LECTURE Topic 4 POTENTIAL September 19, 2005

Alternate Lecture Titles  Back to Physics 2048  You can run but you can’t hide!

The PHY 2048 Brain Partition h m A B To move the mass m from the ground to a point a distance h above the ground requires that work be done on the particle. W is the work done by an external force. mgh represents this amount of work and is the POTENTIAL ENERGY of the mass at position h above the ground. The reference level, in this case, was chosen as the ground but since we only deal with differences between Potential Energy Values, we could have chosen another reference. Reference “0”

Let’s Recall Some more PHY2048 h m A B A mass is dropped from a height h above the ground. What is it’s velocity when it strikes the ground? We use conservation of energy to compute the answer. Result is independent of the mass m.

Using a different reference. y=h m A B y y=b ( reference level) y=0 Still falls to here.

Energy Methods  Often easier to apply than to solve directly Newton’s law equations.  Only works for conservative forces.  One has to be careful with SIGNS. VERY CAREFUL !

I need some help.

THINK ABOUT THIS!!!  When an object is moved from one point to another in an Electric Field, It takes energy (work) to move it. This work can be done by an external force (you). FIELD You can also think of this as the FIELD doing the negative of this amount of work on the particle.

Let’s look at it: move a mass from y i to y f yfyiyfyi External Field Change in potential energy due to external force: Negative of the work done BY THE FIELD. Keep it!

Move It!  Move the charge at constant velocity so it is in mechanical equilibrium all the time.  Ignore the acceleration at the beginning because you have to do the same amount of negative work to stop it when you get there.

And also remember: The net work done by a conservative (field) force on a particle moving around a closed path is ZERO!

A nice landscape  mg h Work done by external force = mgh How much work here by gravitational field?

The gravitational case: 

Someone else’s path 

IMPORTANT  The work necessary for an external agent to move a charge from an initial point to a final point is INDEPENDENT OF THE PATH CHOSEN!

The Electric Field  Is a conservative field. No frictional losses, etc.  Is created by charges.  When one (external agent) moves a test charge from one point in a field to another, the external agent must do work.  This work is equal to the increase in potential energy of the charge.  It is also the NEGATIVE of the work done BY THE FIELD in moving the charge from the same points.

A few things to remember…  A conservative force is NOT a Republican.  An External Agent is NOT 007.

Electric Potential Energy  When an electrostatic force acts between two or more charged particles, we can assign an ELECTRIC POTENTIAL ENERGY U to the system.

U=PE Example: NOTATION U=PE A B dd E q F Work done by FIELD is Fd Negative of the work done by the FIELD is -Fd Change in Potential Energy is also –Fd. The charge sort-of “fell” to lower potential energy. HIGH U LOWER U

Gravity  mg Negative of the work done by the FIELD is –mg  h =  U Bottom Line: Things tend to fall down and lower their potential energy. The change, U f – U i is NEGATIVE!

Electrons have those *&#^ negative signs.  Electrons sometimes seem to be more difficult to deal with because of their negative charge.  They “seem” to go from low potential energy to high.  They DO!  They always fall AGAINST the field!  Strange little things. But if YOU were negative, you would be a little strange too!

An Important Example Designed to Create Confusion or Understanding … Your Choice! E e A sad and confused Electron. Initial position Final position d The change in potential energy of the electron is the negative of the work done by the field in moving the electron from the initial position to the final position. FORCE negative charge Force against The direction of E

An important point  In calculating the change in potential energy, we do not allow the charge to gain any kinetic energy.  We do this by holding it back.  That is why we do EXTERNAL work.  When we just release a charge in an electric field, it WILL gain kinetic energy … as you will find out in the problems!  Remember the demo!

AN IMPORTANT DEFINITION  Just as the ELECTRIC FIELD was defined as the FORCE per UNIT CHARGE: We define ELECTRICAL POTENTIAL as the POTENTIAL ENERGY PER UNIT CHARGE: VECTOR SCALAR

UNITS OF POTENTIAL

Watch those (-) signs!! The electric potential difference  V between two points I and f in the electric field is equal to the energy PER UNIT CHARGE between the points: Where W is the work done BY THE FIELD in moving the charge from One point to the other.

Let’s move a charge from one point to another via an external force.  The external force does work on the particle.  The ELECTRIC FIELD also does work on the particle.  We move the particle from point i to point f.  The change in kinetic energy is equal to the work done by the applied forces.

Furthermore… If we move a particle through a potential difference of  V, the work from an external “person” necessary to do this is q  V

Example Electric Field = 2 N/C 1  C d= 100 meters

One Step More

The Equipotential Surface DEFINED BY It takes NO work to move a charged particle between two points at the same potential. The locus of all possible points that require NO WORK to move the charge to is actually a surface.

Example: A Set of Equipotenital Surfaces

Back To Yesteryear

Field Lines and Equipotentials Equipotential Surface Electric Field

Components Equipotential Surface Electric Field E normal E parallel xx Work to move a charge a distance  x along the equipotential surface Is Q x E parallel X  x

BUT  This an EQUIPOTENTIAL Surface  No work is needed since V=0 for such a surface.  Consequently E parallel =0  E must be perpendicular to the equipotential surface

Therefore E E E V=constant

Field Lines are Perpendicular to the Equipotential Lines

Equipotential

Consider Two Equipotential Surfaces – Close together V V+dV ds a b Work to move a charge q from a to b: E

Where

Typical Situation

A Brief Review of Concept  The creation of a charged particle distribution creates ELECTRICAL POTENTIAL ENERGY = U.  If a system changes from an initial state i to a final state f, the electrostatic forces do work W on the system  This is the NEGATIVE of the work done by the field.

Calculation An external force F is necessary to move the charge q from i to f. The work done by this external force is also equal to the change in potential energy of the charged particle. Note the (-) sign is because F and E are in opposite directions. Continuous case: q i f E

For Convenience  It is often convenient to set up a particular reference potential.  For charged particles interacting with each other, we take U=0 when the particles are infinitely apart.  Consequently U=(-) of the work done by the field in moving a particle from infinity to the point in question.

Keep in Mind  Force and Displacement are VECTORS!  Potential is a SCALAR.

UNITS  1 VOLT = 1 Joule/Coulomb  For the electric field, the units of N/C can be converted to:  1 (N/C) = 1 (N/C) x 1(V/(J/C)) x 1J/(1 NM)  Or 1 N/C = 1 V/m  So an acceptable unit for the electric field is now Volts/meter. N/C is still correct as well.

In Atomic Physics  It is useful to define an energy in eV or electron volts.  One eV is the additional energy that an proton charge would get if it were accelerated through a potential difference of one volt.  1 eV = e x 1V = (1.6 x C) x 1(J/C) = 1.6 x Joules.

Coulomb Stuff Consider a unit charge (+) being brought from infinity to a distance r from a Charge q: q r To move a unit test charge from infinity to the point at a distance r from the charge q, the external force must do an amount of work that we now can calculate. x

The math…. Remember This !!!

For point charges

Example: Find potential at P q1 q 2 q 3 q 4 d r P q1=12nC q2=-24nC q3=31nC q4=17nC  q=36 x C V=350 Volts (check the arithmetic!!)

Brief Discussion on Math

Integration There are many applications of integration in physics. We use it to add things up over an area, a line, or perhaps a volume. We often use density functions. The Examples we will use will be charge densities.

Line of Charge dq dx

Area Charge Density  dx dy dA=dxdy

The Job dx dy dA=dxdy R

Another Way to do dA (note rhyme) r  dd rd  dA= dr rd   r drd  dr

Take a look

Still another view r dr r r r r

When you have to integrate over dA Pick a friendly coordinate system.. Use the appropriate dA Don’t forget the function  or may be functions of position. The coordinate system that you choose should match the symmetry of these two functions.

An Example finite line of charge d r x dx P At P Using table of integrals

Example (from text) z R Which was the result we obtained earlier disk  =charge per unit area

The Potential From a Dipole d P  r(+) r(-) + - r

Dipole - 2 d P  r(+) r(-) + - r

Where is this going?

A Few Problems

A particular 12 V car battery can send a total charge of 81 A · h (ampere-hours) through a circuit, from one terminal to the other. (a) How many coulombs of charge does this represent? (b) If this entire charge undergoes a potential difference of 12 V, how much energy is involved? Sometimes you need to look things up … 1 ampere is 1 coulomb per second. 81 (coulombs/sec) hour = 81 x (C/s) x 3600 sec = 2.9 e +5 qV=2.9 e 05 x 12=3.5 e+6

An infinite nonconducting sheet has a surface charge density = 0.10 µC/m2 on one side. How far apart are equipotential surfaces whose potentials differ by 54 V? d 54V

In a given lightning flash, the potential difference between a cloud and the ground is 2.3 x 10 9 V and the quantity of charge transferred is 43 C. (a) What is the change in energy of that transferred charge? (GJ) (b) If all the energy released by the transfer could be used to accelerate a 1000 kg automobile from rest, what would be the automobile's final speed? m/s Energy = q  V= 2.3 e+09 x 43C=98.9 GJ E=(1/2)Mv2 v= sqr(2E/M)= 14,100 m/s

Electrical Circuits  Do Something!  Include a power source  Includes electrical components  resistors  capacitors  inductors  transistors  diodes  tubes  switches  wires

Battery  Maintains a potential difference between its two terminals. The potential Difference – the “voltage” is maintained by an electrochemical reaction which takes place inside of the battery.  Has two terminals + is held at the higher potential - is held at the lower potential

Symbol DC Power Source Low Potential High Potential

SWITCH schematic symbol

POTENTIAL PART 5 Capacitance

What’s Portending?  No new WebAssign, as promised. Old ones are still in the queue.  Exam #1 On MONDAY.. Charge Electric Field Gauss Potential  Today.. Capacitors

Encore By Special Request Where for art thou, oh Potential?

In the figure, point P is at the center of the rectangle. With V = 0 at infinity, what is the net electric potential in terms of q/d at P due to the six charged particles?

Continuing s Text gets 8.49 … one of us is right!

Derive an expression in terms of q2/a for the work required to set up the four-charge configuration in the figure, assuming the charges are initially infinitely far apart

W=q  V

Add them up..

Chapter 26  Capacitors

Capacitor  Composed of two metal plates.  Each plate is charged one positive one negative  Stores Charge  Can store a LOT of charge and can be dangerous!

Two Charged Plates (Neglect Fringing Fields) d Air or Vacuum Area A - Q +Q E V=Potential Difference Symbol

More on Capacitors d Air or Vacuum Area A - Q +Q E V=Potential Difference Gaussian Surface Same result from other plate!

Device  The Potential Difference is APPLIED by a battery or a circuit.  The charge q on the capacitor is found to be proportional to the applied voltage.  The proportionality constant is C and is referred to as the CAPACITANCE of the device.

UNITS  A capacitor which acquires a charge of 1 coulomb on each plate with the application of one volt is defined to have a capacitance of 1 FARAD  One Farad is one Coulomb/Volt

Continuing…  The capacitance of a parallel plate capacitor depends only on the Area and separation between the plates.  C is dependent only on the geometry of the device!

Units of  0 pico

Simple Capacitor Circuits  Batteries Apply potential differences  Capacitors  Wires Wires are METALS. Continuous strands of wire are all at the same potential. Separate strands of wire connected to circuit elements may be at DIFFERENT potentials.

Size Matters!  A Random Access Memory stores information on small capacitors which are either charged (bit=1) or uncharged (bit=0).  Voltage across one of these capacitors ie either zero or the power source voltage (5.3 volts in this example).  Typical capacitance is 55 fF (femto= )  Question: How many electrons are stored on one of these capacitors in the +1 state?

Small is better in the IC world!

TWO Types of Connections SERIES PARALLEL

Parallel Connection V V C Equivalent =C E C 1 C 2 C 3

Series Connection V C 1 C 2 q -q The charge on each capacitor is the same !

Series Connection Continued V C 1 C 2 q -q V 1 V 2

More General

Example C 1 C 2 V C3C3 C1=12.0  f C2= 5.3  f C3= 4.5  d (12+5.3)pf series (12+5.3)pf

More on the Big C  We move a charge dq from the (-) plate to the (+) one.  The (-) plate becomes more (-)  The (+) plate becomes more (+).  dW=Fd=dq x E x d +q -q E=  0 A/d +dq

So….

Not All Capacitors are Created Equal  Parallel Plate  Cylindric al  Spherica l

Spherical Capacitor

Calculate Potential Difference V (-) sign because E and ds are in OPPOSITE directions.

Continuing… Lost (-) sign due to switch of limits.

Materials  Consist of atoms or molecules bonded together.  Some atoms and molecules do not have dipole moments when isolated.  Some do.  Two types to consider: Polar Non-Polar

Polar Materials (Water)

Apply an Electric Field Some LOCAL ordering Large Scale Ordering

Adding things up Net effect REDUCES the field

Non-Polar Material

Effective Charge is REDUCED

We can measure the C of a capacitor (later) C 0 = Vacuum or air Value C = With dielectric in place C=  C 0 (we show this later)

How to Check This Charge to V 0 and then disconnect from The battery. C0C0 V0V0 Connect the two together V C 0 will lose some charge to the capacitor with the dielectric. We can measure V with a voltmeter (later).

Checking the idea.. V Note: When two Capacitors are the same (No dielectric), then V=V 0 /2.

Some  values Material Dielectric Strength Breakdown KV/mm Air13 Polystyrene2.624 Paper3.516 Pyrex4.714 Strontium Titanate 3108

Messing with Capacitor + V - + V The battery means that the potential difference across the capacitor remains constant. For this case, we insert the dielectric but hold the voltage constant, q=CV since C   C 0 q    C 0 V THE EXTRA CHARGE COMES FROM THE BATTERY! Remember – We hold V constant with the battery.

Another Case  We charge the capacitor to a voltage V 0.  We disconnect the battery.  We slip a dielectric in between the two plates.  We look at the voltage across the capacitor to see what happens.

Case II – No Battery q0q0 qq q=C 0 V o When the dielectric is inserted, no charge is added so the charge must be the same. V0VV0V

Another Way to Think About This  There is an original charge q on the capacitor.  If you slide the dielectric into the capacitor, you are adding no additional STORED charge. Just moving some charge around in the dielectric material.  If you short the capacitors with your fingers, only the original charge on the capacitor can burn your fingers to a crisp!  The charge in q=CV must therefore be the free charge on the metal plates of the capacitor.

A Closer Look at this stuff.. Consider this virgin capacitor. No dielectric experience. Applied Voltage via a battery. C0C V0V0 q -q

Remove the Battery V0V0 q -q The Voltage across the capacitor remains V 0 q remains the same as well. The capacitor is fat (charged), dumb and happy.

Slip in a Dielectric Almost, but not quite, filling the space V0V0 q -q q’ +q’ E0E0 E E’ from induced charges Gaussian Surface

A little sheet from the past q -q -q’ +q’ 0 2xE sheet 0

Some more sheet…

A Few slides back Case II – No Battery q0q0 qq q=C 0 V o When the dielectric is inserted, no charge is added so the charge must be the same. V0VV0V

From this last equation

A Bit more…..

Important Result  Electric Field is Reduced by the presence of the material.  The material reduces the field by a factor .  is the DIELECTRIC CONSTANT of the material

Another look +-+- VoVo

Add Dielectric to Capacitor  Original Structure  Disconnect Battery  Slip in Dielectric +-+- VoVo V0V0 Note: Charge on plate does not change!

What happens? +-+-  i   i  oooo Potential Difference is REDUCED by insertion of dielectric. Charge on plate is Unchanged! Capacitance increases by a factor of  as we showed previously

SUMMARY OF RESULTS

APPLICATION OF GAUSS’ LAW

New Gauss for Dielectrics

The Insertion Process With A Battery +-+- VoVo F