Electric Potential IIIElectric Potential III

Physics 2102

Jonathan Dowling

Physics 2102 Physics 2102 Lecture: 10 WED 04 FEBLecture: 10 WED 04 FEB

Ch24.11–12

PHYS2102 FIRST MIDTERM EXAM!6–7PM THU 05 FEB 2009

Dowling’s Sec. 2 in Lockett Hall, Room 6

YOU MUST BRING YOUR STUDENT ID!YOU MUST WRITE UNITS TO GET FULL CREDIT!

The exam will cover chapters 21 through 24, as covered in homework sets 1, 2, and 3. The formula sheet for the exam can be found here:http://www.phys.lsu.edu/classes/spring2009/phys2102/formulasheet.pdf KNOW SI PREFIXES n, cm, k, etc.

THERE WILL BE A REVIEW SESSION 6–7PM WED 04 FEB 2009 in Williams 103

Continuous Charge Continuous Charge DistributionsDistributions

• Divide the charge distribution into differential elements

• Write down an expression for potential from a typical element — treat as point charge

• Integrate!

• Simple example: circular rod of radius r, total charge Q; find V at center.

dq

r

€

V = dV∫ =kdq

r∫

=k

rdq∫ =

k

rQ

€

Units : Nm2

C2

C

m

⎡

⎣ ⎢

⎤

⎦ ⎥=

Nm

C

⎡ ⎣ ⎢

⎤ ⎦ ⎥=

J

C

⎡ ⎣ ⎢

⎤ ⎦ ⎥≡ V[ ]

Potential of Continuous Charge Potential of Continuous Charge Distribution: Line of Charge Distribution: Line of Charge

/Q Lλ = dxdq λ=

∫∫ −+==L

xaL

dxk

r

kdqV

0)(

λ

[ ]LxaLk 0)ln( −+−= λ

⎥⎦⎤

⎢⎣⎡ +=a

aLkV lnλ ⎥⎦

⎤⎢⎣⎡ +=a

aLkV lnλ

•Uniformly charged rod

•Total charge Q•Length L•What is V at position P?

Px

L a

dx

Units: [Nm2/C2][C/m]=[Nm/C]=[J/C]=[V]

Electric Field & Potential: Electric Field & Potential: A A

Simple Relationship!Simple Relationship!Notice the following:• Point charge:

E = kQ/r2

V = kQ/r

• Dipole (far away):E = kp/r3

V = kp/r2

• E is given by a DERIVATIVE of V!

• Of course!

dx

dVEx −=

dx

dVEx −=

Focus only on a simple case: electric field that points along +x axis but whose magnitude varies with x.

Note: • MINUS sign!• Units for E: VOLTS/METER (V/m)

Note: • MINUS sign!• Units for E: VOLTS/METER (V/m)

f

i

V E dsΔ =− •∫r r

f

i

V E dsΔ =− •∫r r

E from V: Example E from V: Example /Q Lλ =

€

V = kλ lnL + a

a

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= kλ ln L + a[ ] − ln a[ ]{ }

€

V = kλ lnL + a

a

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= kλ ln L + a[ ] − ln a[ ]{ }

•Uniformly charged rod

•Total charge Q•Length L•We Found V at P!•Find E from V?

Px

L a

dx

€

E = −dV

da= −kλ

d

daln L + a[ ] − ln a[ ]{ }

= −kλ1

L + a−

1

a

⎧ ⎨ ⎩

⎫ ⎬ ⎭

= −kλa− L + a( )L + a( )a

⎧ ⎨ ⎩

⎫ ⎬ ⎭= kλ

L

L + a( )a

⎧ ⎨ ⎩

⎫ ⎬ ⎭

€

E = −dV

da= −kλ

d

daln L + a[ ] − ln a[ ]{ }

= −kλ1

L + a−

1

a

⎧ ⎨ ⎩

⎫ ⎬ ⎭

= −kλa− L + a( )L + a( )a

⎧ ⎨ ⎩

⎫ ⎬ ⎭= kλ

L

L + a( )a

⎧ ⎨ ⎩

⎫ ⎬ ⎭

Units:

€

Nm2

C2

C

m

m

m2

⎡

⎣ ⎢

⎤

⎦ ⎥=

N

C

⎡ ⎣ ⎢

⎤ ⎦ ⎥=

V

m

⎡ ⎣ ⎢

⎤ ⎦ ⎥ √ Electric Field!

Electric Field & Potential: Electric Field & Potential: Question Question

• Hollow metal sphere of radius R has a charge +q

• Which of the following is the electric potential V as a function of distance r from center of sphere?

+q

Vr1≈

rr=R

(a)

Vr1≈

rr=R

(c)

Vr1≈

rr=R

(b)

+q

Outside the sphere:• Replace by point charge!Inside the sphere:• E = 0 (Gauss’ Law) • E = –dV/dr = 0 IFF V=constant

2

dVE

drd Qk

dr r

Qkr

=−

⎡ ⎤=− ⎢ ⎥⎣ ⎦

= 2

dVE

drd Qk

dr r

Qkr

=−

⎡ ⎤=− ⎢ ⎥⎣ ⎦

=V

r1≈

Electric Field & Potential: Electric Field & Potential: Example Example

E2

1r

≈

Equipotentials and Equipotentials and ConductorsConductors

• Conducting surfaces are EQUIPOTENTIALs

• At surface of conductor, E is normal (perpendicular) to surface

• Hence, no work needed to move a charge from one point on a conductor surface to another

• Equipotentials are normal to E, so they follow the shape of the conductor near the surface.

EV

Conductors Change the Conductors Change the Field Around Them!Field Around Them!

An Uncharged Conductor:

A Uniform Electric Field:

An Uncharged Conductor in the Initially Uniform Electric Field:

Sharp ConductorsSharp Conductors• Charge density is higher at conductor surfaces that have small radius of curvature

• E = for a conductor, hence STRONGER electric fields at sharply curved surfaces!

• Used for attracting or getting rid of charge: – lightning rods– Van de Graaf -- metal brush transfers charge from rubber belt

– Mars pathfinder mission -- tungsten points used to get rid of accumulated charge on rover (electric breakdown on Mars occurs at ~100 V/m)

(NASA)

Ben Franklin Invents the Lightning Rod!

Summary:Summary:• Electric potential: work needed to bring +1C from infinity; units = V = Volt

• Electric potential uniquely defined for every point in space -- independent of path!

• Electric potential is a scalar -- add contributions from individual point charges

• We calculated the electric potential produced by a single charge: V = kq/r, and by continuous charge distributions : V = ∫kdq/r

• Electric field and electric potential: E= -dV/dx• Electric potential energy: work used to build the system, charge by charge. Use W = U = qV for each charge.

• Conductors: the charges move to make their surface equipotentials.

• Charge density and electric field are higher on sharp points of conductors.