51 ch21b coulombslaw electric field

8/12/2019 51 Ch21B CoulombsLaw Electric Field

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Chapter 21 Electric Field andCoulombs Law (again)

Electric fields and forces (sec. 21.4)

Electric field calculations (sec. 21.5)

Vector addition (quick review)

C 2012 J. Becker


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Coulombs Law

Coulombs Law lets us calculate the FORCEbetween two ELECTRIC CHARGES.


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Coulombs LawCoulombs Law lets us calculate the force

between MANY charges. We calculatethe forces one at a time and ADD themAS VECTORS.

(This is called superposition.)

THE FORCE ON q3CAUSED BY q1AND q2.


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Figure 21.14

SYMMETRY!


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Fig. 21.15 A charged bodycreates an electric field.

Coulomb force of repulsion

between two charged bodies atA and B, (having charges Q and qorespectively) has magnitude:

F = k |Q qo|/r2= qo[ k Q/r2 ]

where we have factored outthe small charge qo.

We can write the force interms of an electric field E:

Therefore we can write forF= qoE

the electric field

E = [ k Q / r2

]


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Calculate E1, E2, and ETOTALat point C:

q= 12 nC

See Fig. 21.23:Electric fieldatC set up by charges q1and q1

(an electric dipole)

At C E1= 6.4 (10)3N/C

E2= 6.4 (10)3N/CET= 4.9 (10)

3N/Cin the +x-direction

A

C

SeeLab #2

Need TABLE of ALLvector component

VALUES.

E1

E2

ET


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Fig. 21.24 Consider symmetry! Ey= 0

Xo

dq

o

dEx= dE cos =[k dq/(xo2

+a2)

][xo/(xo2

+ a2

)1/2

]Ex= k xodq/[xo2+ a2]3/2where xo and a stay

constant as we add all the dqs (dq = Q)in the integration: Ex= k xoQ/[xo2+a2]3/2

|dE| = k dq/ r2

cos = xo/ r

dEx = Ex


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Fig. 21.25 Electric field at Pcaused by a line

of charge uniformly distributed along y-axis.

Consider symmetry! Ey= 0

Xo

y

|dE| = k dq/ r2

dq


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|dE| = k dq/ r2 and r = (xo2+ y2)1/2cos = xo/ r and cos = dEx/ dE

dEx= dE cos

Ex= dEx= dE cosEx= [k dq/r2] [xo/ r]Ex= [k dq/(xo2+y2)] [xo/(xo2+ y2)1/2]

Linear charge density = ll= charge / length = Q / 2a = dq / dydq = ldy


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Ex= [k dq/(xo2+y2)] [xo/(xo2+ y2)1/2]Ex= [k ldy/(xo2+y2)] [xo/(xo2+ y2)1/2]Ex= k l xo[dy/(xo2+y2)] [1 /(xo2+ y2)1/2]

Ex= k l xo[dy/(xo2+y2) 3/2]Tabulated integral: (Integration variable z)

dz / (c2+z2) 3/2= z / c2 (c2+z2) 1/2dy / (c2+y2) 3/2= y / c2 (c2+y2) 1/2dy / (Xo2+y2) 3/2= y / Xo2 (Xo2+y2) 1/2


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Ex= k l xo -a [dy/(xo2+y2) 3/2]Ex= k(Q/2a)

Xo [y /Xo2 (Xo2+y2) 1/2] -aa

Ex= k (Q /2a)Xo [(a (-a)) / Xo2 (Xo2+a2) 1/2]

Ex= k (Q /2a)Xo [2a / Xo2 (Xo2+a2) 1/2]

Ex= k (Q / Xo)[1 / (Xo2+a2) 1/2]

a


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Fig. 21.47 Calculate the electric fieldat the

protoncausedby the distributed charge +Q.

Tabulated integral:dz / (c-z) 2= 1 / (c-z)

b

lis uniform (= constant)+Q

l l


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Fig. 21.48 Calculate the electric fieldat -qcaused by +Q, and thenthe forceon q: F=qE

Tabulated integrals:dz / (z2+ a2)3/2 = z / a2(z2+ a2) for calculation of Ex

zdz / (z2+ a2)3/2= -1 / (z2+ a2) for calculation of Ey

lis uniform (= constant)


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Fig. 21.32 Net forceon an ELECTRIC DIPOLE

is zero, but torque( )is into the page.

= r x F

= p x E

ELECTRIC DIPOLEMOMENT is

p = qd


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seewww.physics.sjsu.edu/Becker/physics51

Review

V i i h


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Vectors are quantities thathave both magnitudeanddirection.

An example of a vectorquantity is velocity. Avelocity has both magnitude(speed) and direction, say60 miles per hour in aDIRECTION due west.

(A scalar quantity isdifferent; it has onlymagnitude mass, time,tem erature, etc.)


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Note: The dotproduct of two

vectors is a scalarquantity.

cosx x y y z z

A B AB A B A B A B

The scalar (or dot) productof two

vectors is defined as


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sinA B AB

The vector (or cross) productof

two vectors is a vector where thedirection of the vector product isgiven by the right-hand rule.

The MAGNITUDE of the vectorproduct is given by:


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PROFESSIONAL FORMAT

51 ch21b coulombslaw electric field

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