Electromagnetism – part one: electrostatics. Physics 1220/1320. Lecture Electricity, chapter 21-26. Electricity. Consider a force like gravity but a billion-billion-billion-billion times stronger And with two kinds of active matter: electrons and protons
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billion-billion-billion-billion times stronger
And with two kinds of active matter:
electrons and protons
And one kind of neutral matter: neutrons
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?
Conservation & quantization of charge
Electrons and protons carry charge.
Positive (proton), negative (electron)
Why does the electron not fall into
Why does the nucleus not fly apart?
Why does the electron not fly apart?
Consequence of QM uncertainty
More forces, total of four
Short ranged – limit for nucleus size
Uranium almost ready to fly apart
b) Why are there not
more ‘-’ charges?
c) Why do like charges
collect at opposite
d) Why does the metal
sphere not stay charged forever?
Force on a charge by other charges ~ ___________
e = 1.602176462(63) 10-19C i.e. even nC extremely good statistics
Modern Physics: why value?
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 q1, q2, and Q:
F1 on Q acc. to Coulomb’s law
Component F1x of F1 in x:
What changes when F2(Q) is
What changes when q1
How does the force ‘migrate’ to be felt by the other charge?
: Concept of fields
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
of distances and
Find fields for
of distances and
at all charge positions.
Direction E at black point equidistant from chargesis 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.
Find flux through each surface for q = 30°
and total flux through cube
What changes for case b?
2q on inner
4q on outer
1 = +q, 2 = -q, 3= +2q
Find the flux through
the surfaces S1-S5
Dust particle m= 5 10-9 [kg], charge q0 = 2nC
Energy conservation: Ka+Ua = Kb+Ub
Surface sphere: V = k q/R
Random walk does not mean ‘no progression’
Random motion fast: 106m/s
Drift speed slow: 10-4m/s
e- typically moves only few cm
Positive current direction:= direction flow + charge
charge through A per unit time
Work done by E on moving charges heat
(average vibrational energy increased i.e. temperature)
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 vd, in dt moves vddt,
volume Avddt, number of particles in volume n Avddt
What is charge that flows out of volume?
Properties of material matter too:
For metals, at T = const. J= nqvd ~ E
Proportionality constant r is resistivityr = E/JOhm’s law
Reciprocal of r is conductivity
Unit r: [Wm] ‘Ohm’ = [(V/m) / (A/m2)] = [Vm/A]
Ask for total current in and
potential at ends of conductor:
Relate value of current to
Potential difference between ends.
Const. r I ~ V
‘resistance’ R = V/I [W]
R = V/I
V = R I
I = V/R
E, V, R of a wire
Typical wire: copper, rCu = 1.72 x 10-8 Wm
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 = rJ = rI/A = V/m (a) V/m (b)
V = EL = mV (a), mV (b), V (c)
R = V/I = V/ A = W
outer radii a and b
Current flow radially! Not lengthwise!
Cross section is not constant:
2paL to 2pbL find resistance of
thin shell, then integrate
Area shell: 2prL with current path dr
and resistance dR between surfaces
dV = I dR to overcome
And Vtot= SdV
R = int(dR) = r/(2pL) intab(dr/r)
= r/(2pL) ln(b/a)
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
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
First, we consider ideal sources (emf) : Vab = E = IR
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,
Vab = E – Ir and I = E/(R+r)
(Ideal wires and am-meters have
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: EI, rate of dissipation I2r –
difference = power output source
Careful: opposite to capacitor series/parallel rules!
Find Req and I and V across
/through each resistor!
Find I25 and I20
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 Req
[Similarly point b:
Find E from loop 2:
Lets do the loop counterclockwise: E = V
Use junction rule
at a, b and c
3 unknown currents
Need 3 eqn
Loop rule to 3 loops:
Let’s set R1=R3=R4=1W
I1 = A, I2 = A, I3 =
E ~ /Q/ Vab ~ /Q/
But ratio Q/Vab is constant
Capacitance is measure of ability of a capacitor to store energy!
(because higher C = higher Q per Vab = higher energy
Value of C depends on geometry (distance plates, size plates, and material
property of plate material)
E = s/e0
For uniform field E and given plate distance d
Vab = E d
= 1/e0 (Qd)/A
Units: [F] = [C2/Nm] = [C2/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
In a parallel connection, the potential difference for all individual capacitors is the same. The charges are not the same.
Or Ceq =
Group task: What is the equivalent capacitance
of the circuit below?
Find equivalent for C series at right
Find equivalent for C parallel
left to right.
Solution: parallel branch …
series branch …
Find equivalent for series.
C1 = 6.9 mF, C2= 4.6mF
Reducing the furthest right leg
Combines parallel with nearest C2:
Leaving a situation identical to what we have just worked out:
So the overall CEQ = mF
Charge on C1 and C2:
For Vab = 420[V], Vcd=?
Vac= Vbd = Vcd=
V = Q/C
W = 1/C int0Q [q dq] = Q2/2C
Need Vab for C, need E for Vab:
Take Gaussian surface as sphere
and find enclosed charge
Dielectric constant K
For constant charge:
And V = V0/K
Dielectrics are never
perfect insulators: material
If V changes with K, E must decrease too: E = E0/K
This can be visually understood considering that materials
are made up of atoms and molecules:
E0 = s/e0
E = s/e
Empty space: K=1, e=e0
As q increases
i decreases 0
Charging a capacitor
Take exponential of both sides:
Discharging a capacitor
charged fully after 1[hr]? Y/N