Electromagnetic Fields - WikiDydwikidyd.iem.pw.edu.pl/.../EF(2f)electrostatics/lecture4.pdf ·...

Electromagnetic Fields

Lecture 4

Electrostatics 2Energy, capacitance and capacitors , screens

Electromagnetic Fields, Lecture 4, slide 2

Energy

Work in electric field

W=∫LF⋅d L=q∫L

E⋅d L=qU AB

Assume electrostatic field described with a scalar electric potential φ such that φ(∞)=0

Work needed to move a single charge q from infinity to some point at which φ=V equals

W=qV

dW=F⋅dl=qE⋅dl

W is a potential energy because the work applied to movethe charge can be taken back from the system: the chargeHas some potential to do a work. We can define potential as

=Wq


Set of three charges

Consider two charges q1 and q

2

Moving q1 from infinity to destination requires no work,

because no field exists yet. Moving of q

1 requires work

W = q1 = q1⋅q2

4 r12

q1

q2


Set of n charges

We have used a simple trick: each term of sum is taken twice. That's why wehave multiplied the whole sum by 0.5.

q1 q

2

q3

The total energy of q1-q

2-q

3 system can be rewritten as

W =12 q1q24 r12

q2q14 r12

q1q34 r13

q3q14 r13

q2 q34 r23

q3q24 r23

qn

...

The first term can be seen as the energy of q1 in field produced by

q2 and q

3 at point, where q

1 is, the second term – as the energy of q

2

in field produced by q1 and q

3 and so on:

W=12

q11q22q33 =12∑i=1

3qii

Rearranging the above sum:

W =12 [q1 q2

4 r12

q34 r13 q2

q14 r12

q3

4 r23 q3q1

4 r13

q24 r23 ]

...The same way we may obtain the general formula:

W=12∑i=1

nqii


Set of three charges

The potential at q3 caused by q

1 and q

2 may be calculated as

superposition of fields due q1 and due q

2:

W =q1q24 r12

q1

q2

q3

Consider three charges: q1 , q

2 and q

3

q1 -q

2 system has potential energy (calculated on the

previous slide)


2 and q

3

q1 -q


previous slide)


2 and q

3

q1 -q


previous slide)

3 =14 q1r13

q2r23

The total energy of q1-q

2-q

3:

W =14 q1q2r12

q1q3r13

q2q3r23


Energy of a “continuous” charge

dl

...

...

...W=

12∫L

dl

W=12∫ dq

W=12∫L

dl

Line

ds...

...

...

...ds...

...

...

...

W=12∬S

ds

dV

...

W=12∭V

dV

...

...

Volume

Surface


The surface integral is zero: if the size of V increases,

the potential is approaching zero as rapidly as 1/r, D is approaching

zero at least as rapidly as 1/r2, while the surface increases as r2.

Do we need charges at all?

dV

...

W=12∭V

dV

...

... ∇⋅D = ∇⋅D∇⋅DMath – divergence of a vector multiplied by a scalar:

=∇⋅D −∇⋅D

∇⋅D=

Gauss's law:

W=12∭V

∇⋅D dV−12∭V

∇D dVApplying Gauss's theorem for the first integral and using definition of electric potential in the second one:

W=12∯∂V

D dS12∭V

E⋅D dV

W=∭VwdV , w=

E⋅D2

Density of energy

V


Capacitance

+ ++ +

+

+ +

++

++

++

+

---

--

--

-

-

-

-

-

-

-

1=const

2=const

Consider two conductors charged embedded in a homogeneous dielectric and carrying charges ofsame size Q but opposite signs. The total charge is 0.The charge is carried on the surface as a surface charge densityThe electric field on the surface, outside of the conductors is perpendicular to the surfaceThe electric field points from the positively charged conductor to the negatively chargedSummarizing: work must be done to carry a positivecharge from A to B.In other words: the system is capable to store some amount of energy. We may define capacitance C:

C =QU,

B

E,D

E,D A

where Q is the charge and U – the voltage between the conductors.

U=2−1

Capacitanceis measured

in farads1 F = 1 C / 1 V


Parallel-plate capacitor

Consider two infinite parallel plates separated bydielectric layer of thickness d. Plates are chargedwith positive and negative charge of density σ.According to the interface conditions

D1n−D2n= , E1t−E2t=0,Thus the electric field in the dielectric layer must beperpendicular to plates and equal to:

U=d

, Q= S , C=

QU

= Sd.

E=

.

The voltage between plates is

U=∫left

right

dl=d

.

S

Consider now the capacitance of plates with finite area S

C= Sd


Spherical capacitor

Consider two concentric spheres separated bydielectric shell of thickness r

2-r

1. Spheres are carrying

opposite charges ±Q.From the Gauss's law the field inside the shell is

E=Q

4 r2.

C=4 1r1

−1r2

=4 r1r2r2−r1

.

The voltage between spheres is

U=Q4

∫r1

r2 1r2dr=

Q4 1r1−

1r2 .

The capacitance:

C=4 r1r2r2−r1

r2

r1+Q

–Q


Cylindrical capacitor

Consider two concentric cylinders of length l separated bydielectric shell of thickness r

2-r

1. Cylinders are carrying

opposite charges ±Q.Neglecting the ends of cylinders we can calculate the field inside the dielectric from the Gauss law:

E=Q

2 r l.

C=2 l

ln r2r1 .

The voltage between cylinders is

U=Q

2 l∫r1

r2 1rdr=

Q2 l

ln r2r1 .The capacitance:

r2r1

+Q

–Q

l

C=2 l

ln r2r1


Real life is not as simple!

d2

A'

A

The real capacitors are not infinite. The disturbance of fieldnear the ends of plates makes real-life calculations muchmore complicated, than simple cases shown on the previousslides.

Consider field of more practical capacitor model, shownon the right figure (only ¼ of the capacitor cross-section wassimulated). It is clearly visible, that electric field gets weakerin the end-zone. This is confirmed by the plot of D

x along the

axis of symmetry AA':


Real life is not as simple!

The real plates are not infinitely thin. Thickness of the plate plays important role when field at the ends of the plate is to be calculated. Compare two parallel-plate capacitor models :

A “very thin”plate

A realistic,thick plate, th=1mm

Note the differencein maximal value of |D|

(continued)

Theory: d=1 cm s= 9 cm2 C=796.8769 fF. Simulations: th=0 – C =805.8 fF, th=1 mm – C = 801.6 fF, th=0.2 mm – C = 802.86 fF


Energy stored in capacitorConsider parallel-plate capacitor (d,S,ε) carrying ±Q.The electric field in the dielectric layer is

W=∫Vw dV=w⋅S d=

U 2

2d2S d=

= Sd

⋅U2

2=CU 2

2.

E=

, =

QS

The volume density of the energy is thenS

The total energy

W=CU 2

2

w=E⋅D2

= E2

2=

U 2

2d2.

E=Ud.

or (knowing the voltage U)


Consider cylindrical capacitor (l , r2,r

1,ε). The electric field

in dielectric isE=

Q2 r l

.

W=C2U 2

82 l2

∫r1

r2 1

r2⋅2 r l dr=

=C2U 2

4 l∫r1

r2 1rdr=

C2U2

2⋅

ln r2r1 2 l

=CU 2

2.

Note that Q=CU.

The total energy:

r2r1

+Q

–Q

l

W=CU 2

2

(continued)Energy stored in capacitor

Thus the volume density of energy is

w= E2

2=

C2U 2

822 l2r2

.Volume element

1/C


Mutual capacitanceConsider 3-wire power cable shown below. Analyzing field in the cable cross-section we clearlysee the potential difference between any pair of wires. We can think of mutual capacitance C

ij of

each pair of cables and self-capacitance Ci of each cable with respect to the sheath.

3

1

2 C12

C13

C23

C1

C3

C2

Analysis of self and mutual capacitances is important if signals are transmitted. Capacitances are one of the causes of cross-talks between wires.


Cross-talkCompare Fig. A and Fig. B. Note the difference of D around the right wire of the 2-wires-over-plate model. Remember that D

n=σ on the conductor surface!

Changes of UAO

(the left wire) can be observed in the right wire!

C3

Considering the circuit model (which is much convenient for cross-talk analysis) we must know the mutual capacitance between wires. It is the most important factor influencing the cross-talk.

UAO

Fig. B: UA0

=2V, UB0

=0.8825 V

Fig. A: UA0

=1V, UB0

=0.4411 V

The pictures looksimilar, but....

check the values!


Cross-talk – can we do anything?We try to put a conducting plate between the wires...

C3

Fig. B: UA0

=2V, UB0

=0.8825 V

Fig. A: UA0

=1V, no plate U

B0=0.4411 V

Fig. B: UA0

=1V, with plate U

B0=0.20483 V

Plate is NOTconnected

to the groud

It seemsto be better...no cross-talk?


Cross-talk – can we do anything?Well... the problem still exists!

Fig. A: UA0

=1V, no plate U

B0=0.4411 V

Fig. A: UA0

=1V, with plate U

B0=0.20483 V

Plate is NOTconnected

to the groud

The wires still can “see” each other.Change of U

A0 can be observed

in the second wire as UB0

.

(continued)

Fig. A: UA0

=1V, no plate U

B0=0.4411 V

Fig. B: UA0

=2V, with plate U

B0=0.41167 V


Cross-talk – solutionWe connect the “beetwen-wires” plate to the ground.

Fig. A: UA0

=1V, no plate U

B0=0.4411 V

Fig. A: UA0

=1V, with grounded plate U

B0=1.36 μV

Plate is nowCONNECTED

to the groud

In the numerical model we stillobserve change of U

B0 when U

A0

changes, but the value of UB0

is nowbelow the numerical errors level.

Fig. A: UA0

=1V, no plate U

B0=0.4411 V

Fig. B: UA0

=2V, with plate U

B0=2.72 μV


Conducting screens

Compare the figures. Both show electric field inside a parallel-plate capacitor connected to ±1 V. The distance between plates is 4 cm, giving field strength of 500 V/m (left figure). Inserting a conductingpipe (the dotted line on the right figure) between the plates reduces field inside the fipe to zero (right figure) – there is no electric field inside a conductor.

A hollow conductor allows to screen an external electrostatic field completely!

zeros!


Conducting screens

We do not need to use closed surfaces as screens. Even big holes in conductor surface are allowedand field inside the conducting surface is still reduced (comparing to the not-screened domain)

A metallic mesh can effectively screen large objects.

E < 9 V/m E < 25 V/m

(continued)


Conducting screens

Application – probably the most common one.

(continued)


Dielectric screens

A layer of different dielectric can also screen electrostatic fieldA layer of different dielectric can also screen electrostatic field

0 10 0Value of E is more than 10 times smaller.


Dielectric screens

A dielectric screen may have greater or lower permeability.A dielectric screen may have greater or lower permeability.

0100Value of E is more than 10 times smaller.

Electromagnetic Fields - WikiDydwikidyd.iem.pw.edu.pl/.../EF(2f)electrostatics/lecture4.pdf ·...

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