FI 2201 Electromagnetism Electrostatics - Multisite ITB

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FI 2201 Electromagnetism

Alexander A. Iskandar, Ph.D.

Physics of Magnetism and Photonics Research Group

Electrostatics

BOUNDARY CONDITIONS AND WORK-ENERGY IN ELECTROSTATICS

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Summary of Electrostatics

• The relations between charge density ρ, electric field and electric potential V in electrostatics can be

summarized in the following diagram :

Er

ρ

∫

^′ ^′

=

V ρ r τ

πε ^d

) r ) (

r ( V

r r 4 0

1

0 2

ε

−ρ

=

∇V 0

0

=

×

∇

=

⋅

∇ Er Er ε

ρ

∫

=

V

rρ r τ πε ^ˆ^d

E ₂

4 0

r 1

Electromagnetism

Alexander A. Iskandar 3

Er V

ε0

V Er=−∇

∫

^⋅

−

= ^rE d )

r ( V

O

rl r r

0

Boundary Condition

• Consider a uniformly charged sphere, determine the electric field inside and outside of the sphere.

( )

^Q ^Q

∫

^r

( )

^r ^rˆ

r E Q

r Q E

a d

E_out _out _out ₂

0 0

2

4 4

1 ⋅ = π =ε → = πε

∫

^r ^r ^r

S

( )

^rˆ

R E Qr

R r Qr E

a d

E_in _in _in

3 0 3

0 3 2

4 4

2 ⋅ = π =ε → = πε

∫

^r ^r ^r

S

E S2

S1

R

• The electric field is continuous across the boundary.

4

R r

E

~r^{– 2}

~r Spherical concentric

Gaussian surface

Alexander A. Iskandar

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Boundary Condition

• Consider a uniformly charged spherical shell, determine the electric field inside and outside of the shell.

( )

^Q ^Q

∫

^r

( )

^r ^rˆ

r E Q

r Q E

a d

E_out _out _out ₂

0 0

2

4 4

1 ⋅ = π =ε → = πε

∫

^r ^r ^r

S

( )

⁴ ² ⁰ ⁰

2

=

→

=

∫

Ê^rⁱⁿ⋅^dâ^r Êⁱⁿ ^π^r Ê^rⁱⁿ

S

S2

S1

R

E

• The electric field is discontinuousacross the boundary.

5

Spherical concentric Gaussian surface

R r

E

~r^{– 2}

• Suppose we have a charged surface with charge density σ, and we wanted to know how the electric field

immediately above and belowthis charge density.

Boundary Condition

• Use Gauss theorem with a pill-boxGaussian surface :

σ

ε ^A ^above

Er

below

Er

Very thin, ε →0; so that the upper and lower surface will be immediately above and below the surface charge density

Very small area; so that even if the charge density is not constant, within this small area we can assume that it isconstant hence the resulting electric field will also be a constant

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• Thus the surface integral only yields

Boundary Condition

0 0

0 ε

σ ε

σ

ε ⁼ ^→ ⁻ ⁼

=

⋅ ^⊥ ^⊥

∫

Ê^r ^dâ^r ^Qênc Â Êâbove Ê^below

σ

ε ^A

⊥ above

E

⊥

Eb l

||

above

E

||

below

E

0 0

0 ε ε

S ε

• The normal componentsof the electric field is

discontinuous by the amount σ/ε₀at the boundary where the surface charge density is not zero.

7 below

E

• For the tangential components, use the Stokes’ theorem with a very thin, ε →0, closed loop integration, so that the upper and lower line segments will be immediately above

Boundary Condition

and below the surface charge density

σ

ε l

⊥ above

E

E⊥

||

above

E E||

• We find that the tangential components(parallel components) are always continuous.

8

0 0 → ^|| − ^|| =

=

⋅

×

∇

=

⋅

∫

Ê ^d Ê ^dâ Êâbove Ê^below

S C

r r rl

r

⊥ below

E

below

E

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• The boundary condition of the electric field components can be combined into

Boundary Condition

E ˆ

Er r σ

where is the unit vector perpendicular to the surface, pointing from “below” to “above”.

n E

E_above _below ε0

=

− nˆ

Electromagnetism 9

• Meanwhile the potential is continuousacross the

boundary, as the path length shrink to zero, the integral goes to zero

Boundary Condition

below above

b

, a below

above V E d V V

V − =−

∫

⋅ =⁰ → =

C

rl r

σ

ε a b

σ

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• However, the gradient of the potential inherits the discontinuity of the electric field

Boundary Condition

σ σ

r r

σ

ε a b

n V

V n

E

E_above _below ˆ _above _below ˆ

0

0 ε

σ ε

σ → ∇ −∇ =−

=

− r r

• Or,

11

σ

ε0

= σ

∂

−∂

∂

n V n

V_below _above

Work and Energy in Electrostatics

• Consider a collection of fixed source charges, imagine you are bringing a charge Qfrom point ato b.

• How muchHow much minimumminimumwork will you have to do ?work will you have to do ?

a

b Q q₁

q₂ q_i

[

V(b) V(a)

]

Q d E Q d F W

b

, a b

, a

−

=

⋅

−

=

⋅

=

∫ ∫

C C

rl l r

r r

• If we want to take a charge from the potential reference point, then the work we need to do is

[

^V⁽^r⁾ ^V⁽^r ⁾

]

^QV⁽^r⁾

Q

W = − _ref =

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Work and Energy in Electrostatics

• Alternatively, we can say that the potential difference between two points is equal to the work per unit charge required to carry a particle between those two points.

• Note the difference :

Electrical Potential is a property of the field (depends on location).

Electrical Potential Energyis a property of a charged object in an electric field

Q ) W a ( V ) b ( V

V = − =

∆

electric field.

Work and Energy in Electrostatics

• To calculate the energy to assemble point charges

distribution, we imagine taking these charges one-by-one from infinity and placed them at their position. _q

3

⎟

⎠

⎞

⎜⎜

⎝

⎛ +

=

⎟⎟

⎠

⎞

⎜⎜

⎝

= ⎛

=

2 1 3 3

12 1 2 0 2

1

4 1 4

1 0

r r

r

q q q

W

q q W

W

πε

q₁ q₂

q₃

r13

r23

r12

• Thus, the total work that has to be done is

14

⎜ ⎠

⎝ ¹³ ²³ 4^πε0

r r

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ + + +

= + + +

= K K

23 3 2 13

3 1 12

2 1 0 3

2

1 4

1 ^q

r

^q ^q

r

^q ^q

r

^q

W W W

W πε

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Work and Energy in Electrostatics

• Or, the total work that has to be done is

q₃

∑∑

= ⁿ ⁿ qⁱq^j

W 1

q₁ q₂

q₃

r13

r23

r12

∑∑

= >

=

i j i ij

W

0 1

4^πε

r

Work and Energy in Electrostatics

• Alternatively, we can just count all pairs of charges twice and then divide by 2

⎜ ⎞

n ⎛ n

n n qq 1 1 q

1

• The term inside the bracket is the electric potential of charge q_jat position , hence

∑ ∑

∑∑

=

≠=

= ≠= ⎟⎟

⎜ ⎠

⎜⎜

⎝

=

= ⁿ

i n

i j

j ij

j i

n

i n

i j

j ij

j

i q

q q W q

1 1 0

1 1

0 4

1 2

1 8

1

r

^πε

r

πε

rri

∑

= ⁿ qV(r )

W 1 r

• This is the amount of work that has to be done to assemble the charge configuration, this energy can be obtained back if we dismantle the system.

∑

i= i iV(r ) q W

2 1

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Work and Energy in Electrostatics

• For a continuous charge distribution, we generalize the previous result to get

∫

⁽ ⁾^V⁽ ⁾^d

W 1 r r

• We can rewrite this results in terms of electric field as follows

• Using integration by parts to transfer the differentiation we

∫

=

V

τ ρ(r)V(r)d W 2

∫ ( )

^∇^⋅

=

→

⋅

∇

=

V

ε τ ε

ρ Er W Er V d

2

0 0

g g y p

get

since

( )

_⎥

⎦

⎢ ⎤

⎣

⎡ + ⋅

⎥=

⎦

⎢ ⎤

⎣

⎡− ⋅ ∇ + ⋅

=

∫ ∫ ∫ ∫

S V

a d E V d E a

d E V d V E

W ε0 r τ r r ε0 2 τ r r

2 2

E

V r

−

=

∇

Work and Energy in Electrostatics

• Which volume that we need to integrate ? From the first integration that we started with

∫

= ρ(r)V(r)dτ

W 1 r r

it is clear that we can take any larger volume as long as it contains ρ, the result will be the same since ρis only defined on the given original volume.

• Thus, when we take a larger and larger volume, the first volume integral below will grow (E²is always positive) and

∫

=

V

τ ρ(r)V(r)d W 2

g g ( y p )

hence the second closed surface integral should decrease, in order that their sum stays constant

⎥⎦

⎢ ⎤

⎣

⎡ + ⋅

=

∫ ∫

S V

a d E V d E

W ε₀ ₂ τ r r 2

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Work and Energy in Electrostatics

• In fact, as the radius r increases,

the potential Vgoes like 1/r

while the electric field Egoes like 1/r²,

and the surface element goes like r².

• In total, the term in the surface integral decays like 1/r.

• Thus we can just as well integrate for all space, then the surface term will vanishes and we obtain

∫

=ε τ

d E

W ⁰ ²

• Example 2.8(worked out on the black board)

∫

space all

τ d E W 2

Work and Energy in Electrostatics

• For the case of a point charge, there seems to be an inconsistency in the last results ?

• Consider a point charge, for which the charge density isConsider a point charge, for which the charge density is

• How much work is involved in assembly of this charge ?

• On the one hand,

( )

r qδ r ρ=

sin )

8 ( )

( ) 2 (

1 ₂

0

=

∫

^r ^V ^r ^d ^q

∫

^r ^q_r ^r ^drd ^d

W δ θ θ φ

τ πε ρ

V V

r r

r

• But on the one hand,

0 )

2 ₀ ₀ (

2 =

= ^q_ε

∫

^∞^δ ^r ^r^dr

( )

⁼⁻ ^→ ^∞

=

∞ ∞

∫

0 0 2

0 2 4 2 0 2 2 0

0 1

8 1

4

2 2 r

dr q r r d q

E

W πε πε

τ ε ε

space all

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Work and Energy in Electrostatics

• The reason is related to the troublesome divergence of .

• Restore the surface integral to the expression of potential energy in terms of field:

ˆ r2

r energy in terms of field:

• With the given point charge

[ ( ) ]

∫

^⎥⁼ ⁺^∇^⋅

⎦

⎢ ⎤

⎣

⎡ + ⋅

=

V S

V

ε τ ε τ

d E V E

a d E V d E

W ₀ ₂ r r ₀ ₂ r

2 2

( ) ₍ ₎ ₍ ₎

4 1

4 ²

2 2 0 2 4

0 2

2 ⎟

⎠

⎜ ⎞

⎝

⎛

∂ + ∂

=

⋅

∇

+ r

q r r q r r

E q V

E πε πε

r

• No inconsistency. But use with care, and keep that surface integral in mind.

( ) ( )

(

4 ₀

)

² ¹⁴ ¹⁴ ⁰ ⁰

2

0 0

=

→

⎟=

⎠

⎜ ⎞

⎝

⎛ −

= W

r r q πε

∫

=ε allspaceE dτ

W ₀ 2 ²

Work and Energy in Electrostatics

• Let us pause for a while and reconsider what we have.

• The first relation to calculate work is given again below 1 n

1

∫

it seems that the resulting value could be positive as well as negative.

• While the second relation will always yields a positive value

∑

=

i

i iV(r ) q W

2 1

1 r ⁼

∫

V

τ ρ(r)V(r)d

W r r

2 1

• This discrepancy, just like the last example, resulted because in the first relation above, we do not take into account the energy to make those charges themselves.

∫

=

space all

ε τ

d E

W ⁰ ²

2

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Work and Energy in Electrostatics

• While the second relation is more complete since it is given from the electric fields.

• In fact, it is more appropriate, in radiation considerationIn fact, it is more appropriate, in radiation consideration that, to say that it is in the field that the electromagnetic energy is stored.

• However, when dealing with point charges, we prefer to use the discrete version of the work-energy relation, since we do not bother the energy attributed to making the point

we do not bother the energy attributed to making the point charges itself. The point charges are given.

Work and Energy in Electrostatics

• Furthermore, in the two expression of the first relation to calculate work, the potential are actually different.

1 n

1

∫

The potential represent the potential due to all charges, except q_i; while the potential represents the full potential.

• Use these equations with caution!!

∑

=

i

i iV(r ) q W

2 1

1 r ⁼

∫

V

τ ρ(r)V(r)d

W r r

2 1

) r ( V r_i

) r ( V r

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Work and Energy in Electrostatics

• Note also that in calculating this work, linear superposition principle can not be applied.

( ) _∫

∫

^ε

ε E d E E d W W E E d

W ⁼^ε⁰

_∫

^E² ^d^τ ⁼^ε⁰

_∫ (

^Er ⁺^Er

)

² ^d^τ ⁼^W ⁺^W ⁺^ε

_∫

^Er ^⋅^Er ^d^τ

W ⁰ ² ⁰ ₁ ₂ ₁ ₂ ₀ ₁ ₂

2 2