1/25/2009
1
±
The electric field of a charged disk
¾
The figure below shows a disk of radius
R
having total charge
Q
which is uniformly distributed on the surface.
¾
The surface charge density is
2
R
Q
A
Q
π
η
=
=
²
Our objective is to calculate the onaxis electric field.
¾
Following the problem solving strategy, we
need to
divide the continuous charge of the
disk (total charge Q) into N very narrow
rings of charge
∆
Q
. Because we’ve already
learn how to find the electric field of a ring.
¾
As shown in the fig the radius of ring
i
is
r
i
,
has a width
∆
r, area
∆
A
i
and charge
∆
Q
i
.
¾
The electric field of ring
i, at a point
along the
zaxis
is given by,
¾
The
onaxis
electric field of the whole disk is the sum of the electric
field of all of the rings:
2
/
3
2
2
0
)
(
4
1
)
(
i
i
z
i
r
z
Q
z
E
+
∆
∈
=
∑
∑
∆
=
=
N
i
N
z
i
z
disk
Q
z
E
E
2
3
2
2
)
(
)
(
⇒
+
∈
=
2
/
3
2
2
0
)
(
4
1
)
(
R
z
zQ
E
z
ring
¾
Since we deal with a surface rather than a line. Thus the charge
∆
Q
on
each ring is
¾
The area
∆
A
i
can be found by “
unrolling the
ring to form a narrow rectangle of length
2
π
r
(
the perimeter
)
and width
∆
r”.
=
=
+
∈
i
i
i
r
z
1
/
0
1
)
(
4
i
i
A
Q
A
Q
∆
=
∆
⇒
∆
∆
=
i
¾
Thus, the area of ring
i
is,
¾
Thus,
r
r
Q
Thus
r
r
A
i
i
i
∆
=
∆
⇒
∆
=
∆
2
,
2
∑
=
+
∆
∈
=
N
i
i
i
z
disk
r
z
r
r
z
E
1
2
/
3
2
2
0
)
(
2
)
(
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View Full Document1/25/2009
2
¾
We want to let
then the width
. Now we can replace
the sum with an integral ranging from
. Now we can
write the electric field as,
¾
If we change the variable as
∞
→
N
dr
r
→
∆
R
r
to
r
=
=
0
∫
+
∈
=
R
i
z
disk
r
z
rdr
z
E
0
2
/
3
2
2
0
)
(
2
)
(
η
If we change the variable as,
¾
Using the new variable, the lower integration limit is
¾
And the upper limit is
Note: in a definite integral, if you change the variables, then you must
du
rdr
rdr
du
r
z
u
2
1
2
2
2
=
⇒∴
=
⇒
+
=
2
0
z
u
r
=
⇒
=
2
2
R
z
u
R
r
+
=
⇒
=
change the limits.
¾
The electric field with new variables becomes,
+
−
∈
=
−
∈
=
∈
=
+
+
∫
2
2
0
2
/
1
0
z
z
2
/
3
0
1
1
2
2
4
2
1
2
)
(

2
2
2
2
2
2
R
z
z
z
u
z
u
du
z
E
R
z
z
R
z
disk
¾
If we multiply and divide the R.H.S of the equation by
z
we get
,
±
Now, let’s check the
limiting case (an approximation).
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 Fall '08
 TAWFIQ
 Physics, Charge, Electric charge, Uniform Electric Fields

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