Physics 1220/1320
Electromagnetism – part one: electrostatics
Lecture Electricity, chapter 21-26


Consider a force like gravity but a billion-billion-billion-billion times stronger
And with two kinds of active matter: electrons and protons
And one kind of neutral matter: neutrons


The phenomenon of charge
Problem
Invisibility
Common problem in physics: have to believe in invisible stuff and find ways to demonstrate its existence.
Danger of sth invisible
If we rub electrons onto the plastic, is it feasible to say that we rub protons on it in the second experiment?
No! If we move protons, we move electrons with them. But what if in the second experiment we still moved electrons – in the other direction?

Why matter is usually electrically neutral:

Like charges repel, unlike charges attract

Mixed + and – are pulled together by enormous attraction

These huge forces balance each other almost out so that matter is neutral

Two important laws:
Conservation & quantization of charge

Electrons and protons carry charge.
Neutrons don’t.
Positive (proton), negative (electron)
Consider:
Why does the electron not fall into the nucleus?
Why does the nucleus not fly apart?
Why does the electron not fly apart?
Consequence of QM uncertainty relation
More forces, total of four
Short ranged – limit for nucleus size
Uranium almost ready to fly apart

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
Materials which conduct electricity well are called ______________
Materials which prohibited the flow of electricity are called ________________
‘_____’ or ‘______’ is a conductor with an infinite reservoir of charge
____________ are in between and can be conveniently ‘switched’
_____________are ideal conductors without losses


Induction : Conductors and Insulators
Induction -
Appears visibly in conductors a) Are charges present?
b) Why are there not more ‘-’ charges?
c) Why do like charges collect at opposite side?
d) Why does the metal sphere not stay charged forever?

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

Concept of point charges
Applies strictly in vacuum although in air deviations are small
Applies for charges at rest (electrostatics)
Force on a charge by other charges ~ ___________
~ ___________
~ ___________
Significant constants: e = 1.602176462(63) 10 -19 C i.e. even nC extremely good statistics
(SI) 1/4 pe
0
Modern Physics: why value?
how constant?

Principle of superposition of forces:
If more than one charge exerts a force on a target charge, how do the forces combine?
Luckily, they add as vector sums!
Consider charges q
1
, q
2
, and Q:
Find F
1
F
1 on Q acc. to Coulomb’s law
Component F
1x of F
1 in x:
What changes when F
2
(Q) is determined?
What changes when q1 is negative?

How does the force ‘migrate’ to be felt by the other charge?
: Concept of fields

Charges –q and 4q are placed as shown.
Of the five positions indicated at
1-far left, 2 – ¼ distance, 3 – middle, 4 – ¾ distance and 5 – same distance off to the right, the position at which E is zero is: 1, 2, 3, 4, 5

Find force of all combinations of distances and charge arrangements

Find fields for all combinations of distances and charge arrangements at all charge positions .

Direction E at black point equidistant from charges is indicated by a vector. It shows that: a) A and B are + b) A and B are c) A + B – d) A – B + e) A = 0 B -

For the visualization of electric fields, the concept of field lines is used.

H
2
O :
O 2- (ion)
H 1+ H 1+

Group Task:
Find flux through each surface for q
= 30 ° and total flux through cube
What changes for case b?
n
1
: n
2
: n
3
: n
4
: n
5
,n
6
:

Basic message:

2q on inner
4q on outer shell http://www.falstad.com/vector3de/

http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html
http://www.falstad.com/vector3de/

Group Task
For charges
1 = +q, 2 = -q, 3= +2q
Find the flux through the surfaces S1-S5

Electric Potential V units volt: [V] = [J/C]

Potential difference:
[V/m]
Potential
Difference

Dust particle m= 5 10 -9 [kg], charge q
0
= 2nC
Energy conservation: K a
+U a
= K b
+U b

Outside sphere: V = k q/r
Surface sphere: V = k q/R
Inside sphere:

Potential Gradient

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Path of moving charges: circuit
Transporting energy http://math.furman.edu/~dcs/java/rw.html
Random walk does not mean ‘no progression’
Current
Random motion fast: 10 6 m/s
Drift speed slow: 10 -4 m/s e typically moves only few cm
Positive current direction:= direction flow + charge

Work done by E on moving charges  heat
(average vibrational energy increased i.e. temperature)
Current through A:= dQ/dt charge through A per unit time
Unit [A] ‘Ampere’
[A] = [C/s]
Current and current density do not depend on sign of charge
 Replace q by /q/
Concentration of charges n [m -3 ] , all move with v volume Av d dt, number of particles in volume n Av d d
, in dt moves v dt d dt,
What is charge that flows out of volume?

Properties of material matter too:
For metals, at T = const. J= nqv d
~ E
Proportionality constant r is resistivity r = E/J Ohm’s law
Reciprocal of r is conductivity
Unit r : [ W m] ‘Ohm’ = [(V/m) / (A/m 2 )] = [Vm/A]

Types of resistivity

Resistance
Ask for total current in and potential at ends of conductor:
Relate value of current to
Potential difference between ends.
For uniform J,E
I = JA and V =EL with Ohm’s law E= r J
 V/L = r I/A
Const. r  I ~ V
‘resistance’ R = V/I [ W ] r vs. R
R = r L/A
R = V/I
V = R I
I = V/R

Group Task
E, V, R of a wire
Typical wire: copper, r
Cu
= 1.72 x 10 -8
W m cross sectional area of 1mm diameter wire is 8.2x10
-7 m -2 current a) 1A b) 1kA for points a) 1mm b) 1m c) 100m apart
E = r
J = r
I/A = V/m (a) V/m (b)
V = EL = m
V (a), mV (b), V (c)
R = V/I = V/ A =
W

Resistance of hollow cylinder length L, inner and outer radii a and b
Current flow radially! Not lengthwise!
Cross section is not constant:
2 p aL to 2 p bL  find resistance of thin shell, then integrate
Area shell: 2 p rL with current path dr and resistance dR between surfaces dR= r dr/(2 p rL)
And V tot
 dV = I dR to overcome
=
S dV
R = int(dR) = r/(2p
L) int ab
(dr/r)
= r/(2p
L) ln(b/a )

Electromotive Force
Steady currents require circuits: closed loops of conducting material otherwise current dies down after short time
Charges which do a full loop must have unchanged potential energy
Resistance always reduces U
A part of circuit is needed which increases U again
This is done by the emf.
Note that it is NOT a force but a potential!
First, we consider ideal sources (emf) : V ab
= E = IR

I is not used up while flowing from + to –
I is the same everywhere in the circuit
Emf can be battery (chemical), photovoltaic (sun energy/chemical), from other circuit (electrical), every unit which can create em energy
EMF sources usually possess Internal Resistance. Then,
V ab
=
E – Ir and I = E
/(R+r)

Circuit Diagrams
(Ideal wires and am-meters have
Zero resistance)
No I through voltmeter
(infinite R) … i.e. no current at all
Voltage is always measured in parallel, amps in series

Rate of conversion to electric energy:
E
I, rate of dissipation I 2 r – difference = power output source

Careful: opposite to capacitor series/parallel rules!

Group Task:
Find R eq and I and V across
/through each resistor!

Group task:
Find I
25 and I
20

Kirchhoff’s Rules
A more general approach to analyze resistor networks
Is to look at them as loops and junctions:
This method works even when our rules to reduce a circuit to its R eq fails.

‘Charging a battery’
Loop rule: 3 loops to choose from
‘1’: decide direction in loop – clockwise
(sets signs!)
 r =
W
Find
E from loop 2:
Lets do the loop counterclockwise:
Circuit with more than one loop!
Apply both rules.
Junction rule, point a:
 I = A
[Similarly point b:
]
 E
= V

Group task: Find values of I1,I2,and I3!
5 currents
Use junction rule at a, b and c
3 unknown currents
Need 3 eqn
Loop rule to 3 loops: cad-source cbd-source cab
(3 bec.of no.unknowns)
Let’s set R
1
=R
3
And R
2
=R
5
=2
W
=R
4
=1
W
I
1
= A, I
2
= A, I
3
=

E ~ /Q/  V ab
Double Q:
~ /Q/
But ratio Q/V ab is constant
Capacitance
Capacitance is measure of ability of a capacitor to store energy!
(because higher C = higher Q per V ab
= higher energy
Value of C depends on geometry (distance plates, size plates, and material property of plate material)

Plate Capacitors
E = s/e
0
= Q/A e
0
For uniform field E and given plate distance d
V ab
= E d
= 1/ e
0
(Qd)/A
Units: [F] = [C 2 /Nm] = [C 2 /J] … typically micro or nano Farad

In a series connection, the magnitude of charge on all plates is the same. The potential differences are not the same.
Or using concept of equivalent capacitance
1/C eq
=
In a parallel connection, the potential difference for all individual capacitors is the same. The charges are not the same.
Or C eq
=

Equivalent capacitance is used to simplify networks.
Group task: What is the equivalent capacitance of the circuit below?
Step 1:
Find equivalent for C series at right
Hand.
Step 2:
Find equivalent for C parallel left to right.
Step 3:
Find equivalent for series.
Solution: parallel branch … series branch …
 total:

Capacitor Networks
24.63
C1 = 6.9 m
F, C2= 4.6
m
F
Reducing the furthest right leg
(branch):
C=(
Combines parallel with nearest C2:
C =
Leaving a situation identical to what we have just worked out:
So the overall C
EQ
= m
F
Charge on C1 and C2:
Q
C1
=
Q
C2
=
For V ab
V ac
= 420[V], V
= V bd
= cd
=?
V cd
=

V = Q/C
W = 1/C int
0Q
[q dq] = Q 2 /2C

Energy Storage in Capacitors
24.24: plate C 920 pF, charge on each plate 2.55 m
C a) V between plates:
V = b)For constant charge, if d is doubled, what will V be?
c) How much work to double d?
If d is doubled, …
Work equals amount of extra energy needed which is mJ


Spherical capacitor
Need V ab for C, need E for V ab
:
Take Gaussian surface as sphere and find enclosed charge

Cylindrical capacitor

Dielectrics
Dielectric constant K
K= C/C
0
For constant charge:
Q=C
0
V
0
=CV
And V = V
0
/K
Dielectrics are never perfect insulators: material leaks

Induced charge: Polarization
If V changes with K, E must decrease too: E = E
0
/K
This can be visually understood considering that materials are made up of atoms and molecules:

Induced charge: Polarization – Molecular View
Dielectric breakdown

Change with dielectric:
E
0
= s/e
0

E = s/e
Empty space: K=1, e=e
0

RC Circuits
Charging a capacitor
From now on instantaneous
Quantities I and V in small fonts v ab v bc
= i R
= q/C
Kirchhoff:

As q increases towards Q f
, i decreases  0
 integral:
Take exponential of both sides :

Characteristic time constant t
= RC !
Discharging a capacitor

Ex 26.37 - C= 455 [pF] charged with 65.5 [nC] per plate
C then connected to ‘V’ with R= 1.28 [M W
] a) i initial
?
b) What is the time constant of RC?
a) i = q/(RC)
[C/(
W
F] =! [A] b) t
= RC =
Group Task:
Find t charged fully after 1[hr]? Y/N

Ex 26.80/82 C= 2.36 [ m
F] uncharged, then connected in series to
R= 4.26 [
W
] and
E
=120 [V], r=0 a) Rate at which energy is dissipated at R b) Rate at which energy is stored in C Group Task c) Power output source a) P
R
= b) P
C
= c) P t
= d) What is the answer to the questions after ‘a long time’?
 all zero