Physics 201 - galileo.phys.virginia.edu

Physics 201Professor P. Q. Hung

311B, Physics Building

Physics 201 – p. 1/28

Electric Potential and Energy

Summary of last lecture

Electric Potential for a constant electric field:

VB − VA = − ~E.~s

Physics 201 – p. 2/28


Summary of last lecture

Electric Potential for a constant electric field:

VB − VA = − ~E.~s

Electric Potential for a point charge:

V = k qr

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Equipotential surfaces and Electric fields:Example

A parallel plate capacitor is composed on twocharged plates, one positive and the other onenegative. A constant electric field points from thepositive plate to the negative plate. Suppose thepotential difference between the 2 plates is 64Vand that they are separated by 0.032m. By thiswe mean ∆V = V− − V+ = −64V . Between thetwo plates, one can draw equipotential planesparallel to the plates. What is the separationbetween 2 of such planes if 3.0V ?

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Equipotential surfaces and Electric fields

Equipotential Surface:

Surface where the electric potential is thesame everywhere.

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What are the equipotential surfaces around apoint charge?

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What are the equipotential surfaces around apoint charge?

What are the equipotential surfaces betweentwo charged parallel plates (one + and one -)(parallel-plate capacitor)?

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Equipotential surfaces for a point charge

V = k qr

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Equipotential surfaces for two point charges

V = k qr

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Equipotential surfaces for a parallel-platecapacitor

Recall Active Example 19.3: | ~E| = σǫ0

: Constantelectric field.

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Equipotential surfaces and Electric fields:Meanings

When a particle moves from one point toanother point on a equipotential surface, thenet electric force does no work. SinceVB − VA = −WAB/q, “equipotential” meansthat VB = VA ⇒ WAB = 0.

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The electric field always point in the directionperpendicular to the equipotential surface. Ifnot, it would have a component parallel to thesurface which means that the surface is nolonger equipotential since the electric fieldpoints in the direction of decreasing potential

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A conductor’s surface is an equipotentialsurface because the electric field has to beperpendicular to the conductor’s surface.Since there cannot be an electric field insidethe conductor, the electric potential inside hasthe same value as that on the surface.

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Charged conductor of arbitrary shape

Charged conducting sphere: 1) V = 4πkσRon the surface and inside; 2) E = 4πkσ on thesurface and perpendicular to it and zeroinside.

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Two spheres with different radii, R1 and R2

and same potential ⇒ σ1R1 = σ2R2 ⇒σ1 = σ2R2/R1

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Two spheres with different radii, R1 and R2

and same potential ⇒ σ1R1 = σ2R2 ⇒σ1 = σ2R2/R1

If R1 ≪ R2 ⇒ σ1 ≫ σ2 ⇒ E1 ≫ E2 ⇒ Sharpends of a charged conductor have largerelectric fields.

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Equipotential surfaces and Electric fields:Medical applications

The body is not an ideal conductor ⇒differences in potential from one place toanother

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Equipotential surfaces and Electric fields:Medical applications

The body is not an ideal conductor ⇒differences in potential from one place toanother

Differences in potential ⇒ Electrocardiographand electroencephalograph

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Equipotential surfaces and Electric fields: Movingfrom one surface to another

Take two equipotential surfaces which arevery close to each other so that the electricfield is more or less constant:

∆V = −E∆ s ⇒ E = −∆V∆ s

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∆V = −E∆ s ⇒ E = −∆V∆ s

∆V∆ s is a potential gradient. The minus signsays that the electric field points in thedirection of decreasing potential.

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∆V = −E∆ s ⇒ E = −∆V∆ s

∆V∆ s is a potential gradient. The minus signsays that the electric field points in thedirection of decreasing potential.

Change in the electric potential ⇒ electricfield.

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Equipotential surfaces and Electric fields:Example

A parallel plate capacitor is composed on twocharged plates, one positive and the other onenegative. A constant electric field points from thepositive plate to the negative plate. Suppose thepotential difference between the 2 plates is 64Vand that they are separated by 0.032m. By thiswe mean ∆V = V− − V+ = −64V . Between thetwo plates, one can draw equipotential planesparallel to the plates. What is the separationbetween 2 of such planes if 3.0V ?

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Equipotential surfaces and Electric fields:Solution to Example

Between 2 parallel charged plates, theelectric field is constant. One obtains

E = −∆V∆s = − −64 V

0.032m

= 2.0 × 103 V/m

The electric field is pointing in the direction ofdecreasing potential.

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Equipotential surfaces and Electric fields:Solution to Example

Let ∆d be the separation between these twoequipotential surfaces. Since E is constant,one has

∆d = −∆VE = − −3.0V

2.0×103 V/m

= 1.5 × 10−3m

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Capacitors

Example of a capacitor: twooppositely-charged parallel plates .

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Capacitors

Example of a capacitor: twooppositely-charged parallel plates .

Experiment: Increase the charge Q on eachplate ⇒ Direct increase in the potentialdifference. ⇒ Q is directly proportional to ∆V .

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Capacitors

The constant of proportionality is called thecapacitance, i.e. the capacity of the device tostore charge.

Q = C|∆V | (6)

C is called the capacitance (a positivenumber always).Unit: 1 farad(F ) = 1C/V .

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Capacitors

Numerous applications: For example, a RAMchip consists of millions of transitor-capacitorunits. Capacitor is charged: 1. Capacitor isuncharged: 0.

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Capacitors: Calculating capacitances

Parallel-plate capacitor of separation d andarea A:Three pieces of information:E = σ/ǫ0 = Q/(ǫ0A), |∆V | = Ed, andC = Q/|∆V |.⇒

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Capacitors: Calculating capacitances

Parallel-plate capacitor of separation d andarea A:Three pieces of information:E = σ/ǫ0 = Q/(ǫ0A), |∆V | = Ed, andC = Q/|∆V |.⇒

C = QEd = ǫ0A

d

Purely geometrical!

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Capacitors: ExampleWhen a potential difference of 150 V is applied tothe plates of a parallel- plate capacitor, the platescarry a surface charge density of 30.0nC/cm2.What is the spacing between the plates?

Use C = ǫ0Ad . ⇒ d = ǫ0A

C .

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Capacitors: ExampleWhen a potential difference of 150 V is applied tothe plates of a parallel- plate capacitor, the platescarry a surface charge density of 30.0nC/cm2.What is the spacing between the plates?

Use C = ǫ0Ad . ⇒ d = ǫ0A

C .

C = Q/|∆V | = σA/|∆V | ⇒

d = ǫ0|∆V |σ = 4.42 × 10−6m

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Capacitors: Dielectric

Electric dipole moments: centers of positiveand negative charges do not coincide. Somematerial have permanent electric dipolemoments

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Electric dipole moments: centers of positiveand negative charges do not coincide. Somematerial have permanent electric dipolemoments

Insert a slab of such material in between twooppositely charged plates.

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Between the plates, the negative sides areattracted to the positive plate and the positivesides are attracted to the negative plate,creating an electric field which points in theopposite direction to that of the externalelectric field E0, and partially cancelling itinside. ⇒ Inside E, will be smaller than theelectric field without the slab ⇒ Reduction inthe electric field ⇒ Decrease in the voltage ⇒Increase in the capacitance when the chargeis kept fixed.

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Capacitors: Capacitance in presence ofDielectric

Dielectric constant: κ:

κ = E0

E

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κ = E0

E

V0 = E0d ⇒ V0 = κEd ⇒ V0 = κV .

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κ = E0

E

V0 = E0d ⇒ V0 = κEd ⇒ V0 = κV .

Capacitance in the presence of a dielectric:

C = QV = Q

(V0/κ) = κC0

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Energy storage

W = 12QV : Total work done to completely

charge up a capacitor

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Energy storage



U = 12QV = 1

2CV 2 = Q2

2C :Stored as potentialenergy

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Energy storage



U = 12QV = 1

2CV 2 = Q2

2C :Stored as potentialenergy

u = U/(Ad) = 12κǫ0E

2: Energy density storedbetween the plates

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Some applications

Computer key:Parallel-plate capacitor filled with a dielectric.One end is fixed and the other end movable(attached to a key). Push the key down ⇒decrease the separation ⇒ increase thecapacitance detected by an electronic circuit⇒ signal is sent to the computer.

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Some applications

Computer key:Parallel-plate capacitor filled with a dielectric.One end is fixed and the other end movable(attached to a key). Push the key down ⇒decrease the separation ⇒ increase thecapacitance detected by an electronic circuit⇒ signal is sent to the computer.

Neurons: Discussed in extraprob1-202.pdf.

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