Download Lesson 4A

Lesson 4A
How to set up an integral to find electric field of continuous charge
distribution
A continuous distribution of charge occurs when a line, a surface, or a solid is
completely filled with charge. Charge density is used to describe the distribution. In
general, the density can vary from point to point. In the case of a line, the linear
charge density is defined by
l=
Dq
D
where D is an infinitesimally small length on the line and Dq the charge on this
small length. The unit of l is C m . For a surface, we define surface charge density
s=
Dq
DA
where DA is an infinitesimally small area on the surface. The unit is C m 2 . For a
solid, the volume charge density is defined by
r=
Dq
DV
where DV is an infinitesimally small volume in the solid. The unit is C m3 . If the
charge distributions are inhomogeneous, the quantities l, s , r are different at
different locations, and become functions of positions.
To find the total charge on a line, a surface, or a solid, the method of integration can
be used. The same method applies in finding the electric field at any point in space
due to such distributions. The following example illustrates how the method works.
Example: An electrically charged thin rod lies on the x-axis with one end at the origin
and the other end at the location x = 2m . The linear charge density is given by the
function l ( x ) = Ax 2 where A is a constant. Find
(a) the total charge
(b) the electric field at the point P on the x-axis with coordinate x = 3m , in terms
of the total charge
(c) the electric field at the origin
1
Solution:
(a) Imagine the rod being divided into tiny pieces.
Identify the position of a typical piece by a variable. In this case, the x-coordinate x
is the obvious choice.
Call the length of the tiny piece Dx
Write down the charge on the piece as
Dq = lDx = l ( x ) Dx
The total charge is the sum:
q = å lDx
Before replacing the sum by an integral, decide what the limits of integration are.
The limits give the range of x from one end of the rod to the other. They are
therefore x = 0 and x = 2 . So we write
q=
2
ò l dx
0
Now evaluate the integral:
q=
2
ò Ax
0
2
dx =
A 32 A 3
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x = ( 2 - 0) = A
0
3
3
3
2
The physical meaning of the quantity A is not clear. But now we can express it in
terms of the total charge:
A=
3q
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(b) To find the electric field at the point P.
Imagine the tiny piece Dx as a point charge Dq and consider the electric field it
produces at the point. This electric field has only x-component.
First note
distance between the tiny piece and P = 3- x
The x-component of the electric field is therefore
DEx =
kDq
(3- x )
2
=
k lDx
(3- x )
2
Summing the contribution:
Ex = å DEx =
2
ò
0
k l dx
(3- x )
3
2
= kA ò
0
x 2 dx
(3- x )
2
To evaluate the integral, make the substitution
u = 3- x
x = 3- u
dx = -du
Also note that the limits for u is now u = 3 (for x = 0 ) and u =1 (for x = 2 )
3
1
E x  kA
3  u 2  du   kA3 u 2  6u  9du  kA3 1  6 
u
3

2
1
u
 
2
1
u
9
u2

du

3

9

 1 1 
 kA u  6nu    kA3  1  6n3  n1  9    8  6n3kA  1.41kA
u1

 3 1 

Since the physical meaning of A is not clear, it is better to express E x in terms of the
total charge q . The result is
E x  1.41k
3q
 0.53k  4.8  10 9 N C
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(c) To find the electric field at the origin O, first note
distance between the tiny piece Dx and O = x
Electric field at O due to the tiny piece is
dEx = -
kDq
x2
Note the minus sign.
The electric field due to the rod is therefore
2
2
kl dx
x 2 dx
3q
3kq
Ex = å DEx = - ò 2 = -kA ò 2 = -kA ò dx = -2kA = -2k = x
x
8
4
0
0
0
2
4