- 1. Physic department University of duhok 2011/3/29
2. Physic department 3. r fa Q1 Q2 2 2 a r qq1 F 04 1 = 2 2 b r
qq1 F 04 1 = 0 12 2 28 85 10= . c m N 1221 - FF = 229 /.109 cmNk =
04 1 = 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. 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 6. Example I1. 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 1014 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 electrons 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 = == 7. Example 2 Three charges are positioned as
shown. Find the force acting on the 2 C charge. 8. Three charges
are positioned as shown. Find the force acting on the 2 C charge.
9. L q 2 = cos 4 1 2 2 0 = r dx dE 22 cos xz z r z + == ( ) 220 0
2220 0 2/3220 2 4 1 4 2 2 4 1 Lzz L xzz xz dx xz z E L Lx x + = + =
+ = = = 10. ( ) + 2/322 xz dx ( ) dx xz z E Lx x = = + = 0 2/3220 2
4 1 How to evaluate this integral Let tanzx = then ( ) 222222
sec1tan zzzx =+=+ and dzdx 2 sec= Substitute these into the
integral: ( ) ( ) c z d z d zz dz xz dx +==== + sin 1 cos 1 sec 11
sec sec 2 2 22/322 2 2/322 From diagram, 22 sin zx x + = ( )
2222/322 1 zx x zxz dx + = + Hence Lq 2= The line looks like a
point charge ,so the field reduces to that of a ( )2 04/ zq point
charge 11. ( ) 2200 222 0 0 2/3220 222 0 2 0 4 11 4 1 1 4 1
cos;;cos 4 1 Lz L zxz x z z dx xz z r z xzr r dx E Lx x L Lx x z +
= + = + = =+== = = = = ( ) + = + = + == = = 2200 220 0 2/32200 2 0
11 4 11 4 1 4 1 sin 4 1 Lzzxz xz xdx r dx E Lx x LL x Net electric
field E = Ex + Ez + + + = zx 1 4 1 22220 Lz L Lz z z E For z
>> L and Lq = z 4 1 2 0 z L E 12. x dldq = r 2/3 22 22 2
((/cos /coscos /coscos ladlqxkFd lakdqqFd rkdqqFd += += = l Ex
apositive charge rod length l and liner charge distributid lenda
find the force the test charge along the x axis through middle of
the road dL dq This qustion from first test Univercity of duhok by
azad sleman 13. 1 2 21 // . / qrqqKE forceF fieldEE qEF qFE = = = =
= 2 q E=k r - - - - - - - - - - - - - + + + + + + + + + + + + + - F
E F ma qE.= = r rr A charged particle in an electric field
experiences a force, and if it is free to move, an acceleration. +q
r q0 E r +q 14. -q -q r r q0 E r 02 04 1 r r q E r = 15. = =+++= 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 4 1 EEEEE rr == ==+++= 102 1 1 0 1 1 r r q 4 1
q F E 0 r == i02 i i 0 i i r r q 4 1 q F E 0 == 202 2 2 0 2 2 r r q
4 1 q F E 0 == qn qi q3q2 q1 r2 ri rn r1 r3 F r nF r iF r 3F r 2F r
1F r q0 16. Draw and label forces (only those on Q3(. Draw
components of forces which are not along axes. x y Q2=+50C Q3=+65C
Q1=-86C 52cm 60 cm 30cm =30 F31 F32Draw a representative sketch.
Draw and label relevant quantities. Draw axes, showing origin and
directions. Step 1: Diagram 17. 1 2 2 12 q q F k 12 r = Do I have
to put in the absolute value signs? x y Q2=+50C Q3=+65C Q1=-86C
52cm 60 cm 30cm =30 F31 F32 Step 2: Starting Equation 18. 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=+50C Q3=+65C
Q1=-86C 52cm r31 =60 cm r32=30cm =30 F31 F32 Step 3: Replace
Generic Quantities by Specifics 19. 3 1 2 31 Q Q F k , 31 r
attractive = r 3 1 2 31 Q Q F k cos 31,x r = + 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 = (+sign comes from diagram( x y Q2=+50C Q3=+65C
Q1=-86C =30 F31 F32 Step 3 (continued( r32=30cm r31 =60 cm 52cm 20.
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=+50C Q3=+65C Q1=-86C 52cm 60 cm
30cm =30 F31 F32 Step 3: Complete the Math 21. 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 22. 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 wont be required to
use this version of the equation. F = qE r r 23. 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 dx. is the
linear density of charge (amount of charge per unit length). may be
a function of position. Think length. times the length of line
segment is the total charge on the line segment. l x dx dx 24. 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' 4 r' 4 r' r $ $ r r'$ dE Im 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
25. The electric field at P due to the entire line of charge is 2 0
1(x) dx E = r' . 4 r' r $ The integration is carried out over the
entire length of the line, which need not be straight. Also, 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 26. If
charge is distributed over a two-dimensional surface, the amount of
charge dq on an infinitesimal piece of the surface is dS, where is
the surface density of charge (amount of charge per unit area). x y
area = dS charge dq = dS 27. dE The electric field at P due to the
charge dq is 2 2 0 0 1 dq 1 dS dE = r' = r' 4 r' 4 r' r $ $ x y P r
r'$ 28. 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' 4 r' r $ E 29. x z P
r r'$ 2 0 V 1 (x, y,z) dV E = r' . 4 r' 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 30. 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. 31. Summarizing: 2 0
1 dx E = r' . 4 r' r $ 2 0 S 1 dS E = r' . 4 r' r $ 2 0 V 1 dV E =
r' . 4 r' r $ Charge distributed along a line: Charge distributed
over a surface: Charge distributed inside a volume: If the charge
distribution is uniform, then , , and can be taken outside the
integrals. 32. The Electric Field Due to a Continuous Charge
Distribution (worked examples) Gauss law Adobe Acrobat Document
Adobe Acrobat Document e.fieldGauss lawe.flux GausssLaw Gauss law
Adobe Acrobat Document Microsoft Office WordDocument Adobe Acrobat
Document ELECTRICITYAND MAGNETISM P10D Coulombs 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 33. Physics 2102 Final Exam Review Physics 2102
Jonathan Dowling Microsoft Office Word 97 - 2003 Document Microsoft
Office Word Document Adobe Acrobat Document 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, . 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 = dx, so,
as with the finite length charge, we use the angle, with 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)
34. Example: A rod of length L has a uniform charge per unit length
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 P x y d L dE x dx dQ = dx 2 2 dq dx dE = k k x x = Note: dE is
in the x direction. dE is the magnitude of dE. Ive 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 1 E = dE = -k
i = -k i = -k i x x x + r r ( ) ( ) ( ) 1 1 d d L L kQ E = -k i =
-k i= -k i= - i d L d d d L d d L d d L + + + + + + + r 35. L P a 0
a La + x dxdQ = x Q V 04 d d = dx + == La a x x dVV 04 d La a x + =
ln 4 0 ( )[ ]aLa lnln 4 0 += a La + ln x Example: A rod of length L
has a uniform charge per unit length 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. 36. cosdEEd 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 = dy Ex = [k dy /(xo2 +y2 )] [xo /(xo2 +
y2 )1/2 [[ Ex = k xo [dy /(xo2 +y2 )] [1 /(xo2 + y2 )1/2 [ Ex = k
xo [dy /(xo2 +y2 ) 3/2 [ Example: A rod of length y has a uniform
charge per unit length 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 37.
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 xo -a a [dy /(xo2
+y2 ) 3/2 [ Ex = k(Q/2a) Xo [y /Xo2 (Xo2 +y2 ) 1/2 ] -a a aQ 2/= 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 38. 2/1222 2/322 2/322
2/322 222 2 22 ).(/ )/(2.2/ )/( )(/ )( /coscos // xaxkqx xaaakqx
xzdyqxk azdzqxkdf azr rdzkdF rkdzqrkdqqF a a += += + += += = == dz
dzdq = In this qustion we have not (a,-a)we want force z Answer 1
Answer 2 39. 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. 40. radius R, which carries a uniform line charge . R q 2 = cos
4 1 2 0 = r dl Ez r z zRr =+= cos;222 ( ) + = dl Rz z Ez 2/32204 1
Rdl 2= ( ) ( ) 2/3220 2 4 1 Rz zR Ez + = 41. 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 2= ( )2 / aq = 22 2 2
a qrdr rdr a q dAdq === ( ) ( ) 2/3222 0 2/3220 2 4 1 4 1 rz rdr a
qz dq rz z dE + = + = ( ) dr rz r a qz E a + = 0 2/3222 0 2 4 1 ( )
dr rz r a qz E a + = 0 2/3222 0 2 4 1 42. 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 S o
When z R, R r 0 0>r < R rkE lrlE /2 /)2( = = L 60. a b +2Q
cylnder o enc o enc l rlE rlA lq q dAE = = = = )2( 2 b>a ar <
R rkE lrlE /2 /)2( = = r conductor insulator shell 61. a b rabkE
abllrlE /)(2 /)(()2( 22 22 = += 62. rabkE rarkE rkE /)(2 /)(2 /2 22
22 = = = insulator b a 0r shell 0>r < R ab 63. 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 64.
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 65. 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 Gausss 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 r =
= = ( ) 3 2 0 0 4 4 3 inQ r E r = = 34 Right side: 3 inQ V r = = (
) 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 =
= = = = 66. 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 r = = = Gausss 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 = = = = =
67. 2 3 We found for , and for , e e Q r a E k r k Q r a E r a >
= < = a Q Lets plot this: E ra For spherical those answer For
conductoe is smilal 68. An insulating sphere of radius a has a
uniform charge density 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 r = = = rr 0 0
Right side: inQ Q = ( )2 2 2 0 0 1 4 or 4 e Q Q Q E r E k r r = = =
69. A conducting spherical shell of inner radius a and outer radius
b with a net charge -Q is centered on point charge +2Q. Use Gausss
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 Gausss 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 70. 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
71. 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 72. 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 = 73. 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 The E
is then brought out of the integral. 74. Gauss Law How does it
work? o encq rE =(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 2 2 4 (4) == This is the
equation for a POINT CHARGE! 75. 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. 76.
Example: use Gauss Law to calculate the electric field due to a
long line of charge, with linear charge density . Example: use
Gauss Law to calculate the electric field due to an infinite sheet
of charge, with surface charge density . 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 = line 0 E .
2 r = 77. Gauss Law and cylindrical symmetry rLA qLQ L Q
MacroRECALL cylinder enc 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 2 (2) (2) = = ==
This is the same equation we got doing extended charge
distributions. 78. 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 2 2, 2 = == ==+ = This is the
same equation we got doing extended charge distributions. 79. 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
= == = = , + + + + + + E =0 80. 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 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 o = r E o 2 = o E 2 = Created byaza d 81. The top 5
reasons why we make you learn Gauss Law: 5. You can solve
(high-symmetry) problems with it. 4. Its good for you. Its fun!
What more can you ask! 3. Its 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 82. Starting
with Gausss 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 Gausss law then gives: 0 0 inQ q E dA E dA = = = rr
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 = = =
= = 83. Conductors in Electrostatic Equilibrium The electric field
is zero everywhere inside the conductor Any net charge resides on
the conductors surface The electric field just outside a charged
conductor is perpendicular to the conductors surface By
electrostatic equilibrium we mean a situation where there is no net
motion of charge within the conductor 84. When electric charges are
at rest, the electric field within a conductor is zero. 85. The
electric field is always perpendicular to the surface of a
conductor if it werent, the charges would move along the surface.
86. 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 /0). Use Gausss 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 Gausss 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
conductors surface. 0 inQ E dA = rr 87. We can also use Gausss law
to determine the electric field just outside the surface of a
charged conductor. Assume the surface charge density is . 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 conductors surface, and the flux through the
cylindrical surface must be zero. Consequently the net flux through
the cylinder is EA and Gausss law gives: 0 0 0 orin E Q A EA E = =
= = 88. Summary Two methods for calculating electric field Coulombs
Law Gausss Law Gausss Law: Easy, elegant method for symmetric
charge distributions Coulombs Law: Other cases Gausss Law and
Coulombs Law are equivalent for electric fields produced by static
charges
