Presentation on theme: "POTENTIAL February 1, 2009 This Week We complete Gauss’s Law We begin the topic of POTENTIAL – Chapter 25. Problem Session Wednesday Morning Examination."— Presentation transcript:
POTENTIAL February 1, 2009
This Week We complete Gauss’s Law We begin the topic of POTENTIAL – Chapter 25. Problem Session Wednesday Morning Examination #1 is on Friday – 2/13/09 ◦Exam includes Chapters 23,24 and as much of 25 as we cover today (Monday) Next Week – More Potential
A particle with charge Q is located at the center of a cube of edge L. In addition, six other identical negative charged particles -q are positioned symmetrically around Q as shown in the figure below. Determine the electric flux through one face of the cube.
A slab of insulating material has a nonuniform positive charge density ρ = Cx 2, where x is measured from the center of the slab, as shown in the figure below, and C is a constant. The slab is infinite in the y and z directions. Derive expressions for the electric field for the following regions.
A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly distributed throughout its volume. Concentric with this sphere is a conducting spherical shell with inner radius b and outer radius c, and having a net charge -Q, as shown in the figure below.
(a) Construct a spherical Gaussian surface of radius r > c and find the net charge enclosed by this surface. (b) What is the direction of the electric field at r > c? (c) Find the electric field at r > c. (d) Find the electric field in the region with radius r where c > r > b. (e) Construct a spherical Gaussian surface of radius r, where c > r > b, and find the net charge enclosed by this surface. (f) Construct a spherical Gaussian surface of radius r, where b > r > a, and find the net charge enclosed by this surface. (g) Find the electric field in the region b > r > a. (h) Construct a spherical Gaussian surface of radius r < a, and find an expression for the net charge enclosed by this surface, as a function of r. Note that the charge inside this surface is less than 3Q. (i) Find the electric field in the region r < a. (j) Determine the charge on the inner surface of the conducting shell. (k) Determine the charge on the outer surface of the conducting shell. (l) Make a plot of the magnitude of the electric field versus r. A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly distributed throughout its volume. Concentric with this sphere is a conducting spherical shell with inner radius b and outer radius c, and having a net charge -Q, as shown in the figure below.
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B A q
We have a region in space where there is an Electric Field There is a particle of charge q at some location. The particle must be moved to another spot within the field. Work must be done in order to accomplish this.
Electric Potential We will be dealing with Work Energy & Conservation Work must be done to move a charge in an electric field. Let’s do a weird demo ….
I need some help.
What we will do …. For the moment, assume the charge has MASS. (It may not.) Assume the charge is initially stationary. The charge is to be moved to the left. The charge is to be moved at CONSTANT velocity. + charge E Mr. ExternalMrs. Fields
Start and Stop ENERGY is required to bring the charge up to speed (if it has mass). ENERGY is required to bring the particle back to rest (if it has mass). The sum of these two is ZERO.
Each does the negative amount of work than the other does.
So, when we move a charge in an Electric Field.. 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.
Summary-- 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 negative You can also think of this as the FIELD doing the negative of this amount of work on the particle.
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. The change in potential energy of a charge is the amount of work that is done by an external force in moving the charge from its initial position to its new position. It is the negative of the work done by the FIELD in moving the particle from the initial to the final position.
Definition – Potential Energy PE or U is the work done by an external agent in moving a charge from a REFERENCE POSITION to a different position. A Reference ZERO is placed at the most convenient position Like the ground level in many gravitational potential energy problems.
Example: E Zero Level q F d Work by External Agent W external = F d = qEd= U Work done by the Field is: W field = -qEd = -W external
A uniform electric field of magnitude 290 V/m is directed in the positive x direction. A +13.0 µC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm).(a) What is the change in the potential energy of the charge field system? [-0.000754] J
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
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 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
Consider Two Plates OOPS …
Look at the path issue
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 defined an absolute level of potential. To do this, we needed to define a REFERENCE or ZERO level for potential. For a uniform field, it didn’t matter where we placed 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.
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 xx 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
Consider Two Equipotential Surfaces – Close together V V+dV ds a b Work to move a charge q from a to b: E
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 -19 C) 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: 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
Just Do It!
OK, doing it! Set the REFERENCE LEVEL OF POTENTIAL at INFINITY so (1/r A )=0.
For point charges
For a DISTRIBUTION of charge:
Ponder – What is the potential a distance d from an infinite plane whose charge per unit area is ?
Given two 2.00-μC charges, as shown in Figure P25.16, and a positive test charge q = 1.28 × 10–18 C at the origin, (a) what is the net force exerted by the two 2.00-μC charges on the test charge q? (b) What is the electric field at the origin due to the two 2.00-μC charges? (c) What is the electrical potential at the origin due to the two 2.00-μC charges?
The three charges in Figure P25.19 are at the vertices of an isosceles triangle. Calculate the electric potential at the midpoint of the base, taking q = 7.00 μC.
A disk of radius R has a non-uniform surface charge density σ = Cr, where C is a constant and r is measured from the center of the disk. Find (by direct integration) the potential at P.
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 10 -9 C V=350 Volts (check the arithmetic!!)
An Example finite line of charge d r x dx P At P Using table of integrals What about a rod that goes from –L to +L??
Example z R Which was the result we obtained earlier disk =charge per unit area
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 23 4 56 HRW 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