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Electromagnetism Physics 15b

Lecture #3 Gauss’s Law

Electric Potential

Purcell 1.13–2.9

What We Did Last Time Defined electric field by F = qE   Can be expressed by field lines

Defined flux of electric field   Note the sign convention: positive if coming out

Gauss’s Law

  Useful for solving E fields with symmetries   Spherical charge distribution

Φ = E ⋅ da∫

E ⋅ da∫ = 4πq

E =

Qr2 r for r ≥ R

QrR3 r for r < R

⎧

⎨⎪⎪

⎩⎪⎪

E

r R

QR2

R

r

S E

2

Today’s Goals Continue with Gauss’s Law   Apply to infinite sheets of charge

Discuss energy in the electric field   Empty space with E has energy?

Define electric potential ϕ   by line-integrating electric field   Closely related to energy   Vector calculus connects electric potential

to electric field and vice versa

Derive differential form of Gauss’s Law   Connect electric field and charge density   More vector calculus

J.C.F. Gauss, 1777-1855

Infinite Sheet of Charge Problem: Calculate the electric field at a distance z from a positively charged infinite plane

  Surface charge density:

Use Gauss again   Which surface to use?

  What symmetry do we have?   Consider a cylinder

  Area A and height 2z

E field must be vertical   How do we know that?

z E

σ =

chargearea

3

Infinite Sheet of Charge Total flux Φtotal = Φtop + Φside + Φbottom

  Side is parallel to E No flux   Top and bottom are symmetric Same flux

Charge inside the cylinder is

Using Gauss   Don’t forget the direction!

z E

The result is worth remembering:

Infinite sheet of charge produces uniform E field of 2πσ above and below

E ⋅da = 0

Φtotal = 2Φtop = 2AE(z)

qinside = Aσ

E(z) = 2πσ

E = +2πσ z for z > 0−2πσ z for z < 0

⎧⎨⎪

⎩⎪

Place two oppositely-charged large sheets in parallel   Consider an area A of them

E fields from the two sheets overlap and add up   Between the sheets: E = 4πσ   Cancel each other outside

Two sheets also attract each other (obviously)   Top sheet feels

  The force on area A of the top sheet is

−σ

Pair of Charged Sheets

+σ

Etop Ebottom z

Ebottom = −2πσ z

F = σAEbottom = −2πσ 2Az = −E2

8πAz

4

Imagine we move the top sheet upward by a distance d

  We must do work

  The energy of the system increases by W

Q: Where exactly is this energy?   Note that the volume of the space

between the sheets increased by Ad   This is also where E field exists

Space with E holds energy with a volume density Total electrostatic energy of a system is

−σ

Pair of Charged Sheets

+σ

Etop Ebottom z W = Fd =E2

8πAd

u =E2

8π

U =E2

8πdV∫ Will come back to this…

Line Integral of Electric Field Electrostatic force is conservative   I said this in Lecture 1 without proof

Given F = qE, the above statement is equivalent to

  Thanks to the Superposition Principle, we have only to prove this for the electric field generated by a single point charge

Line integral E ⋅ds

P1

P2∫ is path independent

Line integral q

r 2 r ⋅dsr1

r2∫ is path independent

q

r1

r2 r

ds

E =

qr 2 r

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Line Integral of Electric Field The dot product is the radial component of the movement, i.e. dr

  The integral depends only on r1 and r2, i.e., is path-independent

Generalize using Superposition:

qr 2 r ⋅ds

r1

r2∫ =qr 2 dr

r1

r2∫ = q1r1−

1r2

⎛

⎝⎜⎞

⎠⎟

q

r1

r2 r

ds

E =

qr 2 r

r ⋅ds

dr

Line integral E ⋅dsP1

P2∫ for any electrostatic

field E has the same value for all pathsfrom P1 to P2

Corollary For the special case of P1 = P2, the path becomes a loop

This is equivalent to the path-independence   Consider two paths (A and B) from P1 to P2

  Path independence means

  Corollary above means

  Will use this later when we do the “curl”

Line integral E ⋅ds∫ of an electrostatic field around

any closed path is zero

dsB P1

P2

E ⋅dsAP1

P2∫ = E ⋅dsBP1

P2∫

E ⋅dsAP1

P2∫ + E ⋅ (−dsB )P2

P1∫ = 0

dsA

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Electric Potential The line integral is useful enough to have a name

  To move a charge q from P1 to P2, you must do work

  As a result, the energy of the system increases by qϕ21

We can fix P1 to a reference point and re-interpret this as a scalar function of the position r

E ⋅ds

P1

P2∫

φ21 ≡ − E ⋅ds

P1

P2∫

Electric potential at position r

Note the negative sign!

W = −qE ⋅ds

P1

P2∫ = qφ21

φ(r) ≡ − E ⋅ds

0

r

∫

Electric potential difference between P1 and P2

Reference Point Reference point for the electric potential is arbitrary   If you choose e.g., point B instead of point A as the reference

  Electric potential is defined up to an arbitrary constant   Just like an indefinite integral

The potential difference is physical (linked to work and energy) and must be free from arbitrary constant

  The constant cancels in the difference

φref=B(r) = − E ⋅ds

B

r

∫ = − E ⋅dsA

r

∫ − E ⋅dsB

A

∫ = φref=A(r) + const.

φ21 = − E ⋅ds

P1

P2∫ = − E ⋅ds0

P2∫ + E ⋅ds0

P1∫ = φ(P2) −φ(P1)

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Unit Dimension of electric potential is (electric field)×(length)   Since electric field is (force)/(charge), this equals to (force)×(length)/

(charge) = (energy)/(charge)

Unit of electric potential is erg/esu = statvolt

In SI, the unit of electric potential is volt = joule/coulomb 300 volt = 1 statvolt

1 coulomb = 3×109 esu   Since the rest of the world uses SI, we will convert to SI when we

deal with real-world (esp. EE) problems

Electric Field ↔ Potential Recall in calculus:

We can “reverse” the line integral as well

  Gradient is “how quickly the function ϕ varies in space”

  In x-y-z coordinate system:

NB: gradient is a vector field   It points the direction of maximum rate of increase (= uphill)

  Hence E points downhill of ϕ

F(x) = f (x)dx∫

f (x) = d

dxF(x)

φ = − E ⋅ds∫ E = −∇φ gradient

∇φ = x ∂φ

∂x+ y ∂φ

∂y+ z ∂φ

∂z

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Visualizing Gradient in 2-d For a potential

∇φ(x,y) = x cos(x)sin(y) + y sin x cos y φ(x,y) = sin x ⋅sin y

φ(x,y)

∇φ(x,y)

© Prof. G. Sciolla, MIT

Equipotential Surfaces The same potential can be expressed with lines of constant values of ϕ   In 3-d, we find surfaces of

constant potential = equipotential surfaces

  For a single point charge, equipotential surfaces are concentric spheres

Electric field is perpendicular to the equipotential surfaces   This is generally true for gradient

of any function

φ(x,y) = sin x ⋅sin y

∇φ(x,y)

© Prof. G. Sciolla, MIT

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Charge Distribution Electric potential due to a charge distribution is

  Continuous case maybe 1-d, 2-d, or 3-d

Since potential is a scalar, it is often easier to calculate than the electric field   Once you have ϕ, you can always get E from −∇ϕ

An example is in order

φ =

qj

rjj =1

N

∑ φ =

dqr∫discrete continuous

φ =

λr

d∫ φ =

σr

da∫ φ =

ρr

dv∫

Charged Disk Problem: Calculate the electric potential produced by a thin, uniformly charged disk on its axis   Disk radius = a, surface charge density σ

Slice the disk into rings, then into the red bits   Charge on the red bit is σsdsdθ   The potential due to the red bit is   Integrate:

z

a

s ds dθ

dφ =

σsdsdθ

z2 + s2

φ = σ dθ sds

z2 + s20

a

∫0

2π

∫

= 2πσ z2 + s2⎡⎣⎢

⎤⎦⎥s=0

s=a

= 2πσ z2 + a2 − z( ) NB: absolute value in case z < 0

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Electric Field Can we calculate E from ?   We need to take gradient, which means we need

to know the x-, y-, and z-dependence

  We calculated ϕ only on the z axis

We are in luck — we know E is parallel to the z axis by symmetry

z

a

s ds dθ

φ = 2πσ z2 + a2 − z( )

E = −∇φ = −z ∂φ∂z

= −2πσ ∂∂z

z2 + a2 − z( ) z= 2πσ −

z

z2 + a2±1

⎛

⎝⎜⎞

⎠⎟z for z > 0

z < 0

Check the Solution

Dimension: [ϕ] = (charge)/(length), [E] = (charge)/(length)2

  NB: [σ] = (charge)/(length)2

Far away from the disk (|z| >> a)

  Same as a point charge with Q = πa2σ

Close to the disk (|z| << a)

  Same as the infinite sheet of charge

φ = 2πσ z2 + a2 − z( )

E = 2πσ −

z

z2 + a2±1

⎛

⎝⎜⎞

⎠⎟z for z > 0

z < 0

φ = 2πσ z 1+ (a z)2 −1( ) z a⎯ →⎯⎯ 2πσ z 1+ 1

2 (a z)2 −1( ) = πa2σz

Ez a⎯ →⎯⎯ ±2πσ z

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Energy and Potential

Does the factor 1/2 make sense?   Imagine increasing ρ slowly everywhere as   The potential is proportional to ρ, i.e.,   Work to go from s to s + ds is

  Integrate

U =

12

qjqk

rjkk ≠ j∑

j =1

N

∑ =12

qj

qk

rjkk ≠ j∑

j =1

N

∑ =12

qj φ jkk ≠ j∑

j =1

N

∑ Potential at the j-th charge due to the

other charges

U =

12

ρφ dv∫

′ρ = sρ s = 0→1

′φ = sφ

dW = (d ′ρ ) ′φ dv∫ = (dsρ)(sφ)dv∫

W = s ds ρφ dv∫0

1

∫ =12

ρφ dv∫ Will come back to this…

We know a system of N charges has a total energy of

Generalizing to continuous distributions   Integrate entire space, or where ρ ≠ 0   No need to avoid j = k because “individual”

charge is infinitesimally small

Shrinking Gauss’s Law Charge is distributed with a volume density ρ(r) Draw a surface S enclosing a volume V

Guass’s Law:

Now, make V so small that ρ is constant inside V

  As we make V smaller, the total flux out of S scales with V

Therefore:

  LHS is “how much E is flowing out per unit volume”   Let’s call it the divergence of E

E ⋅da

S∫ = 4π ρdvV∫ Total charge in V

E ⋅da

S∫ = 4πρV for very small V

limV→0

E ⋅daS∫

V= 4πρ

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Divergence In the small-V limit, the integral depend on volume, but not on the shape We can use a rectangular box   Consider the left (S1) and right (S2) walls

  Add up all walls:

divE ≡ lim

V→0

E ⋅daS∫

V= 4πρ

dx

dy

dz

E(x + dx,y,z) E(x,y,z)S1 S2

E ⋅da

S1∫ = E(x,y,z) ⋅ (−x)dydz

E ⋅da

S2∫ = E(x + dx,y,z) ⋅ xdydz

Sum = Ex(x + dx) − Ex(x)( )dydz

=∂Ex

∂xdxdydz

E ⋅da

S∫ =∂Ex

∂x+∂Ey

∂y+∂Ez

∂z

⎛

⎝⎜

⎞

⎠⎟V = ∇ ⋅E( )V

div E

Gauss’s Law, Local Version We now have Gauss’s Law for a very small volume/surface

  Connects local properties of E with the local charge density   Integrate over a volume and you get Gauss’s Law back

Gauss’s Divergence Theorem (this is math!)

For any vector field F,

  Applying this theorem to Gauss’s Law (of electromagnetism) gives us the integral and differential versions we now know

divE = 4πρ where divE ≡ ∇ ⋅E =

∂Ex

∂x+∂Ey

∂y+∂Ez

∂z

⎛

⎝⎜

⎞

⎠⎟

divFdv

V∫ = F ⋅daA∫

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Coulomb Field Let’s calculate div E for

  We can do this by expressing E in x-y-z :

  Or we can express div in spherical coordinates

  Since only Er is non-zero, we get

This is correct — we have no charge except at r = 0   At r = 0, 1/r2 gives us an infinity   That’s OK because a “point” charge has an infinite density

E =

qr 2 r

E =

qx2 + y 2 + z2

xx + yy + zz

x2 + y 2 + z2

∇ ⋅F =

1r 2

∂(r 2Fr )∂r

+1

r sinθ∂(Fθ sinθ)

∂θ+

1r sinθ

∂Fφ

∂φ

∇ ⋅E =

1r 2

∂(r 2Er )∂r

=1r 2

∂q∂r

= 0

Spherical Charge Let’s give the “point” charge a small radius R   We did this in Lecture 2, and the solution was

  For r < R,

  The charge density of the sphere is

It works everywhere (as long as ρ is finite)

E =

Qr2 r for r ≥ R

QrR3 r for r < R

⎧

⎨⎪⎪

⎩⎪⎪

This part is same as a point charge. We know div E = 0.

Let’s work on this part

∇ ⋅E =

1r 2

∂(r 2Er )∂r

=1r 2

∂∂r

Qr 3

R3

⎛

⎝⎜⎞

⎠⎟=

3QR3

ρ =

Q4π3 R3 ∇ ⋅E = 4πρ

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Energy Again We found earlier and   Are they same?

Consider the divergence of the product Eϕ

  Integrate LHS over very large volume and use Divergence Theorem

  Integral of RHS must be 0, too

U =

12

ρφ dv∫ U =

E 2

8πdV∫

∇ ⋅ (Eφ) = ∂∂x

(Exφ) + ∂∂y

(Eyφ) + ∂∂z

(Ezφ)

=∂Ex

∂xφ + Ex

∂φ∂x

+∂Ey

∂yφ + Ey

∂φ∂y

+∂Ez

∂zφ + Ez

∂φ∂z

= (∇ ⋅E)φ +E ⋅ (∇φ) = 4πρφ − E 2

∇ ⋅ (Eφ)dv

V∫ = (Eφ) ⋅daS∫ = 0 assuming Eφ → 0 at far away

4π ρφ dv∫ − E 2 dv∫ = 0

12

ρφ dv∫ =E 2

8πdv∫

Summary Used Gauss’s Law on infinite sheet of charge   Uniform electric field E = 2πσ above and below the sheet   Electric field has energy with volume density given by

Defined electric potential by line integral

  Electric field is negative gradient of electric potential

  Potential due to charge distribution: or

Differential form of Guass’s Law:   Equivalent to the integral form via Divergence Theorem

u =E2

8π

φ21 = − E ⋅ds

P1

P2∫ = φ(P2) −φ(P1) unit: erg/esu = statvolt

E = −∇φ

φ =

qj

rjj =1

N

∑ φ =

dqr∫

∇ ⋅E = 4πρ

divFdv

V∫ = F ⋅daA∫