Physics 2113 Lecture: 14 - LSU

Physics 2113 Lecture: 14

Electric Potential III

Physics 2113 Jonathan Dowling

Conservative Forces, Work, and Potential Energy

�

W = F r( )∫ dr Work Done (W) is Integral of Force (F)

�

U = −W Potential Energy (U) is Negative of Work Done

�

F r( ) = − dUdr

Hence Force is Negative Derivative of Potential Energy

Coulomb’s Law for Point Charge

P1 P2 q1 q2

P1 P2 q2

�

F12 = kq1q2r2

�

F12 = − dU12

dr

Force [N] = Newton

�

U12 = kq1q2r

�

U12 = − F12dr∫

Potential Energy [J]=Joule

Potential Voltage [J/C]=[V] =Volt

�

V12 = kq2r

�

V12 = − E12dr∫

�

! E 12 = kq2

r2Electric Field [N/C]=[V/m]

�

E12 = − dV12dr

Potential At Center of Ring of Charge •  Divide the charge distribution into

differential elements

•  Write down an expression for potential voltalge 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∫ = kdqr∫

= kr

dq∫ = krQ

�

Units : Nm2

C2Cm

⎡

⎣ ⎢

⎤

⎦ ⎥ =

NmC

⎡ ⎣ ⎢

⎤ ⎦ ⎥ =

JC⎡ ⎣ ⎢

⎤ ⎦ ⎥ ≡ V[ ]

Same result holds along z-axis!

Potential Along Axis of Ring of Charge •  Divide the charge distribution into

differential elements

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

•  Integrate! dq

R’

V = dV∫ = kdqr∫

= kr

dq∫ = krQ

= k Q′R 2 + z2

dq

Potential Voltage of Continuous Charge Distribution: Return of the Rod!

/Q Lλ = dxdq λ=

VP =kdqrrod!

∫ = kλdxxa

L+a

∫

= kλ ln(x)[ ]aL+a

VP = kλ ln L + a( )− ln a( )⎡⎣ ⎤⎦

• Uniformly charged rod • Total charge +Q •  Length L • What is V at position P?

P r = x

L a dq=λdx

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

+ + + + + + + + + + + + + + + + +

Potential Voltage of Continuous Charge Distribution: Return of the Rod!

VP = kλ ln L + a( )− ln a( )⎡⎣ ⎤⎦

= kλ ln L + aa

⎛⎝⎜

⎞⎠⎟ = kλ ln 1+ L / a( )

≅a>>L

kλ La= k λL

a= kQa

• Uniformly charged rod • Total charge +Q •  Length L • What is V at position P? • What about a>>L?

P r = x

L a dq=λdx

In this limit we recover V=kq/r for point charge!

+ + + + + + + + + + + + + + + + +

Potential Voltage of Continuous Charge

• Uniformly charged rod • Total charge +Q •  Length L • What is V at position P? r = x2 + d 2

λ =Q / Ldq = λdx

VP = dV = kdqr∫∫ =

k λdx( )r0

L

∫

= kλ dxx2 + d 20

L

∫ = kλ ln x + x2 + d 2( )⎡⎣⎢

⎤⎦⎥0

L

= kλ ln L + L2 + d 2( )− ln d( )⎡⎣⎢

⎤⎦⎥ = kλ ln L + L2 + d 2

d⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Potential Voltage of Continuous Charge

• What about d>>L?

r = x2 + d 2

λ =Q / Ldq = λdx

VP = kλ ln L + L2 + d 2

d⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= kλ ln Ld+ 1+ L

2

d 2⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

≅ kλ ln 1+ Ld

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥≅ kλL

d= kQd

We recover formulaV=kq/r for a point chargeAgain!

Electric Field & Potential: A 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!

dxdV

Ex −=

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)

f

i

V E dsΔ = − •∫! !

E from V: Example /Q Lλ =

�

V = kλ ln L + aa

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= kλ ln L + a[ ] − ln a[ ]{ }

• Uniformly charged rod

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

P x

L a dq

E = − dVda

= −kλ dda

ln L + a[ ]− ln a[ ]{ }

= kλ 1a− 1L + a

⎧⎨⎩

⎫⎬⎭

Units:

�

Nm2

C2Cmmm2

⎡

⎣ ⎢

⎤

⎦ ⎥ =

NC⎡ ⎣ ⎢

⎤ ⎦ ⎥ =

Vm⎡ ⎣ ⎢

⎤ ⎦ ⎥ √ Electric Field!

Electric Field & Potential: ICPP •  Hollow metal spherical shell of

radius R has a charge on the shell +q

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

V r1≈

r r=R

(a) V r1≈

r r=R

(b)

+

+

+

+

R+ +

+ +

Hint: Inside sphere there are no charges so E=0.But E=dV/dr=0. What can V be?

Δx

Ex = dVdx

≅ ΔVΔx

(exact for constant field)

(a) Since Δx is the same, only |ΔV| matters! |ΔV1| =200, |ΔV2| =220, |ΔV3| =200

|E2| > |E3| = |E1|

The bigger the voltage drop the stronger the field.

++++

++++

++++

−−−−

−−−−

−−−−

E!"

E!"

E!"

(b) = 3

−(c) F = qE = ma

accelerate leftward

ΔV ≡Vf −Vi

Equipotentials and Conductors •  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.

E V

Conductors Change the Field Around Them!

An Uncharged Conductor:

A Uniform Electric Field:

An Uncharged Conductor in the Initially Uniform Electric Field:

Sharp Conductors •  Charge density is higher at

conductor surfaces that have small radius of curvature

•  E =σ/ε0for 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)

Sharp Conductors: Lightning Rods

Ben Franklin Invents the Lightning Rod!

LIGHTNING SAFE CROUCH If caught out of doors during an approaching storm and your skin tingles or hair tries to stand on end, immediately do the "LIGHTNING SAFE CROUCH ”. Squat low to the ground on the balls of your feet, with your feet close together. Place your hands on your knees, with your head between them. Be the smallest target possible, and minimize your contact with the ground.

WE’LLBOTHBEDEADINSECONDS!

QUICK—TAKEASELFIE!

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 : dV = 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.