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Unformatted text preview: Electric Fields  why? Problem:  If want to assign a unique property in space,
force depends on charge: — Thus force is not the right quantity
— Show with different forces in same point  Need to define an Electric Field as this unique property. Electric Field — Definition, cont Calculate:
. Electric field = force (on qn)l unit charge 1:341qu Electric Field — Definition Required Properties:  Charge > electric ﬁeld around it.
— charged object is the source particle  Electric field assigns force to charged object
in space. What about the test charge?  E is the field produced by the source, not the
test charge  E is a propeﬂy of the source, not the test
charge  E exists regardless of the test charge
 Test charge is only a detector of E Electric Field — Test Particle How to find the Electric Field:  Lets have a test positive particle, qu  The test particle is only a probe to:
— (i) find and
— (ii) to get the strength of the electric field  The test charge is assumed to be small enough not
to disturb the charge distribution responsible for the
field Summary, cont. ' Fe=ql§
— This is valid for a point charge only
— One of zero size — For larger objects, the field may vary over the size of
the object  If q is positive, I?" and T3 are in the same direction
 If q is negative,13 and E‘ are in opposite directions Electric Field, Vector Form  Coulomb's Law, point charges:
15,:k, ‘13"; r
 Then, the electric field will be E 2 a f:
E=—“=k if
qa r Superposition Example Components: Ex=kelzcose+ke%c059=2k,%c059
r r r _ q~ ‘1 _
Ey—ke7sm9—ke7sm9—0 geometry... a a
c059=—= 2 2
r
a +y
2a
u. EXZke 2 33,2 Note:fory>>a
(‘1 +y ) 2aq
Ewk, 3
3’ Electric dipole decays faster than a point charge! 10 Superposition with
Electric Fields  At any point (x,y,z) the total electric field
due to a group of source charges equals
the vector sum of electric fields at that point due to all the particles  Method to calculate E for a collection of
point charges. Continuous Charge Distribution  If the distances between charges in a group of
charges is much smaller than the distance
between the group and a point of interest, the
system of charges may be modeled as continuous 11 . E=E,+E, Superposition Example Find the electric field Eldue to ql, Find the electric field T32 due to
C12: — Reminder: fields add as vectors — The direction of the individual
fields is the direction of the force on a positive test charge BTW, this is an electric dipole... n it Continuous Charge Distribution —
recipe for field Procedure: — Dlvlde the charge distribution
into small elements, each of
which contains Aq — calculate the electric field due
to m of these elements at
point P — calculate the totalﬁeld by /
summing the contributions of all [I
the charge elements /‘ Continuous Charge Distribution —
equations for field  For the individual charge elements _. A ‘
411494?i
rt  Because the charge distribution is
continuous E: lim kez A—gikaef d—qf
ri 2
Alb—)0 i r 13 Uniform Ring of Charge Continuous charge problem instead of a bunch of discrete charges. Steps: Discretize the charge and use the eq. for the field Use any available symmetry Do integral instead of sum — still superposition principle 16 Charge Densities  Volume charge density
distributed evenly throughout a volume
 p = Q / V
 Surface charge density — when a charge is
distributed evenly over a surface area
 a = Q IA
 Linear charge density — when a charge is
distributed along a line —x=Q/ —when a charge is Uniform Ring of Charge 1. Discretize: dE=ke 2. Use available symmetry: (1E, = k and geometry c059=ke c059=£ —3x 3. Integrate d x x
E=j ke—gx=ke—3_I dq=ke—3Q
r r r More geometry: r=\laz+x2 E:k Uniform Ring of Charge Ring of radius 'a' has total 'Q' spread out uniformly
throughout the ring. How much is the electric ﬁeld at 'P', a distance 'x' from the
center?
Uniform Ring of Charge x
Eike 2 2 3/2Q
(0 +x )
Useful limits. How much isEat: N _
nxzo 3)E~ke 3x
2)x>>a a
3) X“ a Harmonic force. e.g. on q
1)x=o,E=o _ Q
Q F——keq—3x
a
2) ENke—z 3: Just like a point
charge
18 Problem Solving Strategy  Conceptualize
— Imagine the type of electric field that would be
created by the charges or charge distribution
 Categorize — Analyzing a group of individual charges or a
continuous charge distribution? — Think about symmetry 19 Electric Field Lines  Field lines good to visualize the electric field
 The electric field vector T3 is tangent to these
lines  The number of lines per unit area through a
surface perpendicular to the lines is
proportional to the magnitude of the electric
field in that region 22 Problem Solving Hints, cont  Analyze
— Group of Individual charges: use the superposition
principle
 The resultant ﬁeld is the vector sum of the individual ﬁelds
— Continuous charge distributions: the vector sums for
evaluating the total electric field at some point must be
replaced with vector integrals  Divide the charge distribution into infinitesimal pieces,
calculate the vector sum by integrating over the entire charge
distribution 20 Electric Field Lines, General  density of lines through surface A > through surface B ' IEA>EB \\ v/  In this example the lines at ‘1'\<. different locations point in i _ I: r ‘I differentdirections 7 — Al“ \x 7 ’ _
— This indicates the ﬁeld is non fl  ' _ " uniform _ 1 ‘ j , \,\ ,
\\ \\
A\\ B 23 Problem Solving Hints, final  Analyze, cont
— Symmetry: take advantage of any symmetry in
the system
 Finalize
— Check to see if your field is consistent with the
mental representation and reflects any
symmetry
— Imagine varying parameters to see if the result
changes in a reasonable way 21 Electric Field Lines,
Positive Point Charge  The field lines radiate outward in all directions
— In three dimensions, the
distribution is spherical
— A positive test charge would be
repelled away from the positive
source charge (21) Electric Field Lines, _ _ _ _ Electric Field Lines —
Negative Point Charge Elecmc He'd L'neS — DIPOIG Like Charges  The field lines radiate  Equal but opposite
inward in all directions V charges — A positive test charge would y _
be attracted toward the r " ' The number 0f flaid negative source charge lines leaving the
positive charge equals
the number of lines terminating on the “v “ negative charge  equal and positive charges  The same number of lines
leave each charge since they
are equal in magnitude  At a great distance, the ﬁeld
is approximately equal to that
of a single charge of 2q  Limits of small and big __ distance it.)
M ‘ r N (b) V ’
25 26 27
Electric Field Lines, Electric Field Lines —
Unequal Charges Rules for Drawing
Motion of Charged Particles
 The positive charge is tyvice the a  Lines start on positive and end on negative  Charged particle in an electric field —> electrical
magnitude of the negative ' y/ charge force
charge ,/ g . Two lines leave the positive  in the case Of an EXCEISSIOF one type 0' CharQE. some  Force will cause the particle to accelerate
charge for each line that x.,\\ lines wrll begin or end Infinitely far away according to Newton's Second Law
tehrminate on the negative  The number of lines is proportional to the , 1?" =qE=ma
c arge . . magnitude of the charge _ e _ _
 At a great distance, the field  If E Is uniform, then a Is constant would be approximatelythe  Field lines cannot intersect
same as that due to asingle charge of +q 28 29 3) Uniform field example Electron in a Motion of Particles. cont Uniform Field, Example
Initially at rest particle: , ,  Posltlve char e —> acceleration is along the  The electron Is prolected direction of thg field 1'C°"Sta"t f°'°e . horizontally into a I h I _ _ I h 2.Accelerates to the right uniform electric ﬁeld ( ' 349% gtﬁrgf‘? I: acce eratlon '5 mt e 3.Can use usual equations of  The electron undergoes a "wee '0'" of e_ 'e t t th k_ f motion under constant force to —d°"f’r:war:d away?!“ , ’ ' ’ ’ ’ ’ ' ’ ’ ' “
 acce era Ion Is cons an , e Inema Ic solveforﬁnalveloci  ec argels neg Ive,5_o ‘.—7,’m'  . l L equations can be used I _ ty ﬁgﬁf'era‘m" '5 °PP°5"e ‘ .\ '  Describe its motion... ‘ 7 i++r++++n+h\ i
y Fy=may=qu=—eﬁ
ay=—eEy/me
Fx=0 EX. char E between arallel IBIES .
g p p 32 ax =0 :constant speed nwtwn 33 31 Cathode Ray Tube (CRT)  ex.: oscilloscopes, radar systems, televisions anode  The electrons are m “:33” ' ' ' ' “ ml / (lento. Hm rlrllet Hm gttaitziatizal°.2ie:y“"° 375 i p C I rr/ // A ‘\ why!)
4; [I
/ b
Elixir" ~ Flumed mi
st‘l‘i‘t‘n 34 ...
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This note was uploaded on 11/11/2011 for the course ENGR 231 taught by Professor Olehtretiak during the Fall '10 term at Drexel.
 Fall '10
 OlehTretiak

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