ELECTRIC POTENTIAL February 3. 2006

Wassup? Today Monday, Wednesday Friday – QUIZ and some problems Quick Review of Exam Start New Topic – Electric Potential New WebAssign on board. Read the chapter for Monday Monday, Wednesday More of the same Friday – QUIZ and some problems Monday – Finish Topic (??) + Problems

Potential We will be dealing with Work Energy We do work if we try to move a charge in an electric field. Does the FIELD do work? Does the FIELD have a brain? Muscles?

How Much Work? W=mgd ??? W=qeE ??

Please Recall Work in moving an object from A to B B SCALAR!! A

The PHY 2048 Brain Partition 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. h m A B 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 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 y=h B m y=b (reference level) y=0 Same answer! A Still falls to here.

Energy Methods VERY CAREFUL! 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!

Mrs. FIELDS vs Mr. External I need some help. Push vs Pull Mrs. FIELDS vs Mr. External

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). You can also think of this as the FIELD doing the negative of this amount of work on the particle.

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.

ZERO! And also remember: What the *&@^ does that mean??? The net work done by a conservative (field) force on a particle moving around a closed path is ZERO! What the *&@^ does that mean???

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

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. Is created by charges. 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.

Example: NOTATION U=PE HIGH U LOWER U E  q F A B d d 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.

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

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 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 DV 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 DV, the work from an external “person” necessary to do this is qDV

Example  Electric Field = 2 N/C 1 mC 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 Electric Field Equipotential Surface

Components Eparallel Enormal Electric Field Dx Work to move a charge a distance Dx along the equipotential surface Is Q x Eparallel X Dx Equipotential Surface

BUT This an EQUIPOTENTIAL Surface No work is needed since DV=0 for such a surface. Consequently Eparallel=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 Work to move a charge q from a to b: V V+dV ds b a E Note the (-) sign!

Where I probably won’t ask about this.

Typical Situation

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 sometimes 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 10-19C) x 1(J/C) = 1.6 x 10-19 Joules. Nothing mysterious.

Coulomb Stuff: A NEW REFERENCE Consider a unit charge (+) being brought from infinity to a distance r from a Charge q: x 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.

rfinal<rinitialneg sign The math…. rfinal<rinitialneg sign This thing must be positive anyway.

For point charges

Example: Find potential at P q1 q2 d P r q3 q4 q1=12nC q2=-24nC q3=31nC q4=17nC Sq=36 x 10-9C V=350 Volts (check the arithmetic!!)

An Example finite line of charge d r x dx P What about a rod that goes from –L to +L?? At P Using table of integrals

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

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 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 2.3x 109 V and the quantity of charge transferred is 43 C. 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 = qDV= 2.3 e+09 x 43C=98.9 GJ E=(1/2)Mv2 v= sqr(2E/M)= 14,100 m/s

POTENTIAL PART 5 Capacitance

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

In the figure, point P is at the center of the rectangle 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 1 2 3 4 5 6 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. 1 2 3 4

1 2 3 4 W=qDV

Add them up ..