3. r
fa
Q1 Q2
â
2
2
a
r
qq1
F 04
1
πε
= 2
2
b
r
qq1
F 04
1
πε
=
ε 0
12
2
28 85 10= × −
.
c
m N
1221 - FF
=
229
/.109 cmNk ×=
04
1
πε
=
e
k
Ce
19
106.1
−
×−=
ABOUT COLOUM
Cp
19
106.1
−
×= kgnmass
kgemass
kgpmass
27
31
27
106.1
101.9
106.1
−
−
−
×=→
×=→
×=→
0=n
Neq =
eC
28
1061 ×=
4. K = electrical constant (9.0 x 109
(
F = force of charge
q = amt. of charge on object
d = distance between objects
Answer is (-) when force is attractive
Answer is (+) when force is repulsive
1 2
2
12
q q
F k
12 r
=
Force is a vector quantity
The field strength at any
point in this field is:
E = field strength (Vm-1
)
V = potential difference
(V)
d = plate separation (m)
dVE /=
5.
6.
7. Coulombs Law: 2 Charges
A positive charge of 6.0 x 10 -6
C is 0.030m from a second positive
charge of 3.0 x 10 -6
C. Calculate the force between the charges.
•Fe = k q1 q2
r2
= (8.99 x 109
N m2
/C2
) (6.0 x 10 -6
C) (3.0 x 10 -6
C)
( 0.030m )2
= (8.99 x 109
N m2
/C2
) (18.0 x 10 -12
C)
(9.0 x 10 -4
m2
)
= + 1.8 x 10 -8
N
8. Example I–1. An alpha particle (charge +2.0e) is
sent at high
speed toward a gold nucleus (charge +79e). What is the
electrical
force acting on the alpha particle when it is 2.0 ×10−14 m from
the
gold nucleus?
NE
krqKqE
91
10.2/10.6.1.79.2/21
14192
=
==
−− An electron with a speed of 3.00
× 106 m/s
moves into a uniform electric
field of 1000 N/C. The field is
parallel
to the electron’s motion. How far
does the electron travel before it
is brought to rest?
214
3119
/10*76.1
10*11.9/1000.10*6.1
/
sm
a
meEmfa
e
−−
=
−==
12. ( )
∫
+
2/322
xz
dx
( )
dx
xz
z
E
Lx
x
∫
=
= +
=
0
2/3220
2
4
1 λ
πε
How to evaluate this integral
Let θtanzx = then ( ) θθ 222222
sec1tan zzzx =+=+ and
θθdzdx 2
sec= Substitute these into the integral:
( ) ( )
c
z
d
z
d
zz
dz
xz
dx
+====
+
∫∫∫∫ θθθθ
θθ
θθ
sin
1
cos
1
sec
11
sec
sec
2
2
22/322
2
2/322
From diagram, 22
sin
zx
x
+
=θ
( ) 2222/322
1
zx
x
zxz
dx
+
=
+
∫
Hence
Lq λ2=
The line “looks” like a point charge
,so the field reduces to that of a
( )2
04/ zq πε
point charge
17. بن هه شحن ك له كه ته دحالتي
∑=
=+⋅⋅⋅++=
n
1i
in21 FFFFF
∑=
=+⋅⋅⋅++=
n
i
in FFFFF
1 000
2
0
1
0 qqqqq
02 i
i
i
n
1i 0
n
1i
in21 r
r
q
4π
1
EEEEE
rr
∑∑ ==
==+⋅⋅⋅++=
ε
102
1
1
0
1
1 r
r
q
4π
1
q
F
E
0
r
ε
==
i02
i
i
0
i
i r
r
q
4π
1
q
F
E
0
ε
==
202
2
2
0
2
2 r
r
q
4π
1
q
F
E
0
ε
==
qn
qi
q3q2
q1 r2
ri
rn
r1
r3
F
r
nF
r
iF
r
3F
r
2F
r
1F
r
q0
18. Draw and label forces (only those on Q3(.
Draw components of forces which are not along axes.
x
y
Q2=+50µC
Q3=+65µC
Q1=-86µC
52cm
60
cm
30cm
θ=30º
F31
F32Draw a representative
sketch.
Draw and label relevant
quantities.
Draw axes, showing
origin and directions.
Step 1: Diagram
19. 1 2
2
12
q q
F k
12 r
=
“Do I have to put in the absolute value signs?”
x
y
Q2=+50µC
Q3=+65µC
Q1=-86µC
52cm
60
cm
30cm
θ=30º
F31
F32
Step 2: Starting Equation
20. 3 2
2
32
Q Q
F k ,
32 r
repulsive
=
r
3 2
2
32
Q Q
F k
32,y r
=
F 0
32,x
= (from diagram(
F32,y = 330 N and F32,x = 0 N.
x
y
Q2=+50µC
Q3=+65µC
Q1=-86µC
52cm
r31 =60 cm
r32=30cm
θ=30º
F31
F32
Step 3: Replace Generic Quantities by
Specifics
21. 3 1
2
31
Q Q
F k ,
31 r
attractive
=
r
3 1
2
31
Q Q
F k cos
31,x r
= + θ
You would get F31,x = +120 N and F31,y = -70 N.
(-sign comes from diagram(
3 1
2
31
Q Q
F k sin
31, y r
= − θ
(+sign comes from
diagram(
x
y
Q2=+50µC
Q3=+65µC
Q1=-86µC
θ=30º
F31
F32
Step 3 (continued(
r32=30cm
r31 =60 cm
52cm
22. F3x = F31,x + F32,x = 120 N + 0 N = 120 N
F3y = F31,y + F32,y = -70 N + 330 N = 260 N
You know how to calculate the magnitude F3 and the angle
between F3 and the x-axis.
F3
The net force is the
vector sum of all the
forces on Q3.
x
y
Q2=+50µC
Q3=+65µC
Q1=-86µC
52cm
60
cm
30cm
θ=30º
F31
F32
Step 3: Complete the Math
23. Faraday, beginning in the 1830's, was the leader in developing
the idea of the electric field. Here's the idea:
• A charged particle emanates a "field"
into all space.
• Another charged particle senses the field,
and “knows” that the first one is there.
+
+
-
like
charges
repel
unlike
charges
attract
F12
F21
F31
F13
24. We define the electric field by the force it exerts on a test
charge q0:
0
0
F
E =
q
r
r
This is your second starting equation. By convention the direction of the electric field
is the direction of the force exerted on a POSITIVE test charge. The absence of
absolute value signs around q0 means you must include the sign of q0 in your work.
If the test charge is "too big" it perturbs the electric field, so the
“correct” definition is
0
0
q 0
0
F
E = lim
q→
r
r
Any time you know the electric field, you can use this equation to calculate the force
on a charged particle in that electric field.
You won’t be required to use
this version of the equation.
F = qE
r r
25. If charge is distributed along a straight line segment parallel to
the x-axis, the amount of charge dq on a segment of length dx
is λdx.
λ is the linear density of charge (amount of charge per unit
length). λ may be a function of position.
Think λ ⇔ ⇔ length. λ times the length of line segment is the
total charge on the line segment.
l
x
dx
λ λdx
26. The electric field at point P due to the charge dq is
x
dq
P
2 2
0 0
1 dq 1 dx
dE = r' = r'
4πε r' 4πε r'
λr
$ $
r’
r'$
dE
I’m assuming positively charged objects
in these “distribution of charges” slides.
I would start a homework or test problem with this:
2
dq
dE = k
r
27. The electric field at P due to the entire line of charge is
2
0
1λ(x) dx
E = r' .
4πε r'∫
r
$
The integration is carried out over the entire length of the line, which need
not be straight. Also, λ could be a function of position, and can be taken
outside the integral only if the charge distribution is uniform.
x
dq
P
r’
r'$
E
28. If charge is distributed over a two-dimensional surface, the
amount of charge dq on an infinitesimal piece of the surface is σ
dS, where σ is the surface density of charge (amount of charge
per unit area).
x
y
area = dS
σ
charge dq = σdS
29. dE
The electric field at P due to the charge dq is
2 2
0 0
1 dq 1 dS
dE = r' = r'
4πε r' 4πε r'
σr
$ $
x
y
P
r’
r'$
30. The net electric field at P due to the entire surface of charge is
x
y
P
r’
r'$
2
0 S
1 (x, y) dS
E = r'
4πε r'
σ
∫
r
$
E
31. x
z
P
r’
r'$
2
0 V
1 (x, y,z) dV
E = r' .
4πε r'
ρ
∫
r
$
After you have seen the above, I hope you believe that the net
electric field at P due to a three-dimensional distribution of
charge is…
y
E
32.
33. Calculate E1, E2, and ETOTAL
at point “C”:
q = 12 nC
See Fig. 21.23: Electric field at
“C” set up by charges q1 and q1
(an electric dipole)
At “C” E1= 6.4 (10)3
N/C
E2 = 6.4 (10)3
N/C
EC = 4.9 (10)3
N/Cin
the +x-direction
A
C
Need TABLE of ALL vector
component VALUES.
34. Summarizing:
2
0
1λ dx
E = r' .
4πε r'∫
r
$
2
0 S
1 dS
E = r' .
4πε r'
σ
∫
r
$
2
0 V
1 dV
E = r' .
4πε r'
ρ
∫
r
$
Charge distributed along a line:
Charge distributed over a surface:
Charge distributed inside a volume:
If the charge distribution is uniform, then λ, σ, and ρ can be taken outside the
integrals.
35. The Electric Field
Due to a Continuous Charge
Distribution
(worked examples)
Gauss law
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ELECTRICITYAND MAGNETISM
P10D
Coulomb’s Law
The force of attraction or repulsion between two point charges q1 and q2 is
directly proportional to the product of their charges and inversely proportional
to the square of the distance between them.
where F12 is the force exerted on point charge q1 by point charge q2
when they are separated by a distance r12.
The unit vector is directed from q2 to q1 along the line between the two
charges. The constant is called the permitivity of free space. In SI
units where force is in Newton's (N), distance in meters (m) and
charge in coulombs (C),
Electricty all
36. Physics 2102
Final Exam Review
Physics 2102
Jonathan Dowling
Microsoft Office
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2.2 Applications of Gauss' Law
2.2.1 Electric Field Due to a Line Charge - Cylindrical Symmetry
Let's find the electric field due to a line charge. As we have done this before, much of
the setup of the problem is already done Let's make things a bit tougher by
considering the field due to an infinitely long line of charge as opposed to the one of
finite length which we did before. It's clear here that it's impossible for us to talk
about a finite amount of charge stretched over an infinitely long distance. Instead, we
state that the line has a constant linear charge density, λ.
Realistically, all line charges are finite, but, first, we have done the finite length
problem explicitly, and second, a line of charge which is quite long compared to the
distance from it at which we would like to know the E field is not an uncommon
problem. We can deal with the approximate answer as easily as with the true solution
and compare the differences. Physicists will often engage in thoughts of this kind in
theoretical research as it is important to know where our ideas ``break down''. Those
ideas which can hold up under the most extreme extrapolations without delivering
clearly nonsensical results indicate something deep about our understanding.
The above being said, we still need to do the calculation. Consider the figure below
which shows a view of the line charge and a point P a distance h away from it. We
wish to find the electric field at point P. To set up the integral, we do, as before, the
trick of taking infinitesimally small line segments of charge in pairs so that their
horizontal components cancel and the vertical
(i.e. radial) components add.
Figure 2.4: Calculation of the electric field at the midpoint of a line charge of length l.
Hence we only need to change the definition of dq, the charge on an infinitesimal
segment with length dx, otherwise our approach for the finite line charge is
unchanged for the infinite length case. Now, dq = λdx, so, as with the finite length
charge, we use the angle, θwith respect to the vertical to identify the radial
component, r, as the distance from the infinitesimal charge to point P, i.e.
r = ______ (2.2.1.7)
37. Example: A rod of length L has a uniform charge per unit length
λ and a total charge Q. Calculate the electric field at a point P
along the axis of the rod at a distance d from one end.
Q
= and Q = L
L
λ λ
P x
y
d L
dE x dx dQ = λ dx
2 2
dq dx
dE = k k
x x
λ
=
Note: dE is in the –x direction. dE is the
magnitude of dE. I’ve used the fact that Q>0
(so dq=0) to eliminate the absolute value signs
in the starting equation.
d L
d+L d+L d+L
x 2 2d d d
d
dx dx 1ˆ ˆ ˆE = dE = -k i = -k i = -k i
x x x
+
λ
λ λ − ÷
∫ ∫ ∫
r r
( ) ( ) ( )
1 1 d d L L kQˆ ˆ ˆ ˆE = -k i = -k i= -k i= - i
d L d d d L d d L d d L
− + + λ
λ − + λ ÷ ÷ ÷+ + + +
r
38. L
P
a
0
a
La +
x
dxλdQ =
x
Q
V
04
d
d
πε
=
dxλ
∫∫
+
==
La
a x
x
dVV
04
d
πε
λ
La
a
x
+
= ln
4 0πε
λ
( )[ ]aLa lnln
4 0
−+=
πε
λ
a
La +
ln
x
Example: A rod of length L has a uniform charge per unit length λ
and a total charge Q. Calculate the electric potential at a point P
along the axis of the rod at a distance d from one end.
39. αcosdEEd
X
=
2
dQ
dE=k
r
Ex = ∫ dEx = ∫ dE cos a
Ex = ∫ [k dq /r2
] [xo / r[
Ex = ∫ [k dq /(xo2
+y2
)] [xo /(xo2
+ y2
)1/2
[
dq = λ dy
Ex = ∫ [k λ dy /(xo2
+y2
)] [xo /(xo2
+ y2
)1/2
[[
Ex = k λ xo ∫ [dy /(xo2
+y2
)] [1 /(xo2
+ y2
)1/2
[
Ex = k λ xo ∫ [dy /(xo2
+y2
) 3/2
[
Example: A rod of length y has a uniform charge per unit length λ
and a total charge Q. Calculate the electric field at a point P along
the axis of the rod length from (a to –a ).
Consider symmetry! Ey =
0cos a = dEx / dE
r = (xo2
+ y2
)1/2
40. Tabulated integral: (Integration variable “z”)
∫ dz / (c2
+z2
) 3/2
= z / c2
(c2
+z2
) 1/2
∫ dy / (c2
+y2
) 3/2
= y / c2
(c2
+y2
) 1/2
∫ dy / (Xo2
+y2
) 3/2
= y / Xo2
(Xo2
+y2
) 1/2
Ex = k λ xo ∫ -a
a
[dy /(xo2
+y2
) 3/2
[
Ex = k(Q/2a) Xo [y /Xo2
(Xo2
+y2
) 1/2
] -a
a
aQ 2/=λ
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 / (Xo2
+a2
) 3/2
Ex = k (Q / Xo) [1 / (Xo2
+a2
) 1/2
[Answer 1
Answer 2
42. Fig. 21.48 Calculate the electric field at -q
caused by +Q, and then the force on -q.
Tabulated integral:
∫ dz / (z2
+ a2
)3/2
= z / a2
(z2
+ a2
) 1/2
∫ z dz / (z2
+ a2
)3/2
= -1 / (z2
+ a2
) 1/2
Tabulated integral: ∫ dz / (c-z) 2
= 1 / (c-z)
Fig. 21.47 Calculate the
electric field at +q
caused by the
distributed charge
+Q.
43. radius R, which carries a uniform line charge λ.
R
q
π
λ
2
=
θ
λ
πε
cos
4
1
2
0
∫=
r
dl
Ez
r
z
zRr =+= θcos;222
( )
∫
+
= dl
Rz
z
Ez 2/32204
1 λ
πε
Rdl π2=∫
( )
( ) 2/3220
2
4
1
Rz
zR
Ez
+
=
πλ
πε
44. Find the electric field a distance z above the center of a
flat
circular disk of radius a, which carries a uniform surface
density, segma.
(Models an electrostatic microphone)
A typical element is a ring of radius r and thickness dr, which
has
an area.
rdrdA π2= ( )2
/ aq πσ =
22
2
2
a
qrdr
rdr
a
q
dAdq === π
π
σ
( ) ( ) 2/3222
0
2/3220
2
4
1
4
1
rz
rdr
a
qz
dq
rz
z
dE
+
=
+
=
πεπε
( )
dr
rz
r
a
qz
E
a
∫
+
=
0
2/3222
0
2
4
1
πε
( )
dr
rz
r
a
qz
E
a
∫
+
=
0
2/3222
0
2
4
1
πε
45. Let
22
rzu += then du = 2r dr, and
( )
−
+
−=
−
==
+
+−+=
=
∫∫
zaz
/
u
u
du
dr
az
r
az
z
azu
zu
a
11
2
21
2
22
2/1
2/3
0
2/322
22
2
22
2
+
−=
222
0
1
4
2
az
z
a
q
E
πεS
o
When z << a,
2
04
2
a
q
E
πε
≅
When z << a,
2
04
1
z
q
E
πε
=
46. Example: A ring of radius a has a uniform charge per unit length
and a total positive charge Q. Calculate the electric field at a point
P along the axis of the ring at a distance x0 from its center.
By symmetry, the y- and z-components
of E are zero, and all points on the ring
are a distance r from point P.
P
a
dQ
r
dE
xθ
θx0
2
dQ
dE=k
r
x 2
dQ
dE =k cos
r
θ
No absolute value
signs because Q is
positive.
2 2
0r = x a+
0x
cos
r
θ =
( )
0 0 0 0
x x 3/ 22 3 3 2 2
ring ring ring 0
x x x kx QdQ
E dE k k dQ k Q
r r r r x a
= = = = = ÷
+
∫ ∫ ∫
Or, in general, on the ring axis
( )
x,ring 3/ 22 2
kxQ
E .
x a
=
+
For a given x0, r is a constant
for points on the ring.
47. R r
x
0
P
∫=
Q r
dQ
V
04πε∫= Q
r
d
4
1
0πε
r
Q
04πε
=
22
xRr +=
22
04 xR
Q
V
+
=
πε
r
Q
V
04
d
d
πε
=
Qd
x
A Example: Find the an expression electric potential due to a
uniformly charged ring of radius R and total charge Q at a point P
on the axis of the ring.
.)/(2.)(2/1
).(/
2/3222/322
2/122
−−
−
+=+−−
+=−=
rxkqxxrxkq
rxkqdxdvE
X
48. Example: A disc of radius R has a uniform charge per unit area σ.
Calculate the electric field at a point P along the central axis of the
disc at a distance x0 from its center.
P
r
dQ
x
x0R
The disc is made of concentric rings. The area of a ring at a
radius r is 2πrdr, and the charge on each ring is σ(2πrdr).
We can use the equation on the previous
slide for the electric field due to a ring,
replace a by r, and integrate from r=0 to
r=R.
( )
0
ring 3/22 2
0
kx 2 rdr
dE .
x r
σ π
=
+
Caution! I’ve switched
the “meaning” of r!
( ) ( )
R
0
x x 03/ 2 3/ 202 2 2 2
disc disc 0 0
kx 2 rdr 2r dr
E dE kx
x r x r
σ π
= = = πσ
+ +
∫ ∫ ∫
( )
( )
R1/ 22 2
0 0 0
x 0 1/ 22 2
0 0
0
x r x x
E kx 2k
1/ 2 x x R
−
+
÷ = πσ = πσ −
÷ − +
2 2
0r = x a+
0x
cos
r
θ=
2
dQ
dE=k
r
θ
49. Electric Potential due to non-uniformly charged disk
A disk of radius R has a non-uniform surface charge density
σ=Cr where C is constant and r is distance from the center of
the disk as shown. Find the potential at P.
22
xr
dqk
dV e
+
=
dq = σdA = Cr(2πrdr) and
∫ +
=
R
e
xr
drr
kCV
0
22
2
)2( π
V= C(2πk) {R(R2
+x2
)1/2
+x2
ln(x/[R+ (R2
+x2
)1/2
[){
Standard
integral
بدريخه فيلد الكتريك
50. Electric Potential due to uniformly charged annulus
Calculate the electric potential at point P on the axis of an
annulus, which has uniform charge density σ.
Pictures from Serway & Beichner
drrdAdq
xrKdqdE
πσσ 2where
)(/
2/322
==
+=
b
a
b
a
e
xrxxrk
r
drr
xkE
+
== ∫
2/122
3
)(/.2
2
σπ
σπ
E= 2πσk [ b/(x2
+b2
)1/2
- a/(x2
+a2
)1/2
[
51. Example: Calculate the electric field at a distance x0 from an
infinite plane sheet with a uniform charge density σ.
Treat the infinite sheet as disc of infinite radius.
Let R→∞ and use to get
0
1
k
4
=
πε
sheet
0
E .
2
σ
=
ε
Interesting...does not depend on distance from the sheet.
I’ve been Really Nice and put this on your starting equation sheet. You don’t
have to derive it for your homework!
S∆
S
d E
n
n
Using gauss
law
0
0
/2
/.
εσ
ε
AEA
qdAE
enc
=
=∫
52. Electric Field Lines
Electric field lines help us visualize the electric field and predict
how charged particles would respond the field.
Example: electric field lines for isolated +2e and -e charges.
+-
54. Gauss’ Law
Electric Flux
We have used electric field lines to visualize electric fields and
indicate their strength.
We are now going to count the
number of electric field lines passing
through a surface, and use this
count to determine the electric field.
E
55. The electric flux passing through a surface is the number of
electric field lines that pass through it.
Because electric field lines are drawn
arbitrarily, we quantify electric flux
like this: ΦE=EA, except that…
If the surface is tilted, fewer lines cut
the surface.
E
A
Later we’ll learn about magnetic flux, which is
why I will use the subscript E on electric flux.
E
θ
56. If the electric field is not uniform, or the surface is not flat…
divide the surface into
infinitesimal surface
elements and add the
flux through each…
dA
E i
E i i
A 0
i
lim E A
∆ →
Φ = ×∆∑
E E dAΦ = ×∫
∆A
57. Electric Flux
What is the electric flux of
this cylinder?
0
0)1()1(
90cos180cos0cos
constant,,
21
321
21321
=Φ
+−+=Φ
++=Φ
==Φ+Φ+Φ=Φ=Φ ∑
E
E
E
E
EAEA
EAEAEA
AAE
What does this tell us?
his tells us that there are NO sources or sinks INSIDE the cylindrical
58. If the surface is closed (completely encloses a volume)…
E
…we count* lines going
out as positive and lines
going in as negative…
E E dAΦ = ×∫
Ñ
dA
a surface integral, therefore a
double integral ∫∫
59. Question: you gave me five different equations for electric flux.
Which one do I need to use?
E E dAΦ = ×∫
Ñ
E E dAΦ = ×∫
E E AΦ = ×
E EAcosΦ = θ
E EAΦ =
Answer: use the simplest (easiest!) one that works.
Flat surface, E || A, E constant over surface. Easy!
Flat surface, E not || A, E constant over surface.
Flat surface, E not || A, E constant over surface.
Surface not flat, E not uniform. Avoid, if possible.
Closed surface. Most general. Most complex.
If the surface is closed, you may be able to “break it up” into
simple segments and still use ΦE=E·A for each segment.
60. Mathematically*, we express the idea two slides back as
enclosed
E
o
q
E dAΦ = × =
ε∫
Ñ Gauss’ Law
We will find that Gauss law gives a simple way to calculate
electric fields for charge distributions that exhibit a high degree
of symmetry…
…and save more complex and realistic charge distributions for
advanced classes.
*“Mathematics is the Queen of the Sciences.”—Karl Gauss
61. Gauss’ Law
∫ =•
o
encq
dAE
ε
The electric flux (flow) is in
direct proportion to the charge
that is enclosed within some
type of surface, which we call
Gaussian
The vacuum permittivity constant is
the constant of proportionality in this
case as the flow can be interrupted
should some type of material come
between the flux and the surface
area. Gauss’ Law then is derived
mathematically using 2 known
expressions for flux.
62. r < R
Therefore, for r < R
Case II: r > R
Therefore, for r > R,
R r
0
0>r < R
rkE
lrlE
/2
/)2(
λ
ελπ
=
=
L
66. Question
Use Gauss's Law to find the electric field everywhere due to a
uniformly charged insulator shell, like the one shown below.
The shell has a total charge Q, which is uniformly distributed
throughout its volume.
(c) Use Gauss's Law to find the electric field for radius r < a.
(d) Use Gauss's Law to find the electric field for radius a < r < b.
(e) Use Gauss's Law to find the electric field for radius r > b.
(a) What is the charge on the inner surface of the conductor?
(b) What is the charge on the outer surface of the conductor?
We need to look at this problem in three parts: one, for when
Answer
qenc = 0
r
Insulator shell-sph
67. Here, qenc
is Q, so we have
Again, since the electric field will be constant at every point on
the spherical Gaussian surface, we have
For the case where r > R:
We construct a spherical Gaussian
surface through an outside point r > R,
and apply Gauss's Law. In this case,
R
68. a
Q
Find the electric field at a point inside the sphere.
Now we select a spherical Gaussian surface
with radius r < a. Again the symmetry of the
charge distribution allows us to simply evaluate
the left side of Gauss’s law just as before.r
The charge inside the Gaussian sphere is no longer Q. If we
call the Gaussian sphere volume V’ then
( )2
Left side: 4E dA E dA E dA E rπ× = = =∫ ∫ ∫
Ñ Ñ Ñ
( )
3
2
0 0
4
4
3
inQ r
E r
ρπ
π
ε ε
= =
34
Right side:
3
inQ V rρ ρ π′= =
( )
3
3 32
30 00
4 1
but so
43 43 4
3
e
r Q Q Q
E r E r k r
a ar a
ρπ ρ
ρ
ε πεε π π
= = = = =
69. a
b
-Q
+2Q
Find the field for r > b
From the symmetry of the problem, the field
in this region is radial and everywhere
perpendicular to the spherical Gaussian
surface. Furthermore, the field has the same
value at every point on the Gaussian surface
so the solution then proceeds exactly as in
Ex. 2, but Qin=2Q-Q.
( )2
4E dA E dA E dA E rπ× = = =∫ ∫ ∫
Ñ Ñ Ñ
Gauss’s law now gives:
( )2
2 2
0 0 0 0
2 1
4 or
4
in
e
Q Q Q Q Q Q
E r E k
r r
π
ε ε ε πε
−
= = = = =
70. 2
3
We found for ,
and for ,
e
e
Q
r a E k
r
k Q
r a E r
a
> =
< =
a
Q
Let’s plot this:
E
ra
For spherical those answer
For conductoe is smilal
71. An insulating sphere of radius a has a uniform charge density ρ and a total
positive charge Q. Calculate the electric field outside the sphere.
a
Since the charge distribution is spherically
symmetric we select a spherical Gaussian
surface of radius r > a centered on the
charged sphere. Since the charged sphere
has a positive charge, the field will be
directed radially outward. On the Gaussian
sphere E is always parallel to dA, and is
constant.Q
rE
dA
( )2
Left side: 4E dA E dA E dA E rπ× = = =∫ ∫ ∫
rr
Ñ Ñ Ñ
0 0
Right side: inQ Q
ε ε
=
( )2
2 2
0 0
1
4 or
4
e
Q Q Q
E r E k
r r
π
ε πε
= = =
72. A conducting spherical shell of inner radius a and outer radius b with a
net charge -Q is centered on point charge +2Q. Use Gauss’s law to
find the electric field everywhere, and to determine the charge
distribution on the spherical shell.
a
b
-Q First find the field for 0 < r < a
This is the same as Ex. 2 and is the field due to a
point charge with charge +2Q.
2
2
e
Q
E k
r
=
Now find the field for a < r < b
The field must be zero inside a conductor in equilibrium. Thus from
Gauss’s law Qin is zero. There is a + 2Q from the point charge so we
must have Qa = -2Q on the inner surface of the spherical shell. Since the
net charge on the shell is -Q we can get the charge on the outer surface
from Qnet = Qa + Qb.
Qb= Qnet - Qa = -Q - (-2Q) = + Q.
+2Q
73. Pictures from Serway &
Outside the sphere, we have
k Q
Er=
r2
For r > R
To obtain potential at B, we use
VB= -
r
Er dr = - kQ
r
r2
dr
Potential must be continuous at r = R, => potential at surface
74. Pictures from Serway &
Inside the sphere, we have
k Q
Er=
R3
For r < R
To obtain the potential difference at D, we use
VD - VC= -
r
Er dr= - r dr=
r
R
k Q
R3
r
R
k Q
2R3
)R2
– r2
(
Since
To obtain the absolute value of the potential at D, we add
the potential at C to the potential difference VD - VC:
Check V for r = R
For r < R
75. r< R1
In condector sphereca
And we have two
charge
E=0
r> R2 22
/21/21 rqKqrqKqF −=+=
2
/12 rqKF =
R1< r< R2
2
/1 rKqE =
76. Gauss’ Law – How does it work?
Step 1 – Is there a source of
symmetry?
Consider a POSITIVE POINT CHARGE, Q.
Yes, it is spherical symmetry!
You then draw a shape in such a way
as to obey the symmetry and
ENCLOSE the charge. In this case,
we enclose the charge within a
sphere. This surface is called a
GAUSSIAN SURFACE.
Step 2 – What do you know about the
electric field at all points on this
surface?
It is constant.
∫ =
o
encq
daE
ε
The “E” is then brought out of the integral.
77. Gauss’ Law – How does it work?
o
encq
rE
ε
π =(4) 2
Step 4 – Identify the charge enclosed?
The charge enclosed is Q!
Step 3 – Identify the area of the
Gaussian surface?
In this case, summing
each and every dA gives
us the surface area of a
sphere.
oo r
Q
E
Q
rE
επε
π 2
2
4
(4) =→=
This is the
equation for a
POINT CHARGE!
78. Strategy for Solving Gauss’ Law Problems
• Select a Gaussian surface with symmetry that matches the
charge distribution.
• Draw the Gaussian surface so that the electric field is either
constant or zero at all points on the Gaussian surface.
• Evaluate the surface integral (electric flux).
• Determine the charge inside the Gaussian surface.
• Solve for E.
• Use symmetry to determine the direction of E on the Gaussian
surface.
79. Example: use Gauss’ Law to calculate the electric field due to a
long line of charge, with linear charge density λ.
Example: use Gauss’ Law to calculate the electric field due to
an infinite sheet of charge, with surface charge density σ.
These are easy using Gauss’ Law (remember what a pain they
were in the previous chapter). Study these examples and others
in your text!
sheet
0
E .
2
σ
=
ε
line
0
E .
2 r
λ
=
πε
80. Gauss’ Law and cylindrical symmetry
rLA
qLQ
L
Q
MacroRECALL
cylinder
enc
π
λ
λ
2
:
=
==
=→
Consider a line( or rod) of charge that is very long (infinite
+
+
+
+
+
+
+
+
+
+
+
+
We can ENCLOSE it within a
CYLINDER. Thus our Gaussian surface
is a cylinder.
o
o
o
enc
o
enc
r
E
L
rLE
q
rLE
q
daE
επ
λ
ε
λ
π
ε
π
ε
2
(2)
(2)
=
=
==∫
This is the same equation we got doing extended charge distributions.
81. Gauss’ Law for insulating sheets and disks
A charge is distributed with a uniform charge density over an infinite
plane INSULATING thin sheet. Determine E outside the sheet.
For an insulating sheet the
charge resides INSIDE the
sheet. Thus there is an electric
field on BOTH sides of the plane.
o
o
oo
o
enc
E
A
EA
A
Q
Q
EA
Q
EAEA
q
dAE
ε
σ
ε
σ
σ
εε
ε
2
2,
2
=
==
=→=+
=•∫
This is the same equation we got
doing extended charge distributions.
82. Gauss’ Law for conducting sheets and disks
A charge is distributed with a uniform charge density over an infinite
thick conducting sheet. Determine E outside the sheet.
+
For a thick conducting sheet, the charge exists
on the surface only
o
o
o
o
enc
E
A
EA
A
Q
Q
EA
q
dAE
ε
σ
ε
σ
σ
ε
ε
=
==
=
=•∫
,
+
+
+
+
+
+
E =0
83. In
summaryWhether you use electric charge distributions or Gauss’ Law you get the
SAME electric field functions for symmetrical situations.
∫ =•
=→=
o
enc
oo
q
dAE
r
dq
dE
r
Q
E
ε
πεπε 22
44
Function Point, hoop,
or Sphere
(Volume)
Disk or Sheet
(AREA)
“insulating
and thin”
Line, rod, or
cylinder
(LINEAR)
Equation
2
4 r
Q
E
oπε
=
r
E
oπε
λ
2
=
o
E
ε
σ
2
=
Created
byaza
d
84. The top 5 reasons why we make you learn Gauss’ Law:
5. You can solve (high-symmetry) problems with it.
4. It’s good for you. It’s fun! What more can you ask!
3. It’s easy. Smart physicists go for the easy solutions.
2. If I had to learn it, you do too.
And the number one reason…
…will take a couple of slides to present
85. Starting with Gauss’s law, calculate the electric
field due to an isolated point charge q.
q
E
r dA
We choose a Gaussian surface that is a
sphere of radius r centered on the point
charge. I have chosen the charge to be
positive so the field is radial outward by
symmetry and therefore everywhere
perpendicular to the Gaussian surface.
E dA E dA× =
rr
Gauss’s law then gives:
0 0
inQ q
E dA E dA
ε ε
× = = =∫ ∫
rr
Ñ Ñ Symmetry tells us that the field is
constant on the Gaussian surface.
( )2
2 2
0 0
1
4 so
4
e
q q q
E dA E dA E r E k
r r
π
ε πε
= = = = =∫ ∫Ñ Ñ
86. Conductors in Electrostatic
Equilibrium
The electric field is zero everywhere inside the
conductor
Any net charge resides on the conductor’s
surface
The electric field just outside a charged conductor
is perpendicular to the conductor’s surface
By electrostatic equilibrium we mean a situation where
there is no net motion of charge within the conductor
88. The electric field is always perpendicular to the
surface of a conductor – if it weren’t, the
charges would move along the surface.
89. Any net charge on an isolated conductor must reside on its surface and
the electric field just outside a charged conductor is perpendicular to its
surface (and has magnitude σ/ε0). Use Gauss’s law to show this.
For an arbitrarily shaped conductor we
can draw a Gaussian surface inside the
conductor. Since we have shown that the
electric field inside an isolated conductor
is zero, the field at every point on the
Gaussian surface must be zero.
From Gauss’s law we then conclude that the
net charge inside the Gaussian surface is
zero. Since the surface can be made
arbitrarily close to the surface of the
conductor, any net charge must reside on the
conductor’s surface.
0
inQ
E dA
ε
× =∫
rr
Ñ
90. We can also use Gauss’s law to determine the electric field just outside
the surface of a charged conductor. Assume the surface charge density
is σ. Since the field inside the conductor is
zero there is no flux through the face
of the cylinder inside the conductor. If
E had a component along the surface
of the conductor then the free charges
would move under the action of the
field creating surface currents. Thus E
is perpendicular to the conductor’s
surface, and the flux through the
cylindrical surface must be zero.
Consequently the net flux through the
cylinder is EA and Gauss’s law gives:
0 0 0
orin
E
Q A
EA E
σ σ
ε ε ε
Φ = = = =
91. Summary
Two methods for calculating electric field
Coulomb’s Law
Gauss’s Law
Gauss’s Law: Easy, elegant method for
symmetric charge distributions
Coulomb’s Law: Other cases
Gauss’s Law and Coulomb’s Law are
equivalent for electric fields produced by static
charges
