 r = −E→

E = − dV

drr

Find the electric potential at the origin due to the following charges: +2µC at (3,0); −5µC at (0,5); and +1µC at (4,4).

Sample Problem: Finding Electric Potential due to a Collection of Point Charges

V = 14πε0

qirii

∑ = 14πε0

+2×10−6

3+ −5×10−6

5+ +1×10−6

42 + 42⎛

⎝⎜⎞

⎠⎟= −1410V

Determine the work required to take a point charge q2 from infinity (U=0) to some point a distance R away from point charge q1.Electrical Potential Energy due to a Point Charge

q1 q2 Fe

Rr

W = −Fe

r=∞

r=R

∫ • dr =Fe

r=R

r=∞

∫ • dr = 14πε0

q1q2r 2R

∞

∫ dr =q1q24πε0

drr 2R

∞

∫ →W =q1q24πε0

−1r R

∞⎛

⎝⎜

⎞

⎠⎟ →U = 1

4πε0

q1q2R

Independent of path since Feis a conservative force!

Equipotentials are surfaces with constant potential, similar to altitude lines on a topographic map.

Equipotential lines always run perpendicular to electric field lines.

The work done in moving a particle through space is zero if its path begins and ends anywhere on the same equipotential line since the electric force is conservative.

Electric field points from high potential to low potential.

Equipotentials

+ V1V2

+ -

An electron is released from rest in a uniform electric field of 500 N/C. What is its velocity after it has traveled one meter?

Sample Problem: Speed of an Electron Released in an E Field

ΔK = −ΔUΔU=−q

Eidr

A

B

∫⎯ →⎯⎯⎯⎯ ΔK = qE i dr

A

B

∫ →

ΔK = qE dr = qEΔr→A

B

∫ 12mv

2 = qEΔr→

v = 2qEΔrm

= 2(1.6×10−19 )(500)(1)9.11×10−31

→

v = 1.33×107 ms

Given an electric potential of V(x)=5x2-7x, find the magnitude and direction of the electric field at x=3 m.

Sample Problem: Electric Field from Potential

E = − dV

drr = − d

dx(5x2 − 7x)i = −(10x − 7)i = (7 −10x)i x=3m⎯ →⎯⎯

E = (7 −10(3))i = −23V m i

Two point charges (5 µC and 2 µC) are placed 0.5 meters apart. How much work was required to establish the charge system? What is the electric potential halfway between the two charges?

Work Required to Establish a System of Point Charges

Ue =14πε0

q1q2r

= 14πε0

(5×10−6 )(2×10−6 )0.5

= 0.18J V = 14πε0

q1r+ 14πε0

q2r= 14πε0

5×10−6

0.25+ 14πε0

2×10−6

0.25= 252kV

V =Wq= 1q

Fe • d

rr

∞

∫ =Feq• dr

r

∞

∫E=Feq⎯ →⎯⎯ V =

E • dr

r

∞

∫

ΔV =VB −VA = −E • dl

A

B

∫ = ΔUq

Electric Potential due to Other Charge Distributions

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1.a.b.

AP-C Objectives (from College Board Learning Objectives for AP Physics)Electric potential due to other charge distributions

Calculate the electric potential on the axis of a uniformly charged disk.Derive expressions for electric potential as a function of position for uniformly charged wires, parallel charged plates, coaxial cylinders, and concentric spheres.

Find the electric potential on the axis of a uniformly charged ring of radius R and total charge Q at point P located a distance z from the center of the ring.

Electric Potential on the Axis of a Uniformly Charged Ring

x

y

z

R

Q

P

ri

VP = Vii∑ = 1

4πε0

Qirii

∑ →VP =14πε0

dqr∫

r=ri⎯ →⎯

VP =1

4πε0ridq ri= z2+R2

dq=Q∫⎯ →⎯⎯⎯∫ VP =

Q4πε0

1

z2 + R2

x

y

z

R

P

r

charge of ring is !Q

drdQ

ri " z 2� r 2

Find the electric potential on the axis of a uniformly charged disk of radius R and total charge Q at point P located a distance z from the center of the disk.

Electric Potential on the Axis of a Uniformly Charged Disk

σ = QπR2

→ dQ = 2πrdrσ→

dQ = 2πrdrQπR2

VP = Vii∑ = 1

4πε0

dQrii

∑ →VP =14πε0

2πrdrQπR2rir=0

R

∫ →

VP =2Q

4πε0R2

rdrri0

R

∫ →VP =Q

2πε0R2 (z2 + r 2 )−

12 r dr

0

R

∫ →

VP =Q

2πε0R2

12

(z2 + r 2 )−12 2r dr u=z2+r2

du=2rdr⎯ →⎯⎯0

R

∫

VP =Q

4πε0R2 2(z

2 + r 2 )12⎡

⎣⎤⎦0

R= Q2πε0R

2 ( z2 + r 2 − z)

Find the electric potential both inside and outside a uniformly charged shell of radius R and total charge Q.Electric Potential due to a Spherical Shell of Charge

RQ

Voutside = −E • dl∫ = − 1

4πε0

Qr 2dr

r=∞

r

∫ = −Q4πε0

r −2 drr=∞

r

∫ = −Q4πε0

−1r

⎛⎝⎜

⎞⎠⎟ ∞

r

→

Voutside =Q4πε0

1r− 1∞

⎛⎝⎜

⎞⎠⎟→Voutside =

Q4πε0r

Vinside = −E • dl∫ = − 1

4πε0

Qr 2dr

∞

R

∫ − 0drR

r

∫ = Q4πε0R

V

rR

Q4 0R

Q4 0r

Find the electric field and electric potential inside a uniformly charged solid insulating sphere of radius R and total charge Q.Electric Potential Inside a Uniformly Charged Solid Insulating Sphere

RFirst find the electric field. Choose a sphere as our Gaussian surface.

Strategy

E • d

A∫ =

Qencε0

→ E(4πr 2 ) = ρVε0

= 3Q4πε0R

3

4πr3

3→ E = Q

4πε0R3 r

Next, integrate to find the electric potential. Note that your total integration from infinity to r must be done piece-wise since the electric field is discontinuous.

Strategy

ρ = QV

= Q43 πR

3 →

ρ = 3Q4πR3

V = −E • dr

∞

r

∫ piece−wise⎯ →⎯⎯⎯ V = −E • dr

∞

R

∫ −E • dr

R

r

∫ = − 14πε0

Qr 2dr

∞

R

∫ − 14πε0R

3 r drR

r

∫ →

V = Q4πε0R

− Q4πε0R

3 r dr =R

r

∫Q

4πε0R− Q4πε0R

3

r 2

2− R

2

2⎛⎝⎜

⎞⎠⎟= Q4πε0R

− Qr 2

8πε0R3 +

QR2

8πε0R3 →

V = 3Q8πε0R

− Qr 2

8πε0R3 →V = Q

8πε0R3− r

2

R2⎛⎝⎜

⎞⎠⎟

Conductors

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1.a.

i.

ii.iii.

b.c.

i.ii.

AP-C Objectives (from College Board Learning Objectives for AP Physics)Conductors

Understand the nature of electric fields and electric potential in and around conductors.Explain the mechanics responsible for the absence of electric field inside a conductor, and know that all excess charge must reside on the surface of the conductor.Explain why a conductor must be an equipotential, and apply this principle in analyzing what happens when conductors are connected by wires.Show that the field outside a conductor must be perpendicular to the surface.

Graph the electric field and electric potential inside and outside a charged conducting sphere.Understand induced charge and electrostatic shielding.

Explain why there can be no electric field in a charge-free region completely surrounded by a single conductor.Explain why the electric field outside a closed conducting surface cannot depend on the precise location of charge in the space enclosed by the conductor.

Charges are free to move in conductors. At electrostatic equilibrium, there are no moving charges in a conductor, therefore there is no net force, and the electric field inside the conductor must be zero. Gauss’s Law therefore states that the charge enclosed must be zero. All excess charge on a conductor lies on the surface of the conductor, and the field on the surface of the conductor must be perpendicular to the surface, otherwise the charges would move.

Charges in a Conductor

Looking at just the outer surface of a conductor:

This should make sense… you have the largest electric field where you have the highest surface charge density.

Electric Field at the Surface of a Conductor

E • d

A∫ =

Qencε0

Symmetry existsQ=σA⎯ →⎯⎯⎯⎯ EA = σA

ε0

→ E = σε0

In a hollow conductor, you can determine the location of charge by utilizing Gauss’s Law. Choose a Gaussian surface in the metal of the hollow conductor, making note that the electric field inside the conductor must be zero.

Therefore, the charge must remain on the outer surface. The entire conductor is at equipotential, and field lines must run perpendicular to the conducting surface.

Charge Distribution in a Hollow Conductor

E • d

A∫ =

Qencε0

E=0⎯ →⎯ Q = 0

Any hollow conductor has zero electric field in its interior. This allows for hollow conductors to be utilized to isolate regions completely from electric fields. In this configuration, a hollow conductor is known as a Faraday Cage.

Faraday Cage

Two conducting spheres, A and B, are placed a large distance from each other. The radius of Sphere A is 5 cm, and the radius of Sphere B is 20 cm. A charge Q of 200 nC is placed on Sphere A, while Sphere B is uncharged. The spheres are then connected by a wire. Calculate the charge on each sphere after the wire is connected.

Sample Problem: Conducting Spheres Connected by a Wire

Once connected by a wire, the spheres must be at equipotential.Further, the sum of the charges on each sphere must equal Q.

A BQ

Q

A

B

Q = QA +QB →QB = Q −QA

VA =QA

4πε0RA=

QB4πε0RB

→QARA

=QBRB

=Q −QARB

→

QARB = QRA −QARA →QARB +QARA = QRA →

QA =QRARA + RB

= (200nC)(5cm)(5cm+ 20cm)

= 40nC

QB = 200nC − 40nC = 160nC

Graph the electric field and the electric potential both inside and outside a solid conducting sphere.

Electric Field and Potential Due to a Conducting Sphere

V

rR

Q4 0R

Q4 0r

E

rR

0

0

Q4 r 2

Capacitors

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1.a.

i.b.

i.ii.iii.iv.

c.d.

AP-C Objectives (from College Board Learning Objectives for AP Physics)Capacitors

Understand the definition and function of capacitance.Relate stored charge, voltage, and stored energy for a capacitor.

Understand the physics of a parallel-plate capacitor.Describe the electric field inside the capacitor and relate the strength of the field to the potential difference and separation between the plates.Relate the electric field to the charge density on the plates.Derive an expression for the capacitance of a parallel-plate capacitor.Determine how changes in the geometry of the capacitor will affect its capacitance.

Describe the electric field inside cylindrical and spherical capacitors.Derive an expression for the capacitance of cylindrical and spherical capacitors.

A capacitor is a device which stores electrical energy. Consisting of two conducting plates separated by an insulator, a capacitor holds opposite charges on each plate with a potential difference across the plates.

What is a Capacitor?Capacitance is the ratio of the charge separated on the plates of the capacitor to the potential difference between the plates. Units of capacitance are coulombs/volt, or farads.

Capacitance

C = QV

1. Assume a charge of +Q and -Q on each of the conductors.2. Find the electric field between the conductors.3. Calculate V by integrating the electric field.4. Utilize C=Q/V to solve for capacitance.

Calculating Capacitance

V = −E • dl∫

Determine the capacitance of two identical parallel plates of area A separated by a distance d.Capacitance of Parallel Plates

d

A

1. Assume a charge of +Q and -Q on each plate.2. Electric field due to a plane of charge is σ/𝓔0.

3.

4.

V = −E • dl∫ = Ed E=σ/ε0⎯ →⎯⎯ V = σd

ε0

σ=Q/A⎯ →⎯⎯ V = Qdε0A

C = QV

= QQd / ε0A

=ε0Ad

Note the electric field between the plates: E=V/d is a constant as long as you stay far from the edges of the plates.

Electric Field Between Parallel Plates

d

A

+Q -Q

E

Determine the capacitance of a long, thin hollow conducting cylinder of radius RB surrounding a long solid conducting cylinder of radius RA.Capacitance of a Cylindrical Capacitor

1. Assume a charge of +Q and -Q on each of the cylinders.2. Determine the electric field between the cylinders.E • d

A∫ =

Qencε0

λ=QL⎯ →⎯⎯ E(2πrL) = λL

ε0→ E = λ

2πε0r3. Calculate V by integrating the electric field.

4. Find C using C=Q/V: C = QV

= QQ

2πε0LlnRARB

⎛⎝⎜

⎞⎠⎟

=2πε0L

ln(RA / RB )

V = −E • dl = − λ

2πε0rdr

r=RA

RB∫∫ = − λ2πε0

drrRA

RB∫ = − λ2πε0

lnRBRA

⎛

⎝⎜⎞

⎠⎟= λ2πε0

lnRARB

⎛

⎝⎜⎞

⎠⎟λ=Q/L⎯ →⎯⎯ V = Q

2πε0LlnRARB

⎛

⎝⎜⎞

⎠⎟

λ = QL

Determine the capacitance of a thin hollow conducting shell of radius RB concentric around a solid conducting sphere of radius RA.Capacitance of a Spherical Capacitor

1. Assume a charge of +Q and -Q on each of the cylinders.2. Determine the electric field between the shells.

RARB

E • d

A∫ =

Qencε0

→ E(4πr 2 ) = Qε0

→ E = Q4πε0r

2

3. Calculate V by integrating the electric field.

V = −E • dl∫ = − Q

4πε0r2 drRA

RB∫ = −Q4πε0

r −2 drRA

RB∫ = −Q4πε0

−1r

⎛⎝⎜

⎞⎠⎟RA

RB

= −Q4πε0

1RA

− 1RB

⎛

⎝⎜⎞

⎠⎟= Q4πε0

1RB

− 1RA

⎛

⎝⎜⎞

⎠⎟= Q4πε0

RA − RBRARB

RA<RB⎯ →⎯⎯

V = Q4πε0

RB − RARARB

4. Find C using C=Q/V: C = QV

= QQ4πε0

RB − RARARB

=4πε0RARBRB − RA

Capacitors and Energy

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1.a.

i.ii.

b.i.ii.

c.

AP-C Objectives (from College Board Learning Objectives for AP Physics)Capacitors

Understand the definition and function of capacitance.Relate stored charge, voltage, and stored energy for a capacitor.Recognize situations in which energy stored in a capacitor is converted to other forms.

Understand the physics of a parallel-plate capacitor.Derive and apply expressions for the energy stored in a parallel-plate capacitor as well as the energy density in the field between the plates.Analyze situations in which capacitor plates are moved apart or closer together, or in which a conducting slab is inserted between capacitor plates.

Describe how insertion of a dielectric between the plates of a charged parallel-plate capacitor affects its capacitance and the field strength and voltage between the plates.

Work is done in charging a capacitor, allowing the capacitor to store energy. If you consider two uncharged conductors in close proximity, the potential difference in moving some amount of charge q from the negative to the positive plate is q/C. Moving more charge increases the potential, therefore the electric potential energy of the charge and the capacitor must also increase.

Energy Stored in a Capacitor

Ucap =qCdq

q=0

Q

∫ = 1C

qdq0

Q

∫ = 1CQ2

2= 12Q2

CC=Q/V⎯ →⎯⎯ U = 1

2QV C=Q/V

Q=CV⎯ →⎯⎯ U = 12CV 2

U = 12CV 2 C=ε0A/d

E=V /d→V =Ed⎯ →⎯⎯⎯⎯ U = 12ε0AdE2d 2 = 1

2ε0E

2Ad Ad=Volume⎯ →⎯⎯⎯ UVolume

= ue =12ε0E

2

Consider a parallel plate capacitor of plate area A and plate separation d. We can think of the energy stored in the capacitor as the work required to create the electric field between the plates. Therefore, a capacitor stores energy by creating an electric field. The amount of energy stored as electric field per unit volume between the plates is known as the field energy density ue.

Field Energy Density

Dielectrics are insulating materials which are placed between the plates of a capacitor to increase the device’s capacitance. When a dielectric is placed between the plates of a capacitor, the electric field between the plates is weakened. This is due to the molecules of the dielectric becoming polarized in the electric field created by the potential difference of the capacitor plates, creating an opposite electric field. The greater the amount of polarization, the greater the reduction in the electric field. Therefore, for a fixed charge charge on the plates Q, the voltage decreases, increasing the capacitance (C=Q/V).

Dielectrics

molecule with no E Field in!uence

polarized due to E Field in!uence

E

+-Eind

The amount by which the capacitance is increased when a dielectric is introduced between the plates of a capacitor is known as the dielectric constant (𝝹) of the dielectric. This constant also corresponds to the amount the electric field strength is reduced due to introduction of the dielectric. The more the molecules / atoms of the dielectric are polarized, the greater the dielectric constant.

Dielectric Constant (𝝹) C =κε0Ad

ε=κε0⎯ →⎯⎯ C = εAd

Enet =

Ew/o dielectric

κε = κε0

permittivity of the dielectric

Capacitors in series have the same charge because each plate must obtain charge from the next plate due to conservation of charge.

Capacitors in Series

V

C1

C2

�Q

�Q

Q

Q

V1

V2

Ceq =QV

= QV1 +V2

= QQC1+ QC2

= 11C1+ 1C2

→

1Ceq

= 1C1

+ 1C2

+ ...

Capacitors in parallel have the same voltage across their plates due to conservation of energy.

Capacitors in Parallel

V C1 C2

Ceq =QV

=Q1 +Q2V

=Q1V

+Q2V

= C1 +C2 →

Ceq = C1 +C2 + ...