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Gauss's Law 1
Example 1:Use Gauss’s Law to find the electric field adistance r away from a point charge q.
!
v E "d
v A # =
qenc
$0
!
EdA" =qenc#0
!
E dA" =qenc#0
!
E4"r2 =q#0
!
E =q
4"#0r2
Example 1:
Gauss's Law 3
Example 2:Electric charge is distributed uniformlyalong an infinitely long, thin wire. Thecharge per unit length is !. Find the electricfield.
!
v E "d
v A # =
qenc$0
!
v E "d
v A # +
v E "d
v A # =
qenc$0cylinder ends
r
!
EdA" + 0 =qenc#0
!
E dA" =qenc#0
!
E 2"rl( ) =#l
$0
!
E ="
2#$0r
Example 2:
Gauss's Law 5
Example 3:Positive charge Q is on a solid conductingsphere with radius R. Find the electric fieldat any point inside or outside the sphere.
!
v E "d
v A # =
qenc
$0
!
EdA" =qenc#0
!
E dA" =qenc#0
!
E4"r2 =0#0
!
E = 0
!
r < R
R
Q++ +
+
+
+
++
+
+
+
+
r
Example 3:
!
v E
!
v E "d
v A # =
qenc
$0
!
EdA" =qenc#0
!
E dA" =qenc#0
!
E4"r2 =Q#0
!
="R2
#0r2 " =
Q4$R2
% & '
( ) *
!
r > R
R
++ +
+
+
+
++
+
+
+
+
r
Q
!
E =Q
4"#0r2
Example 3:
!
v E
Gauss's Law 8
Example 3: Charged Spherical Conductor
Er
r
!
Q4"#0R
2
!
E = 0 r < R( )
R
!
E =Q
4"#0r2 r > R( )
!
"#0
$
% &
'
( )
Gauss's Law 9
Example 4:Positive charge Q is distributed uniformlythroughout the volume of an insulatingsphere with radius R. Find the magnitudeof the electric field at a point P a distance rfrom the center of the sphere.
!
v E "d
v A # =
qenc
$0
!
EdA" =qenc#0
!
E dA" =qenc#0
!
E4"r2 =#
43" r 3
$0
!
E =" r3#0
!
qenc = "dV0
r#
!
qenc = " dV0
r#
!
qenc = "V r( )
!
qenc = "43# r 3
!
" =3Q4#R3
$ % &
' ( )
!
qenc =Q r 3
R3
!
=Qr
4"#0R3
R
+ + + + +
+ + + + +
+ + + + + + ++ + + + + + +
+ + + + + + ++ + + + + +
+ + + +
+ + + +
r < R
r
Q
Example 4:
!
v E
!
qenc =Q
!
v E "d
v A # =
qenc
$0
!
EdA" =qenc#0
!
E dA" =qenc#0
!
E4"r2 =Q#0
!
E =Q
4"#0r2
!
="R3
3#0r2 " =
3Q4$R3
% & '
( ) *
R
+ + + + +
+ + + + +
+ + + + + + ++ + + + + + +
+ + + + + + ++ + + + + +
+ + + +
+ + + +
r > R
r
Q
Example 4:
!
v E
Gauss's Law 12
Example 4: Uniformly Charged Spherical Insulator
Er
r
!
Q4"#0R
2
R
!
E =Q
4"#0r2 r > R( )
!
E =Qr
4"#0R3 r < R( )
Gauss's Law 13
Example 5:Positive charge is distributed uniformly along conducting cylinder with radius R. Findthe magnitude of the electric field at a pointP a distance r from the center of the cylinder.Express in terms of surface charge density "and linear charge density !.
lr
r < R
R
!
v E "d
v A # =
qenc
$0
!
v E "d
v A # +
v E "d
v A # =
qenc
$0cylinder ends
!
EdA" + 0 =qenc#0
!
E dA" =qenc#0
!
E2"rl =0#0
!
and E = 0
Example 5:
!
v E
l
r
r > R
R
!
v E "d
v A # =
qenc
$0
!
v E "d
v A # +
v E "d
v A # =
qenc
$0cylinder ends
!
E2"rl =#l
$0
!
E ="
2#$0r
!
" =#2$R
% & '
( ) *
!
="R#0r
!
EdA" + 0 =qenc#0
!
E dA" =qenc#0
Example 5:
!
v E
Gauss's Law 16
Example 5: Charged Cylindrical Conductor
Er
r
!
E = 0 r < R( )
R
!
E ="
2#$0r r > R( )
!
"2#$0R
!
"#0
$
% &
'
( )
Gauss's Law 17
Example 6:Positive charge is distributed uniformlythroughout the volume of a long insulatingcylinder with radius R. Find the magnitudeof the electric field at a point P a distance rfrom the center of the cylinder. Express interms of volume charge density # and linearcharge density !.
!
qenc = "V r( )
!
qenc = "# r2l
!
" =#$R2
% & '
( ) *
!
qenc = "r2
R2 l
!
v E "d
v A # =
qenc
$0
!
v E "d
v A # +
v E "d
v A # =
qenc
$0cylinder ends
!
EdA" + 0 =qenc#0
!
E dA" =qenc#0
!
E2"rl =#" r2l
$0
!
and E ="r2#0
!
="r
2#$0R2
r
r < R
rR
l
Example 6:
!
v E
!
v E "d
v A # =
qenc
$0
!
E2"rl =#l
$0
!
E ="
2#$0r
l
r
r > R
Rr
!
v E "d
v A # +
v E "d
v A # =
qenc
$0cylinder ends
!
="R2
2#0r
!
" =#$R2
% & '
( ) *
Example 6:
!
v E
Gauss's Law 20
Example 6: Uniformly Charged Cylindrical Insulator
Er
r
!
"2#$0R
R
!
E ="
2#$0r r > R( )
!
E ="r
2#$0R2 r < R( )
Gauss's Law 21
Example 7:A point charge with charge -q is located inthe center of a spherical conductive shellwith an inner radius R1 and outer radius R2.The outer shell has a net charge of +3q.Find the magnitude of the electric field at apoint P a distance r from the center of thespherical shell.
!
v E "d
v A # =
qenc
$0
!
EdA" =qenc#0
!
E dA" =qenc#0
!
E4"r2 =qenc#0
!
E =qenc4"#0r
2
+
++
+
++
+
+
-q
+
+
R1R2+ +
+
+q
+
+
+2q+
+
++
+
+
+
+
+
r
Example 7:
!
v E
+
++
+
++
+
+
-q
+
+
R1R2+ +
+
+q
+
+
+2q+
+
++
+
+
+
+
+
!
E =qenc4"#0r
2
!
0 < r < R1( ) E ="q
4#$0r2
!
R1 < r < R2( ) E ="q+ q4#$0r
2 = 0
!
R2 < r( ) E ="q+ q+ 2q
4#$0r2 =
2q4#$0r
2
!
E =q
2"#0r2
Example 7:
r
r
r
!
v E
!
v E
!
v E
Gauss's Law 24
Example 7:
Er
rR1 R2
!
E ="q
4#$0r2 0 < r < R1( )
!
E = 0 R1 < r < R2( )
!
E =q
2"#0r2 R2 < r( )
