Exam One Review

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W05D2

Announcements

Exam 1 March 5 7:30-9:30 pm Conflict Exam 1 March 6 8-10 am in 26-314, or 9-11 am in 34-101 No Class Friday March 6 Sunday Tutoring March 8 2-5 pm in 26-152 Problem Set 5 Due Tuesday March 10 at 9 pm Week 6 Prepset 1 due L01-L04 Mon at 8:30 am Week 6 Prepset 1 due L05-L08 Tues at 8:30 am

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Conflict Forms

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Exam 1 Google Conflict Form for Friday March 6 https://docs.google.com/forms/d/e/1FAIpQLSd9uu_OHWjH8wXzazy4G6A__a4001Dz-8HexHARMEJjvmcGhw/viewform?usp=sf_link Exam 1 Google Conflict Form for Students who are out of town March 5 and March 6 https://docs.google.com/forms/d/e/1FAIpQLSeO_nL6so5G8YCZxIInsJRxibFZgyE5C2NLFRYs_pBnm7PFTg/viewform?usp=sf_link

Exam 1 Room Assignments:

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Exam 1 Thursday March 5 from 7:30-9:30 pm L01 in 26-100 L02 in 26-100 L03 in 34-101 L04 in 50-340 L05 in 50-340 L06 in 50-340 L07 Last names starting with A- L in 34-101 L07 Last names starting with M-Z in 6-120 L08 in 26-152

Exam 1 Topics (1)  Fields (visualizations) (2)  Electric Field •  Discrete and Continuous Charge Distributions •  Symmetric Distributions – Gauss’s Law with

superposition (3) Electric Dipole: Torque and Force (4) Electric Potential Difference: Discrete and Continuous Charge Distributions (5) Energy and Force Applications (6) Electric Field from Electric Potential (7) Conductors and Equipotential Surfaces (8) Capacitors, Capacitance, and Stored Energy

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What You Should Study •  Review Friday Problem Solving (& Solutions) •  Review In Class Problems (& Solutions) •  Review Concept Questions (& Solutions) •  Review Problem Sets (& Solutions) •  Review PowerPoint Presentations •  Review Relevant Parts of Course Notes

(& Included Examples) •  Review Videos •  Try Practice problems on Exam 1 webpage

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Exam One: Review Problems

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Using Gauss’s Law to find Electric Potential from Electric Field

If the charge distribution has “enough” symmetry to use use Gauss’s Law to calculate the electric field, then calculate the electric potential difference between two points A and B using This is a line integral.

B

B AA

V V d− = − ⋅∫E s! !

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Choosing a Gaussian Surface Gauss’s Law holds for all closed surfaces. However it is only useful (to calculate electric field ) for sources with enough of symmetry in which there exists closed surfaces such that

1.  The electric field on some faces of the closed surface is

both perpendicular to the face and has uniform magnitude on that face. Then

2.  The electric field on some faces of the closed surface is parallel to that face. Then

E ⋅dA

face∫∫ = EA

E ⋅dA

face∫∫ = 0

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Summary: Applying Gauss’s Law 1.  Based on the source, identify regions in which to calculate

electric field.

2.  Choose Gaussian surface S: Symmetry

3.  Calculate

4.  Calculate qenc, charge enclosed by surface S. You may need to integrate if charge density is non-uniform

5.  Apply Gauss’s Law to calculate electric field:

E ⋅dA

closedsurface S

∫∫ =qenclosed

ε0

ΦE =

E ⋅dA

S∫∫

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Continuous Charge Distributions

( ) ?P =Er

V

Q = Δqi

i∑

Break distribution into parts:

ΔE = ke

Δqr 2 r

E field at P due to

Superposition:

E = Δ

E∑ → d

E∫

→ dq

V∫∫∫

→ dE = ke

dqr 2 r

Δq

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Calculating Electric Potential Difference for Continuous Distributions

1.  Choose 2.  Choose integration

variables 3.  Identify

4.  Choose field point variables 5.  Calculate source to field

point distance 6.  Define limits of integral 7.  Integrate

V (∞) = 0

′r

r

d ′q =

λd ′s

σd ′a

ρd ′v

⎧

⎨⎪⎪

⎩⎪⎪

!r - !′r = !rS ,P

V (r) −V (∞) = ke

d ′qr - ′rsource

∫16

Deriving E from V: gradient operator

Gradient (del) operator:

∇ ≡ ∂

∂xi + ∂

∂yj + ∂

∂zk

In Cartesian coordinates:

E = − ∂V

∂xi + ∂V

∂yj + ∂V

∂zk

⎛⎝⎜

⎞⎠⎟

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ΔV = −

!E

A

B

∫ ⋅d !s⇒!E = −

!∇V

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Electric Field

The electric field at a point P due to a charged object (source) with charge qs is the force acting on a test point-like charged object with charge qt at that point P, divided by the charge qt :

Es(P) = ke

qs

rst2 rs(P)

Es(P) ≡

Fst (P)

qt

Units: N/C

Superposition Principle

The electric field due to a collection of N point charges is the vector sum of the individual electric fields due to each charge

E =E1 +E2 + . . . . .=

Ei

i=1

N

∑

Electric Potential Function

ΔVAB = kqs

1rB

− 1rA

⎛

⎝⎜⎞

⎠⎟

Choose rA = ∞ and rB = r. Choose V(∞) = 0. Then the electric potential function at a point P a distance r from the point-like charged source (charge qs) is Superposition principle:

Vs(P) =

kqs

r

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B

A

r

d sinfinity

ES (P)

qs > 0

qttest charge

V (P)−V (∞) = ke

q j

rq j ,Pj=1

j=N

∑

A charged particle, charge q and mass m, in an electrostatic electric field will experience a force given by When a charged particle (charge q) is moved across an electric potential difference then the change in potential energy is The change in mechanical energy of the charged particle is

E Field and Potential: Effects

F = q

E

ΔU = qΔV

Wext = ΔEmech = ΔK + ΔU = (1/ 2)m(v f

2 − vi2 )+ qΔV

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Configuration Energy What is the potential energy stored in a configuration of charged objects? Start with all the charged objects at infinity. Choose

(1) Bring in the first charged object.

(2) Bring in the second charged object

(3) Bring in the third charged object

(4) Configuration energy

q1

q2

q3

r12

r13

r23 ΔU2 =U12 = keq1q2 / r12

U (∞) ≡ 0

ΔU3 =U23 +U13 = keq2q3 / r23 + keq1q3 / r13

ΔU1 = 0

ΔU =U12 +U23 +U13 = keq1q2 / r12 + keq2q3 / r23 + keq1q3 / r1323

Electric Dipole Moment: Two Charges

Dipole moment vector points from negative to positive charge, with magnitude p

!p ≡ q+!r+ + q−

!r−= (+q)(acosθ i + asinθ j)+ (−q)(−acosθ i − asinθ j)= 2qa(cosθ i + sinθ j) = p(cosθ i + sinθ j)

+

q

q

ai

j

+

a

0

r+

r

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+

q

q

ij

+0

p

+

q

q

aE

+

a

0

r+

r

ij

k

F+

F

Dipole in External Uniform Field: Torque

E = Ei external field

Fnet =

F+ +F−

= qE+ (−q)

E = 0

Total Net Force:

Torque on Dipole:

tends to align with the external electric field p

τ = r+ ×

F+ +r− ×F−

= r+ × q+

E+ r− × q−

E

= (q+r+ + q−

r− )×E

= p×E

p ≡ q+

r+ + q−r−

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Capacitance Consider an isolated conductor with total charge Q. The conductor is an equipotential surface, with a certain potential V, where V = 0 at infinity. The charge on the conductor Q is linearly proportional to the potential The constant of proportionality C is called the capacitance and only depends on the size and shape of the conductor. SI Units: 1 coulomb/volt = 1 farad

Q = CVQ

conductor at potential V,(V = 0 at infinity)

conductor with charge Q,

+Q Q

Capacitor: Two Conductors

Capacitor: Two isolated conductors with equal and opposite charges ± Q. Potential difference ΔV between the two conductors. The capacitance of the two conducting surfaces is C is always a positive constant that just depends on the size and shape of the conductors

C = Q / ΔV

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Parallel Plate Capacitor

C depends only on geometric factors A and d ΔV = −

E ⋅dS

bottom

top

∫ = Ed = QAε0

d C = Q

ΔV=ε0 Ad

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CQ: Electric Field Grass Seeds

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CQ: Electric Force

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CQ: Dipole

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CQ: Electric Field

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CQ: Electric Potential