Review: Electric Field and Electric Force · Review: Electric Field and Electric Force q 0 q r r r q K e ... Don’t take notes of my solution: I AM FOOLING AROUND! It is very complicated

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