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25 Electric Potential25.1 Electric Potential and Potential
Difference25.2 Potential Difference in a Uniform Electric Field25.3
Electric Potential and Potential Energy Due to Point Charges25.4
Obtaining the Value of the Electric Field from the Electric
Potential25.5 Electric Potential Due to Continuous Charge
Distributions25.6 Electric Potential Due to a Charged Conductor25.7
The Millikan Oil-Drop Experiment25.8 Applications of
Electrostatics4. Fields and potentials of other charge
distributions a) Students should be able to use the principle of
superposition to calculate by integration: ! (1) The electric field
of a straight, uniformly charged wire. ! (2) The electric field and
potential on the axis of a thin ring of charge, or at the center of
a circular arc of charge. ! (3) The electric potential on the axis
of a uniformly charged disk. b) Students should know the fields of
highly symmetric charge distributions, so they can: ! (1) Identify
situations in which the direction of the electric field produced by
a charge distribution can be deduced ! ! from symmetry
considerations. ! (2) Describe qualitatively the patterns and
variation with distance of the electric field of: ! ! (a)
Oppositely-charged parallel plates. ! ! (b) A long,
uniformly-charged wire, or thin cylindrical or spherical shell. !
(3) Use superposition to determine the fields of parallel charged
planes, coaxial cylinders or concentric spheres. ! (4) Derive
expressions for electric potential as a function of position in the
above cases.
! Electric Potential, Part 1 86:57 (4 days?) * The potential
difference between two points A and B, written as Vb – Va, is equal
to the work done in moving a unit charge
from A to B.
* If a uniform E-field exists in a region of space, the Vb – Va
= - E·d, the dot product of the vector E and the vector d from A to
B.
* In a parallel plate capacitor, the potential difference
between the plates is equal to Ed, where E is the electric field
between the plates and d is the separation between the plates. The
positive plate is at a higher potential than the negative
plate.
http://sdsu-physics.org/physics180/physics196/Topics/electricPotential.html
AP Physics C (Electromagnetics) Chapter 25 Class work
http://sdsu-physics.org/physics180/physics196/Topics/electricPotential.htmlhttp://sdsu-physics.org/physics180/physics196/Topics/electricPotential.html
Electric Potential, Part 2 91:50 (5 days?) * The electric
potential produced by a point charge Q at a point a distance r away
is kQ/r.
* For a collection of point charges Q1, Q2, Q3, …, the
electrostatic potential energy is equal to
U = (½)∑(i, j) kQiQj / rij, where rij is the distance between
charges Qi and Qj. The factor of 1/2 is inserted in order to avoid
double counting. In other words, we sum over the pairs of
charges.
* The electric field vector may be obtained from the electric
potential: E = -∇ ??V; i.e., E is the negative of the gradient of
V.
Electric Potential, Part 3 69:12 (3 days?) * For a continuous
charge distribution, V = ∫Q(k dq / r), where the integral is over
the charge distribution, and r is the
distance from dq to the point where V is to be found.
*In the lecture, examples are given, which show how to find the
electric potential produced by a charged ring, a charged disk, and
a finite line of charge.
Electric Potential, Part 4 71:16 (4 days) *For a conductor, any
excess charge must reside on the surface. Second, the electric
field inside the conductor is zero. Third,
E on the surface is perpendicular to the surface, and is given
by E = 𝝈𝑞 / εo , where sigma is the surface charge density. Fourth,
the whole conductor is an equipotential region; i.e., all points on
or within the conductor are at the same potential.
* If a conducting sphere has a total charge Q distributed on its
surface, the potential inside the sphere is given by kQ/R, where R
is the radius of the sphere.
* In a cavity within a conductor, the electric field is zero; if
it were not, the inner surface of the conductor would not be an
equipotential surface. The vanishing of E in the cavity explains
why conductors are used as electrical shields.
AP Physics C (Electromagnetics) Chapter 25 Class work
