Line of Charge
Continuation
Class Objectives
• Find the electric field strength due to a
continuous charge distribution.
• Develop problem solving techniques for
continuous charge distributions.
Student Objectives
• Be able to identify the type charge
distribution for a given problem.
• Be able to use diagrams to describe the
problem.
• Be able to identify and use symmetry in
problems.
• Be able to set out questions clearly and to
develop problem solving strategies.
Line of Charge
• Consider a rod with total charge +Q. What
is the total electric field a distance z along
an axis through its centre?
Line of Charge
• Consider a rod with total charge +Q. What
is the total electric field a distance z along
an axis through its centre?
Z
-L + + + + + + + + + + + + + + + + + + + L
2L
Line of Charge
• Consider a rod with total charge +Q. What
is the total electric field a distance z along
an axis through its centre?

Z
r
x
dx
-L + + + + + + + + + + + + + + + + + + + L
2L
Line of Charge
• Solution
1. Understand the geometry.
2. Determine the easiest way to span the
geometry.
3. Evaluate the contribution due to an
infinitesimal charge element.
4. Exploit symmetry if possible.
5. Set up and solve integral.
Line of Charge
• We break up the line into two regions –L to
0 and 0 to +L.
Line of Charge
• We break up the line into two regions –L to
0 and 0 to +L.
• The easiest way to span the line is to
integrate over x.
• Because of symmetry we could integrate
from x=0 to x=L and multiply the result by
2.
• However we will integrate from –L to L.
Line of Charge
• The line of charge is uniformly spread over
the rod so that, λ=Q/L (charge per unit
length).
Line of Charge
• The line of charge is uniformly spread over
the rod so that, λ=Q/L (charge per unit
length).
• Therefore a charge element dq of width
dx, has charge: dq = λdx.
Line of Charge
• The line of charge is uniformly spread over
the rod so that, λ=Q/L (charge per unit
length).
• Therefore a charge element dq of width
dx, has charge: dq = λdx.

Z
r
x
-L + + + + + + + + + + + + + + + + + + + L
Line of Charge
• The line of charge is uniformly spread over
the rod so that, λ=Q/L (charge per unit
length).
• Therefore a charge element dq of width
dx, has charge: dq = λdx.

dq
Z
r
width dx
x
-L + + + + + + + + + + + + + + + + + + + L
Line of Charge
• Recall: The field at a point due to a point
charge is:
E 
kq
r
2
Line of Charge
• The field set up at the point due to a small
line element dq is dE.
dE

dE 
k dq
r
2
Z
r
x
-L + + + + + + + + + + +
L
Line of Charge
• The field set up at the point due to a small
line element dq is dE.
dE

dE 
k dq
r
2
Z
r
x
-L + + + + + + + + + + +
L
• From symmetry the horizontal components
from the fields produced by elements on
the opposite sides of the centre cancel.
Line of Charge
• Method 1: Integrating from –L to L (ignore
symmetry).
Line of Charge
• Therefore the net electric field is:
dE y 
k  dx  cos 
r
2
Line of Charge
• Therefore the net electric field is:
dE y 
k  dx  cos 
r
2
• Summing over the line from x=-L to x=L
gives the total electric field due to the line.
L
L
k  dx  cos 
Ez 
 dE
L
z


L
r
2
Line of Charge

dE
L
Ez 
 dE z 
L
L

L
k  dx  cos 
r
Z
r
2
x
-L + + + + + + + + + + +
• However we are not yet ready to integrate
since we have two variables in our integral.
L
Line of Charge

dE
L
Ez 
 dE z 
L
L

L
k  dx  cos 
r
Z
r
2
x
-L + + + + + + + + + + +
• However we are not yet ready to integrate
since we have two variables in our integral
x,θ.
L
Line of Charge

dE
L
Ez 
 dE z 
L
L

L
k  dx  cos 
r
Z
r
2
x
-L + + + + + + + + + + +
• However we are not yet ready to integrate
since we have two variables in our integral
x,θ.
• So that our integral is in terms of x we
substitute for cosθ.
z
• From the diagram: cos  
r
L
Line of Charge

dE
L
Ez 
 dE z 
L
L

L
k  dx  cos 
r
Z
r
2
x
-L + + + + + + + + + + +
• However we are not yet ready to integrate
since we have two variables in our integral
x,θ.
• So that our integral is in terms of x we
substitute for cosθ.
z
z
• From the diagram: cos   
r
z
2
 x
2

1
2
L
Line of Charge
• Therefore we can write that:
L
k  dx  z
Ez 

L
z
2
x
2

3
2
Line of Charge
• Therefore we can write that:
L
k  dx  z
Ez 

L
z
2
x
2

• Where z is a constant.
3
2
Line of Charge
• Therefore we can write that:
L
k  dx  z
Ez 

L
z
2
x
2

3
2
• Where z is a constant.
Ez 
z
4  0
L

L
dx
z
2
x
2

3
2
Line of Charge
• Therefore we can write that:
L
k  dx  z
Ez 

L
z
2
x
2

3
2
• Where z is a constant.
Ez 
z
4  0
L

L
dx
z
2
x
2

3
2
Line of Charge
• Therefore we can write that:
L
k  dx  z
Ez 

L
z
2
x
2

3
2
• Where z is a constant.
Ez 
z
4  0
L

L
dx
z
2
x
2

• We now can integrate.
3
2
Line of Charge
• We integrate by substitution.
Line of Charge
• We integrate by substitution.
• Using the substitution x  z tan  we get:
Ez 
z
4  0
d
*

*
z sec 
2
Line of Charge
• We integrate by substitution.
• Using the substitution x  z tan  we get:
Ez 
z
4  0
d
*

*


z sec  4  0 z
2
*
d
 sec 
*
Line of Charge
• We integrate by substitution.
• Using the substitution x  z tan  we get:
Ez 

z
4  0

4  0
d
*

*


z sec  4  0 z
2
*
cos 

z
*
d
*
d
 sec 
*
Line of Charge
• We integrate by substitution.
• Using the substitution x  z tan  we get:
Ez 

z
4  0

4  0
d
*

*


z sec  4  0 z
2
*
cos 

z
*
d

4  0 z
*
d
 sec 
*
sin 
*
*
Line of Charge
• Instead of changing limits we note that:
sin  
x
z
2
 x
2

1
2
Line of Charge
• Instead of changing limits we note that:
sin  
x
z
2
x
2

1
2
• We can therefore write,

x
Ez 

1
4  0 z   z 2  x 2  2


L


  L
Line of Charge
• Instead of changing limits we note that:
sin  
x
z
2
x
2

1
2
• We can therefore write,

x
Ez 

1
4  0 z   z 2  x 2  2


L

2
L
 
4 0 z z 2  L2
  L


1
2
Line of Charge
• Considering the case z>>L,
Ez 
2L
4 0 z
2
Line of Charge
• Considering the case z>>L,
Ez 
2L
4 0 z
2
• Since Q  2  L , then
Ez 
Q
4 0 z
2
Line of Charge
• Considering the case z>>L,
Ez 
2L
4 0 z
2
• Since Q  2  L , then
Ez 
Q
4 0 z
2
• The field due to a point charge.
Line of Charge
• Method 2: Use symmetry and integrate
from 0 to L, multiplying the result by 2.
Line of Charge
• Therefore the net electric field is:
dE
y

2 k  dx  cos 
r
2
Line of Charge
• Therefore the net electric field is:
dE
y
2 k  dx  cos 

r
2
• Summing over the line from x=0 to x=L
gives the total electric field due to the line.
L
Ez 
 dE
0
L
z


0
2 k  dx  cos 
r
2
Line of Charge

dE
L
Ez 

0
L
dE z 

0
2 k  dx  cos 
r
Z
r
2
x
-L + + + + + + + + + + +
• However we are not yet ready to integrate
since we have two variables in our integral.
L
Line of Charge

dE
L
Ez 

0
L
dE z 

0
2 k  dx  cos 
r
Z
r
2
x
-L + + + + + + + + + + +
• However we are not yet ready to integrate
since we have two variables in our integral
x,θ.
L
Line of Charge

dE
L
Ez 

0
L
dE z 

0
2 k  dx  cos 
r
Z
r
2
x
-L + + + + + + + + + + +
• However we are not yet ready to integrate
since we have two variables in our integral
x,θ.
• So that our integral is in terms of x we
substitute for cosθ.
L
Line of Charge

dE
L
Ez 

0
L
dE z 

0
2 k  dx  cos 
r
Z
r
2
x
-L + + + + + + + + + + +
• However we are not yet ready to integrate
since we have two variables in our integral
x,θ.
• So that our integral is in terms of x we
substitute for cosθ.
z
• From the diagram: cos  
r
L
Line of Charge

dE
L
Ez 

0
L
dE z 

0
2 k  dx  cos 
r
Z
r
2
x
-L + + + + + + + + + + +
• However we are not yet ready to integrate
since we have two variables in our integral
x,θ.
• So that our integral is in terms of x we
substitute for cosθ.
z
z
• From the diagram: cos   
r
z
2
 x
2

1
2
L
Line of Charge
• Therefore we can write that:
L
Ez 
2 k  dx  z
 z
0
2
 x
2

3
2
Line of Charge
• Therefore we can write that:
L
Ez 
2 k  dx  z
 z
0
2
 x
2

3
2
• Where z is a constant.
Line of Charge
• Therefore we can write that:
L
Ez 
2 k  dx  z
 z
0
2
 x
2

3
2
• Where z is a constant.
Ez 
2z
4
0
L
 z
0
dx
2
x
2

3
2
Line of Charge
• Therefore we can write that:
L
Ez 
2 k  dx  z
 z
0
2
 x
2

3
2
• Where z is a constant.
Ez 
2z
4
0
L
 z
0
dx
2
x
2

3
2
• We now can integrate.
Line of Charge
• We integrate by substitution.
Line of Charge
• We integrate by substitution.
• Using the substitution x  z tan  we get:
Ez 
2z
4
0
*
z
*
d
2
sec 
Line of Charge
• We integrate by substitution.
• Using the substitution x  z tan  we get:
Ez 
2z
4
0
*
z
*
d
2
sec 

*
2
4
0
z
d
 sec 
*
Line of Charge
• We integrate by substitution.
• Using the substitution x  z tan  we get:
Ez 

*
2z
4
0
z
*
0
2
sec 
*
2
4
d
z

*
cos  d 

*
2
4
0
z
d
 sec 
*
Line of Charge
• We integrate by substitution.
• Using the substitution x  z tan  we get:
Ez 

*
2z
4
0
z
*
2
4
0z
d
2
sec 

*

*
cos  d  
*
2
4
0
z
2
4
0z
d
 sec 
*
sin 
*
*
Line of Charge
• Instead of changing limits we note that:
sin  
x
z
2
 x
2

1
2
Line of Charge
• Instead of changing limits we note that:
sin  
x
z
2
x
2

1
2
• We can therefore write,
Ez

x


2
4 0 z  2
 z x
L
2


1
2



o
Line of Charge
• Instead of changing limits we note that:
sin  
x
z
2
x
2

1
2
• We can therefore write,
Ez

x


2
4 0 z  2
 z x
L
2


1
2



o

2
L

4 0 z z  L
2
2

1
2
Line of Charge
• Considering the case z>>L,
Ez 
2L
4 0 z
2
Line of Charge
• Considering the case z>>L,
Ez 
2L
4 0 z
2
• Since Q  2  L , then
Ez 
Q
4 0 z
2
Line of Charge
• Considering the case z>>L,
Ez 
2L
4 0 z
2
• Since Q  2  L , then
Ez 
Q
4 0 z
2
• The field due to a point charge.