Download 1 Lesson 5 (1) Electric Field of a Line Charge Consider a long thin r

Lesson 5 (1) Electric Field of a Line Charge Consider a long thin rod with a uniform distribution of charge so that the line charge density ! (C / m ) is the same everywhere on the rod. We will calculate the electric field at a point close to the rod. For such a point, the rod can be considered to be infinitely long. Let us place the rod on the y-­‐axis, and the point on the x-­‐axis so that its x-­‐coordinate is x . Divide the y-­‐axis into small segments dy . The charge on this segment is dq = ! dy !
It creates the electric field dE as indicated by the arrow in the figure. The total electric field is the sum of all these vectors from different parts of the rod. Instead of adding an infinitely many small vectors, we first note that because of symmetry, we expect the total electric field to have only x-­‐
component. Thus it is sufficient to calculate E x = ! dE x !
The magnitude of dE is !
k dq
k ! dy
dE = 2 2 = 2 2 x +y
x +y
The x-­‐component is !
!
x
x
dE x = dE cos! = dE
= k ! dy
2
2 32
x 2 + y2
x
+
y
(
)
Therefore, +"
Ex = k! x #
!"
dy
( x 2 + y2 )
32
1 The integral can be evaluated by changing from y to the variable ! . From y = x tan !
dy = x sec 2 ! d! ! 2
! 2
x sec 2 " d"
k"
2k "
Ex = k! x &
=
cos# d# =
&
32
2
2
x
x
!
#
%! 2 x (1+ tan " )
%! 2
"
$
Note that the electric field falls off as 1/distance. Later when we learn Gauss law, we’ll give a simpler derivation of this formula. (2) Electric Field of a Charged Circular Ring For a circular ring of radius a with uniform line charge density ! , the electric field at a point on the axis can be calculated easily, because the point is at equal distance from all the charges on the ring. Suppose the ring is placed on the x-­‐y plane with its center at the origin, so that the z-­‐axis is its axis. Consider a point P on the z-­‐axis with z-­‐coordinate z . Divide the ring into small angular segments d! . The length of such a segment is d! = ad! The charge on the segment is dq = ! d! = ! ad" !
The electric field dE due to the segment has magnitude 2 !
kdq
k ! ad"
dE = 2
= 2
2
z +a
z + a2
By symmetry, we expect the net electric field to have only z-­‐component. !
!
z
k " az
dEz = dE cos ! = dE
=
d# 32
z2 + a2 (z2 + a2 )
The electric field is therefore 2!
k ! ( 2" ) az
k ! az
k ! az 2 "
kqz
Ez = ! dEz = "
d
!
=
d! =
=
"
3
2
3
2
2
2
2
2
2
2 32
2
2 32
z
+
a
z
+
a
z
+
a
0 (z + a )
0
(
)
(
) (
)
where in the last step, we have replaced ! by the total charge q = ! ( 2" a ) . An easy way to obtain this result is to multiply the electric field due to total charge q , all at a distance of z 2 + a 2 from the point concerned, by the constant factor cos ! : kq
kq
z
Ez = 2 2 cos ! = 2 2
2
z +a
z + a z + a2
(3) Electric Field of a Uniformly Charged Circular Disk The electric field due to a uniformly charged circular disk at a point on its axis can also be calculated using the result for a ring. Let a be the radius of the disk, which we place on the x-­‐y plane with its center at the origin. Let ! be the surface charge density. The axis of the disk is the z-­‐
axis. Let the z-­‐coordinate of the point P be z . Divide the disk into thin rings of radius r and thickness dr . The charge on the ring is dq = ! ( 2" rdr ) The electric field due to this ring at the point P has only z-­‐component, and is equal to 3 dEz =
(kdq) z
(z
2
+r
2 32
)
= 2! k" z
rdr
(z
2
+ r2 )
32
The total electric field is therefore a
rdr
Ez = 2! k" z !
2
2 32
0 (r + z )
Change to the variable ! using: r = z tan !
dr = zsec 2 ! d!
! max = tan !1 ( a z )
! max
E x = 2! k" z
!
0
z tan # zsec 2 # d#
z 3 (1+ tan 2 # )
32
#
! max
= 2" k#
! sin! d! = 2" k# (1" cos! ) = 2! k" %%1"
max
0
$
&
((
z2 + a2 '
z
(4) Electric Field due to an Infinite Charged Sheet When the radius a of the circular disk becomes very large compared with z , the electric field of the circular disk at P becomes # z&
Ez ! 2! k" %1" ( ! 2! k" $ a'
This can be considered the electric field due to an infinite sheet charge of uniform density ! at any point, and is independent of the location of the point . The field lines are as shown: The Coulomb constant can be written in the form 1
k=
4!" 0
where ! 0 = 8.85 !10 "12 C 2 / N # m 2 is called the permittivity of free space. The electric field due to an infinite sheet can be written 4 E=
!
2" 0
The electric field due to multiple sheets is the vector sum of the field of the individual sheets. An important case is a pair of parallel sheets with surface charge densities ! and !! respectively. For points on either side of the pair, the electric fields due to the two sheets cancel out. For any point between the sheets, the electric fields due to the two add. Therefore, the electric field is confined to the space between the sheets, and is given by !
E = "0
pointing from the positive to the negative sheet. The field lines deviate from straight lines only near the edges of the sheets. 5