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Physics

PHYSICS 1E03

Waldemar Okon

Fall

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James Hyatt – Physics 1E03 The “source” q produces an electric field that
pushes test charge q0
Crib Sheet
Outline q
I. Electrostatics (4 weeks) Units=N/C E = k e 2rˆ
*E=F/q 0 r
II. DC Circuits (2 weeks, plus labs)
III. Magnetism (3 weeks)
IV. Waves (3 weeks)
-includes electromagnetic waves **The field decrases in strength with
distance
****************************************** **Fields add together with multiple charges.
I. Electrostatics (4 weeks) Vector Sum
Chapter 23
e ≈ +1.602×10-19 C
Insulator: can be charged by rubbing but charges
DO NOT move (ie. glass, rubber, paper)
Conductors: (some) charges move freely (ie.
metals, some liquids)
Semiconductors: electrical properties are
between insulators and conductors (ie. Si, Ge)
r qq 9 2
F 12k e 1 2rˆ ke=8.988x10 N• m 2
r 2 C
Lambda is linear charge density (charge
*K = “Coulomb’s Constant” per length of the object). L is S for a curve
**where r hat is a unit vector parallel to r**
When one q is
doubled and **Integrate between d and d+L. Set q equal to
one is halved, lambda dx, you know that lambda and k are
the angle does constant so they get removed from the integral.
not change. If Integrate 1/x , then plug in d and d+L. (Keep in
the masses mind b-a for integrals)**
were to change,
the angle
would too.
Given m=1g, L=60cm,q1=q2
Solution: FBD, Fg=Ftsinx, find Ft, get
Ftcosx, trig=find distance between q,
Fqq=Ftcosx=(kqq)/r^2
In this example, be sure to consider the **use pythag to note that r = a + x , multply the
force between q a3d q (ie4 watch the
diagonal force as well). integral of dE by costheta (since it is a
constinuous charge distribution). The integral of
dq is Q because of s=r*theta and dq=lambda*ds
Electric Fields and the integral between 0 and 2Pi of
lambdaRdtheta equals lambda 2Pir which equals
Q because 2Pi = theta and r = r** **integrate dE=dEcostheta to get to klambda/r as
a constant is multiplied by the integral of
costhetadtheta**
Gauss
**Note that there are twice as many field lines
on the 2Q as opposed to the Q because # of field
lines are proportional to charge** 2
**Flux is equal to E dot A. A is 4Pi*r -. This
is all equal to q(enclosed) over epsilon.
Rearrange for E. Note that 1/4Pi*epsilon is
2
equal to k. This gives you E = k*q/r **
Uniformly charged sphere:
Since one can decide the Gaussian surface
themselves, let’s analyze what would
happen if the radius of the surface is less
than the original and greater than it.
**E is parallel to dA and is constant on the
Gaussian surface S, therefore can take it out
of the integral as a constant. The integral of
dA is just A. The area of a sphere is 4Pi*r .2
2
Rearrange for E, get E=k*q/r NOTE:
Gaussian radius is little r, actually radius is
big R**
**For when the Gaussian surface is smaller
than the surface of the sphere the equation
ends up being E=kqr/R . This was derived
using the volume charge density. Since the
exact Qenclosed is not known, it is used to
solve for this unknown variable. The E is still
parallel and constant.**
For any uniform distribution of charge Q
Conductors in Electrostatic Equilibrium
within a sphere of radius R: (F=0!):
-There is NO electric field INSIDE of a
conductor
-The field lines are perpendicular to the
surface
-Any and all net charge resides on the
surface ONLY V=J/C
Work
**For P1there is no flux because it is inside W=Fd=qEd, W=K=1/2*m*v^2
of the conductor and there is no charge that
resides within a conductor. For P 2t works
out to just E=k*q/r **
*Any Gaussian surface inside the conductor
encloses zero net charge.
**Electric potential is absolute, no –ve,
EVER!**
Electron Volt
Equipotential: a surface on which V is
constant. They are always perpendicular to
the electric field, a conductor is always an
equipotential, the closer they are, the larger
the E, the farther they are, the weaker the E.
**Derive V with respect to x!! DO NOT try
and derive x. Recall with dA you just take A,
same idea here
**V=Ed as well as dV=Eds, the +ve plates
have the high potential of 20V whereas the-
ve plates have the low potential of 0V**
Electric Potential (V) Solution to above picture
**Note, with squares or multiple sided
things, ensure to ADD THE DIAGONALS!!**
*Potentials add as SCALARS! **Inside of a conductor the electric field is 0
however the electric potential is a constant.
As you approach the edge and leave it it
decreases. Note that the electric field
increases and the V decreases as it
approaches R**
END OF FIRST MIDTERM

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