# Physics 1220/1320 Physics 1220/1320

Electromagnetism – part one: electrostatics

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

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

### Where do the charges come from?

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

### Electric Properties of Matter (I)

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?

### Coulomb’s Law

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?

### Electric Fields

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 -

### Electric field lines

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

H

2

O :

O 2- (ion)

H 1+ H 1+

### Gauss’s Law, Flux

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:

### Law

2q on inner

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

For charges

1 = +q, 2 = -q, 3= +2q

Find the flux through the surfaces S1-S5

### Electric Potential Energy

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

Potential difference:

[V/m]

Potential

Difference

### Calculating velocities from potential differences

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:

### Moving charges: Electric Current

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?

### Resistivity and Resistance

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

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

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

### Energy and Power in Circuits

Rate of conversion to electric energy:

E

I, rate of dissipation I 2 r – difference = power output source

### Resistor networks

Careful: opposite to capacitor series/parallel rules!

### Combining Measuring Instruments

Find R eq and I and V across

/through each resistor!

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

### Capacitor Networks

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

=

### Energy Storage in Capacitors

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

### Other common geometries

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 =

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