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Physics 202, Lecture 3Today’s Topics
Calculate Electric Field With Superposition (Direct Sum/Integral of Coulomb’s Law)
Calculate Electric Field With Gauss’s Law Gauss’s Law Examples
Expected from preview:Calculate E with continuous charge.Surface, closed surface, surface integral, flux, the
Gauss’s Law.2
Review: Electric Field and Electric Force
q0
qr
rrqKe ˆ20=E
qrrqKq e EF ˆ20==
q0: source chargeE: field by q0q: test chargeF = qE force on q by E
Electric Field is a form of matter. It carries energy (later in the semester)
3
How to Calculate Electric Field?
Single point-like :
Multiple charges :(superposition principle)
Continuous Charge Distribution:
Note:For now, we assume charges are not moving.(electrostatic)
rrqkE e ˆ20=
∑= ii
ie r
rqkE ˆ2
rrdqkr
rqkE eii
i
qe ˆˆ 22
0lim ∫∑ =Δ=
→Δ
4
Example: Charged Rod A uniformly charged rod of length L has a total charge Q, find
the electric field: at point A answer: Ex= -keQ/(a(L+a)) , Ey=0 (see board) at point B answer: Ey= 2keQ/(Lb)*sinθ0 , Ex=0 (see board, show method only) at an arbitrary point C. (see board, conceptual only).
A
a
B
bCθ
tanθ0 =_L/b
y
xdx dxdxdq=Q/Ldx
Calculus Requirements in this Course
The level of calculus shown in this example:
1: Shall be understood at conceptual level.2: Will be practiced in HW problems3: Will not be tested in the exam. (some less complicated forms might be in exams)
5
Example: Uniformly Charged Sphere A uniformly charged sphere has a radius a and total
charge Q, find the electric field outside and insidethe sphere.
Solution:
...
Don’t take notes of my solution: I AM FOOLING AROUND!It is very complicated with the superposition method! Gauss’s Law to the rescue!
a
x
y
z
E(x,y,z)=?
Electric Flux The electric flux through a surface element is
defined as the dot product of the electric field andthe surface area vector:
ΔΦE=E•ΔA = EΔAcosθ
The net electric flux through a closed surface
∫ •≡Φ AE dE
ΔAθ
Gauss’s Law Net electric flux through any closed surface (“Gaussian
surface”) equals the total charge enclosed inside theclosed surface divided by the permittivity of free space.
ΦE =
EidA∫ =
qenclε0
ε0: permittivity constant
q1
qi
q2
Gaussian surface(any shape)
: all charges enclosedregardless of positions
electric flux qencl
(4πε0 )−1 = k
Trivia Quiz 1 Compare electric fluxes through closed surfaces
s1,s2,s3:1. Φs1>Φs2>Φs3
2. Φs1=Φs2=Φs3
3. Φs1<Φs2<Φs3
Trivia Quiz 2 What is the electric flux through closed surface S?
1. Φ = 02. Φ = (q1+q2+q3+q4+q5)/ε0
3. Φ = (q1+q2+q3)/ε0
q1
q2q3
S
q4q5
Uniformly Charged Sphere Again
Solution using Gauss’s Law :
The setting is highly symmetrical
Gaussian surface will be concentric sphere of radius r.
ErEA
dAEEdAAdE24π==
==• ∫∫∫
inqAdE0
1ε
=•∫
How to evaluate ?
Note the symmetry: Direction of E: RadialMagnitude of E: Same in all direction
∫ • AdE
Uniformly Charged Sphere: Details
r>a: Outside the Sphere r<a: inside the Sphere
inqrE
0241επ
= inqErAdE
0
2 14ε
π ==•∫
{ ar if: Q
ar if: Q :where
>
<=
3
3
arinq
Q
Uniform Charge Sphere: Final Solution
Note: This has the same form asthe point charge
inside outside
Procedure to Use Gauss’s Law
General principle:Gauss’s law is valid for any charge distributions, but practically it isuseful only in limited situations where the charge distribution ishighly symmetric
Procedure:1. Draw a Gaussian surface passing the field point concerned.
Observe symmetry so that the surface integral is trivial. Direction of E: Either perpendicular or parallel to the surface. Magnitude of E: The same (or be zero) on the surfaces
2. Evaluate the surface integral using arguments of symmetry. AndEquate the surface integral to qin/ε0 and solve for E.
0εinqd =•∫ AE
Three Common Symmetric Cases Spherical(point Q, uniform sphere, shell)
Cylindrical(infinite line/cylinder of Q )
Planar(infinite sheet of Q )
The above symmetric settings give very predictable E Direction: Normal to surfaces of same symmetry Magnitude: Same across surface (of same symmetry)
A EAdE =•∫
Another Example: Thin Spherical Shell Find the E field inside/outside a uniformly charged
thin sphere.Solution: Exercise with your TAs.
Gaussian Surfacefor E(outside)
ResultGaussian Surfacefor E(inside)
Typical Electric Field ObtainableBy Gauss’s Law
3
One More Exercise On Gauss’s Law Two charges +2Q and –Q are placed at locations
shown. Find the electric field at point P. Solution:
1. Draw a Gaussian surface passing P2. Apply Gauss’s law:
3. qin= +2Q +(-Q)=Q4. Surface integral:
5. E=1/4πε0 (Q/r2)
EP=?r
+2Q -Q
P
0εinqd =•∫ AE
Erd 24π=•∫ AE
Which step is wrong? Is this correct? No! step 4