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W04D2 Exam One Review

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Exam One Thursday Feb 27 7:30-9:30 pm

Conflict Exam Friday Feb 28

8:00-10:00 am in 32-082 10-12 noon in 32-082

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Exam Rooms Section L01 Walker Memorial Third Floor 50-340 Section L02 26-152 Section L03 32-123 Section L04 26-100 Section L05 34-101 Section L06 26-100 Section L07 32-123 Section L08 34-101 Section L09 Walker Memorial Third Floor 50-340

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Exam 1 Topics (1)  Fields (visualizations) (2)  Electric Field •  Discrete Point Charge and Continuous

Charge Distributions •  Symmetric Distributions – Gauss’s Law

with superposition (3) Electric Potential Difference (4) Electric Dipole: Torque and Force (5) Energy and Force Applications

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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 Pre-class MITx pages

•  Do Sample Exams (online under Exam Prep)

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Exam One: Review Problems Spring 2013

http://web.mit.edu/8.02t/www/materials/ExamPrep/exam1/Exam1_2013Spr.pdf

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Coulomb's Law Coulomb’s Law: Force on q2 due to interaction between q1 and q2

ke =

14!"0

= 8.9875#109 N m2 /C2

!F12 = ke

q1q2

r122 r12

r12 : unit vector from q1 to q2

r12 =

!r12r12

!!F12 = ke

q1q2

r123

!r12

!r12 : vector from q1 to q2

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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

!

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Fields

Grass Seeds Know how to read

Field Lines Know how to draw

•  Field line density tells you field strength •  Lines have tension (want to be straight) •  Lines are repulsive (want to be far from other lines) •  Lines begin and end on sources (charges) or infinity

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Electric Dipole Two equal but opposite charges +q and –q, separated by a distance 2a

!p ! q2aj= 2qaj

points from negative to positive charge pr

q

-q

2a

Dipole moment two charges

pr

Dipole moment for neutral charge distribution, N point charges

!p ! qii=1

i=N

" !ri

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Dipole in Uniform Field

!E = Ei

!p = 2qa(cos! i + sin! j)

!Fnet =

!F+ +

!F! = q

!E + (!q)

!E = 0Total Net Force:

Torque on Dipole:

tends to align with the electric field pr

!! = !r "

!F

= (2a)(qE)sin(!) =!p !!E

! = rF+ sin(") = pE sin(!)

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Force on a Charge in an electric Field

!F = q

!E

If you put a charged particle, (charge q), in a field:

!F = m!a ! q

!E = m!a

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Continuous Sources: Charge Density

Length L=L

w

L

Area = A = wL

R

L

dLdQ λ=

dAdQ σ=

dVdQ ρ=

!= Q / V (uniform)

Volume =V = !R2L

! = Q / A (uniform)

! = Q / L (uniform)

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( ) ?P =Er

V

Continuous Charge Distributions

Q = !qi

i"

Break distribution into parts:

!!E = ke

!qr 2 r

E field at P due to

Superposition:

!E = !

!E" # d

!E$

! dq

V"""

! d!E = ke

dqr 2 r

!q

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Calculating Electric Field for Continuous Distributions

1.  Choose integration variables. define vector from origin to

2.  Identify

3.  Define vector from origin to charge element

4.  Choose field point variables. Define vector from origin to field point

5.  Calculate source to field point distance

6.  Define limits of integral and integrate

!!r

!r

d !q =

"d !s

#d !a

$d !v

%

&''

(''

!r - !!r

!E(!r) = ke

d !q (!r - !!r )!r - !!r

3source"

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r

θ

θ

2L−

2L+

xd ′

x′

xddq ′= λs

22 xsr ′+=

P

j

i

Line of Charge

r =!rr= ! "x i + sj

( "x 2 + s2 )1/2

!E = d

!E! = ke

dqr 2 r

source!

!E(0,s) = ke

!d "x (# "x i + sj)( "x 2 + s2 )( "x 2 + s2 )1/2

"x =#L/2

"x =L/2

$

Coordinates of point-source dq: Give integration variable, x‘, a “primed” variable name. Coordinates of field point P: Distance form source to field point:

( !x ,0)

dq = !d "x

(0,s)

r = ( !x 2 + s2 )1/2

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Case II: E is uniform vector field directed at angle to planar surface S of area A

!E = EAcos"

Electric Flux

!E =

!E "d!A##

d!A = dAn

n

θ

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Gauss’s Law – The Equation

Electric flux (the surface integral of E over closed surface S) is proportional to charge enclosed by the volume enclosed by S

!E =

!E "d!A

closedsurface S

"## =qenclosed

$0

ΦE

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Gauss’s Law:

!E ! d!A

S""" =

qin

#0

Spherical Symmetry

Cylindrical Symmetry

Planar Symmetry

Gaussian Pillbox

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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

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

!E ! d!A

closedsurface S

""" =qenclosed

#0

!E =

!E "d!A

S"##

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Electrical Work

Work done by electrical force moving object 1 from A to B:

W1 =

!F12 ! d

!s12A

B

"

Electrical force on object 1 due to interaction between charged objects 1 and 2:

!F12 = ke

q1q2

r122 r12

LINE INTEGRAL

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Potential Energy Difference

!U "U B #U A = #

!F12 $ d

!sA

B

% = #W = keqsq1

1rB

# 1rA

&

'()

*+

Suppose charged object 1 is fixed and located at the origin and charge object 2 moves from an initial position A a distance rA from the origin to a final position B, a distance rB from the origin. The potential energy difference due to the interaction is defined to be the negative of the work done object 2 in moving from A to B:

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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

!U2 =U12 = keq1q2 / r12

U (!) " 0

!U3 =U23 +U13 = keq2q3 / r23 + keq1q3 / r13

!U1 = 0

!U =U12 +U23 +U13 = keq1q2 / r12 + keq2q3 / r23 + keq1q3 / r13

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Electric Potential Difference

!V " !U

qt

= #!FqtA

B

$ % d !s = #!E

A

B

$ % d !s

Units: Joules/Coulomb = Volts

Change in potential energy per test object in moving the test object (charge qt) from A to B:

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Charged Particle moving Across an Electric Potential Difference

!U = q!V

To move a charged particle, (charge q), in a field and the particle does not change its kinetic energy then:

!U + !K = 0" q!V + !K = 0

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Potential Created by Pt Charge

d!s = dr r + r d! !

!V =VB "VA = "

!E # d!s

A

B

$

!E = kQ r

r 2

= ! kQ r

r 2 " d!s

A

B

# = !kQ drr 2A

B

#

= kQ 1

rB

! 1rA

"

#$%

&'

Take V = 0 at r = ∞:

VPoint Charge (r) =

kQr

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Continuous Charge Distributions Break distribution into infinitesimal charged elements of charge dq. Electric Potential difference between infinity and P due to dq.

Superposition Principle: Reference Point:

dV !Vdq (P) "Vdq (#) = ke

dqr

V (P) !V (") = ke

dqrsource

#

V (!) = 0 " V (P) = ke

dqrsource

#

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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 =

"d !s

#d !a

$d !v

%

&''

(''

!r - !!r

V (!r) !V (") = ke

d #q!r - !#rsource

$

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Electric Potential: Ring d !q = "d !s = "Rd !#

dV =

14!"0

d #q!r - !#r

=14!"0

$Rd #%R2 + z2

V (z)!V (") = ke#Rd $%R2 + z20

2&

' =ke#R

R2 + z2d $%

a

b

'

V (z)!V (") =ke 2&R#

R2 + z2=

keQ

R2 + z2

V (!) = 0Choose ! = Q / 2"R

!!r = Rr

!r = z k

!r - !!r = R2 + z2

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Concept Questions

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Concept Question: 5 Equal Charges

Six equal positive charges q sit at the vertices of a regular hexagon with sides of length R. We remove the bottom charge. The electric field at the center of the hexagon (point P) is:

1.!E = 2kq

R2 j 2.!E = ! 2kq

R2 j 5.!E = 0

3.!E = kq

R2 j 4.!E = ! kq

R2 j

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Concept Question Answer: 5 Equal Charges

E fields of the side pairs cancel (symmetry) E at center due only to top charge (R away) Field points downward

Alternatively: “Added negative charge” at bottom R away, pulls field down

!E = ! kq

R2 jAnswer 4.

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Concept Question: Electric Field Two charged objects are placed on a line as shown below. The magnitude of the negative charge on the right is greater than the magnitude of the positive charge on the left, . Other than at infinity, where is the electric field zero?

1.  Between the two charged objects. 2.  To the right of the charged object on the right. 3.  To the left of the charged object on the left. 4.  The electric field is nowhere zero. 5.  Not enough info – need to know which is positive.

qR > qL

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Concept Question Answer: Electric Field

Answer: 3. To the left of the positively charged object on the left

Between: field points from positively charged object to negatively charged object On right: field is dominated by negatively charged object On left: electric field due to close smaller positively charged object will cancel electric field due to further larger negatively charged object.

http://web.mit.edu/viz/EM/visualizations/electrostatics/ElectricFieldConfigurations/pcharges/pcharges.htm

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Concept Question: Dipole in Non-Uniform Field

A dipole sits in a non-uniform electric field E

E

Due to the electric field this dipole will feel:

1.  force but no torque 2.  no force but a torque 3.  both a force and a torque 4.  neither a force nor a torque

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Concept Question Answer: Non-Uniform Field

Because the field is non-uniform, the forces on the two equal but opposite point charges do not cancel. As always, the dipole wants to rotate to align with the field – there is a torque on the dipole as well

Answer: 3. both force and torque E

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Concept Question Electric Field of a Rod A rod of length L lies along the x-axis with its left end at the origin. The rod has a uniform charge density λ. Which of the following expressions best describes the electric field at the point P

1.!E(P) = !ke

"d #x( #x + d)3

#x =0

#x = L

$ i

2.!E(P) = ke

!d "x( "x + d)3

"x =0

"x = L

# i

3.!E(P) = !ke

"d #x( #x + d)2

#x =0

#x = L

$ i

4.!E(P) = ke

!d "x( "x + d)2

"x =0

"x = L

# i

5.!E(P) = !ke

"L(d + L)2 i

6.!E(P) = ke

!L(d + L)2 i

7.!E(P) = !ke

"Ld 2 i

8.!E(P) = ke

!Ld 2 i

9. None of the above.

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Concept Question Electric Field of a Rod: Answer

A rod of length L lies along the x-axis with its left end at the origin. The rod has a uniform charge density λ. Which of the following expressions best describes the electric field at the point P

3.!E(P) = !ke

"d #x( #x + d)2

#x =0

#x = L

$ i

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A uniformly charged ring of radius a has total charge Q. Which of the following expressions best describes the electric field at the point P located at the center of the ring?

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Concept Question Electric Field of a Ring

1.!E(P) = !ke

"ad#a3

# =0

# =2$

% i

2.!E(P) = ke

!ad"a3

" =0

" =2#

$ i

3.!E(P) = !ke

Qa2 i

4.!E(P) = +ke

Qa2 i

5.!E(P) =

!0

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Concept Question Electric Field of a Ring: Answer

A uniformly charged ring of radius a has total charge Q. Which of the following expressions best describes the electric field at the point P located at the center of the ring?

5.!E(P) =

!0

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Concept Question: Flux thru Sphere The total flux through the below spherical surface is

+q +q

1.  positive (net outward flux). 2.  negative (net inward flux). 3.  zero. 4.  Not well defined.

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Concept Question Answer: Flux thru Sphere

We know this from Gauss’s Law: No enclosed charge no net flux. Flux in on left cancelled by flux out on right

Answer: 3. The total flux is zero

+q +q

!E =!E "d!A

closedsurface S

"## =qenclosed

$0

Concept Question: Gauss’s Law

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The grass seeds figure shows the electric field of three charges with charges +1, +1, and -1, The Gaussian surface in the figure is a sphere containing two of the charges. The electric flux through the spherical Gaussian surface is 1.  Positive 2.  Negative 3.  Zero 4.  Impossible to determine without more information.

Concept Question Answer: Gauss’s Law

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Answer 3: Zero. The field lines around the two charged objects inside the Gaussian surface are the field lines associated with a dipole, so the charge enclosed in the Gaussian surface is zero. Therefore the electric flux on the surface is zero. Note that the electric field E is clearly NOT zero on the surface of the sphere. It is only the INTEGRAL over the spherical surface of E dotted into dA that is zero.

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Concept Question: Spherical Shell We just saw that in a solid sphere of charge the electric field grows linearly with distance. Inside the charged spherical shell at right (r<a) what does the electric field do?

a

Q

1.  Zero 2.  Uniform but Non-Zero 3.  Still grows linearly 4.  Some other functional form (use Gauss’ Law) 5.  Can’t determine with Gauss Law

Concept Question Answer: Flux thru Sphere

Spherical symmetry Use Gauss’ Law with spherical surface. Any surface inside shell contains no charge No flux E = 0!

Answer: 1. Zero

a

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Concept Question: Superposition

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Three infinite sheets of charge are shown above. The sheet in the middle is negatively charged with charge per unit area , and the other two sheets are positively charged with charge per unit area . Which set of arrows (and zeros) best describes the electric field?

−2σ

+σ

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Concept Question Answer: Superposition

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Answer 2 . The fields of each of the plates are shown in the different regions along with their sum.

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Concept Question: Sign of W

Suppose a fixed positively charged object (charge qs > 0) is at the origin and we move a negatively charged object (charge q1 < 0) from A to B with rA < rB , where r is the distance from the origin.

1.  Work done by the electrostatic force is positive and we do a positive amount of work

2.  Work done by the electrostatic force is positive and we do a negative amount of work

3.  Work done by the electrostatic force is negative and we do a positive amount of work

4.  Work done by the electrostatic force is negative and we do a negative amount of work

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Concept Question Ans.: Sign of W

W is the work done by the electrical force. This is the opposite of the work that we must do in order to move a charged object in an electric field due to source. The electrical force is attractive and we are moving the positively charged object away from the source (opposite the direction of the electric field).

Answer 3: W is negative and we do a positive amount of work

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Group Problem: Work Done by Electrical Force

A point-like charged source object (charge qs) is held fixed. A second point-like charged object (charge q1)is initially at a distance rA from the fixed source and moves to a final distance rB from the fixed source. What is the work done by the electrical force on the moving object? Hint: What coordinate system is best suited for this problem?

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Concept Question: Motion of Charged Objects

Two oppositely charged are released from rest in an electric field.

1.  Both charged objects will move from lower to higher potential energy.

2.  Both charged objects will move from higher to lower potential energy.

3.  The positively charged object will move from higher to lower potential energy; the negatively charged object will move from lower to higher potential energy.

4.  The positively charged object will move from lower to higher potential energy; the negatively charged object will move from higher to lower potential energy.

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Concept Q. Ans.: Motion of Charged Objects

2. Both charged objects will move from higher to lower potential energy so that

!U < 0

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Concept Question: Motion of Charged Objects

Two oppositely charged are released from rest in an electric field.

1.  Both charged objects will move from lower to higher electric potential.

2.  Both charged objects will move from higher to lower electric potential.

3.  The positively charged object will move from higher to lower electric potential; the negatively charged object will move from lower to higher electric potential.

4.  The positively charged object will move from lower to higher electric potential; the negatively charged object will move from higher to lower electric potential.

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Concept Q. Ans.: Motion of Charged Objects

Two oppositely charged are released from rest in an electric field.

3. The positively charged object will move from higher to lower electric potential; the negatively charged object will move from lower to higher electric potential.

For the positively charged object: For the negatively charged object:

!V < 0 " !U = q!V < 0

!V > 0 " !U = q!V < 0

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Concept Question: Two Point Charges

The work done in moving a positively charged object that starts from rest at infinity and ends at rest at the point P midway between two charges of magnitude +Q and –Q 1.  is positive. 2.  is negative. 3.  is zero. 4.  can not be determined – not enough info is given.

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Concept Question Answer: Two Point Charges

The potential at ∞ is zero. The potential at P is zero because equal and opposite potentials are superimposed from the two point charges (remember: V is a scalar, not a vector)

3. Work from ∞ to P is zero.