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## ELECTRIC POTENTIAL

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**ELECTRIC POTENTIAL**September 13, 2006**Goings On For the Next Few Days**• Quiz Today – Gauss/Electric Field • Today – Begin Chapter 25 – Potential • Monday - Potential • Wednesday – EXAMINATION #1 • Friday – More Potential**Picture a Region ofspace Where there is an Electric Field**• Imagine there is a particle of charge q at some location. • Imagine that the particle moves to another spot within the field. • Work must be done in order to accomplish this.**Electric 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?**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.**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.**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**Work done by external force = mgh How much work here by gravitational field? h mg**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.**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 –Fd. The charge sort-of “fell” to lower potential energy.**Another way to look at it:**• Consider an external agent. • The external agent moves a charge from one point to the other. • The work done by the external agent in moving the charge between these points is the change in potential energy. • This is the negative of the work that the FIELD does!**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 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**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. Assume this is zero for now.**Furthermore…**If we move a particle through a potential difference of DV, the work from an external “person” necessary to do this is qDV**Electric Field = 2 N/C** d= 100 meters 1 mC Example**Consider Two Plates**OOPS …**The difference in potential between the accelerating plates**in the electron gun of a TV picture tube is about 25 000 V. If the distance between these plates is 1.50 cm, what is the magnitude of the uniform electric field in this region?**An ion accelerated through a potential difference of 115 V**experiences an increase in kinetic energy of 7.37 × 10–17 J. Calculate the charge on the ion.**Important**• We can define an absolute level of potential. • To do this, we need to define a REFERENCE or ZERO level for potential. • For a uniform field, it doesn’t matter where we place the reference. • For POINT CHARGES, we will see shortly that we must place the level at infinity or the math gets very messy!**An Equipotential Surface is defined as a surface on which**the potential is constant. 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.**Field Lines and Equipotentials**Electric Field Equipotential Surface**Components**Enormal Electric Field Dx Eparallel 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**ds**V+dV V Consider Two EquipotentialSurfaces – Close together Work to move a charge q from a to b: b a E Note the (-) sign!**Over a certain region of space, the electric potential is V**= 5x – 3x2y +2yz2. Find the expressions for the x, y, and z components of the electric field over this region. What is the magnitude of the field at the point P that has coordinates (1, 0, –2) m?**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: INFINITY**Consider a unit charge (+) being brought from infinity to a distance r from a Charge q: x r q 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.