Electric Field
Calculations for Line of
Charge Problems
Montwood High School
AP Physics C
R. Casao
Electric Field Due to a
Charged Rod on the Axis
of the Rod
The Picture!

Divide the length of the rod L into small pieces
of length dx.
– This also divides the total charge Q on the rod into
small elements of charge dq.
– Over the length L of the rod, the charge on each
small piece of length dx is dq.
Q
λ
L

The charge density on the rod is:

Because the charge is uniformly distributed
over the length of the rod, we can set up a
proportion relating the total charge per unit
length to the charge per unit length for each
small piece of the rod.
Q dq
L

dx
Each piece of length dx contains a charge dq
and can be considered to be a point charge.
 Each element of charge dq contributes to the
net electric field E at point P.
 Since the charge on the rod is positive, the
net electric field at point P is directed along
the negative x-axis from point P.
 The distance from point P to each piece dx is
x.
 Electric field equation for a point charge:

E
k Q
r
2
rx
E
k Q
x2
The electric field contribution from each
element of charge dq is designated dE.
 To determine the total electric field E at
point P, add the electric field contribution
from each element of charge dq from the
left end of the rod to the right end of the
rod.

– Integrate from a at the left end of the rod to
(a + L) at the right end of the rod. In other
words, the electric field contributions begin
at a and end at (a + L).
– This means that dq needs to be expressed in
terms of dx because we will integrate along
the x-axis from a to (a + L).

For each point charge dq:
k Q
k  dq
E  2 becomes dE  2
x
x
Q
Q dq
dq
λ
and

so λ 
L
L dx
dx
dq  λ  dx
k  dq
k  λ  dx
Substitute : dE  2
dE 
2
x
x

Integrate both sides of the equation.

a L
a
dE 

a L
a
k  λ  dx
x2

The integral of 1·dE is E.

a L
dE  E
a

For the other side of the integration, any
constants can come out in front of the integral:
E

aL
k  λ  dx
x2
a
 kλ

aL
a
E  kλ

1
x
dx
x2
a
rewrite as : E  k  λ 
aL

a L
2
 dx
x - 2  dx
a
a L
 x - 2 1 
E  k  λ   - 2 1 



a
aL
 1
E  k  λ   
 x a
aL
 x 1 
 kλ


 1 
a
Substitute the (a + L) and the a into the 1/x part of the equation. The equation is:
upper limit expression – lower limit
expression.
 1
 1
E  kλ


a

L
a


1
 1
E  kλ
 
a  L a 
 Common denominator is a∙(a + L):

  1 a   1 a  L 
E  k  λ  
  

 a  L a   a a  L 
 a
aL 
E  kλ


 a  a  L  a  a  L  
a  a  L


L
E  kλ
  kλ

 a  a  L  
 a  a  L  
kλL
E
a  a  L 
Q
 Alternative: if replace λ with
L
k QL
k Q
E

a  a  L   L a  a  L 
Electric Field Off the
Axis of a Finite Line of
Charge
The Picture!

Charge per unit
length:
Q
λ
L
 Divide
the length of the rod L into
small pieces of length dx.
– This also divides the total charge Q on
the rod into small elements of charge dq.
– Over the length L of the rod, the charge
on each small piece of length dx is dq.
 The
charge density on the rod is:
Q
λ
L

Because the charge is uniformly distributed over
the length of the rod, we can set up a proportion
relating the total charge per unit length to the
charge per unit length for each small piece of the
rod.
Q dq

L dx
Each piece of length dx contains a charge dq and
can be considered to be a point charge.
 Each element of charge dq contributes to the net
electric field E at point P.
 The total electric field E at point P is the sum of
the electric fields produced by each element of
charge dq from –x to x.


Electric field equation for a point
charge:
k Q
E
r
2
The electric field vector at point P for each
element of charge dq can be resolved into
an x component (Ex)and a y component
(Ey).
 At point P, the Ex components of the
electric field produced by each symmetric
–x and +x pair will be equal in magnitude
but opposite in direction, therefore, these
components cancel each other.
 At point P, the net electric field will be the
sum of the Ey components of the electric
field produced by each element of charge
dq from –x to x.

cos θ 
Ey
E
E y  E  cos θ

For each element of charge dq from –x to x, the
values of E, r, and q all change as x changes.
The relationships between these variables must
be determined.
r 2  x 2  y2

r
rewrite as : x  y
y
cos θ 
r
Ey
2
2
x 2  y2

1
2
y
 E  cos θ  E 
r
The electric field contribution from each element
of charge dq is designated dEy.
 To determine the total electric field E at point P,
add the Ey contribution from each element of
charge dq from the left end of the rod to the
right end of the rod.

– Integrate from -x at the left end of the rod to
+x at the right end of the rod. In other words,
the electric field contributions begin at -x and
end at +x.
– This means that dq needs to be expressed in
terms of dx because we will integrate from –x
to +x.
y k Q y
k  dq y
E y  E   2  becomes dE y  2 
r
r
r
r
r
Q
Q dq
dq
λ
and

so λ 
and dq  λ  dx
L
L dx
dx
k  λ  dx y k  λ  y  dx
dE y 


2
3
r
r
r

However, the radius r changes too and can
be expressed in terms of x (in other words,
r changes as x changes).
dE y 

k  λ  y  dx
 x 2  y2 




3

k  λ  y  dx
x
2
y
2

3
2
Integrate both sides of the equation from
–x to x: x
x

x
dE y 
k  λ  y  dx
 x
x
2
y
2

3
2

x
x

dE y  E y
Pull the constants k, l, and y out in front of
the integral sign and integrate. The y2 term
cannot be removed from the integration
because it is being added to the x2 term.
Ey  k  λ  y 
x
 x
x
dx
2
y
2

3
2

From a table of integration (#98):
 u
 x

du
2
2
a
dx
2
y
2

3


2
3

2
u
a  u a
2
2
2
x
y  x y
2
2
2
Replacing this in the equation:

x
Ey  k λ  y 
 y2  x 2  y2

x



 x
 kλyx
Ey  
 y2  x 2  y2


x

 kλx
 

 y  x 2  y2
 x 
x



 x
Substitute the limits of integration into the
equation; remember that the equation is: upper
limit expression – lower limit expression.
k l  x
k  l  x
Ey 

2
2
2
y x  y
y   x   y 2
Ey 
k l  x
y  x2  y2
Ey  2

k l  x
y  x2  y2
k l  x
y  x2  y2

In this example, x is equal to 0.5·L as long
as the center of the line of charge is at the
origin.
Q
k  Q  0.5  L
λ
so : E y  2 
2
2
L
y  L  0.5  L   y
Ey 
k Q
y  0.25  L  y
2
2
MIT Visualizations



URL:
http://web.mit.edu/8.02t/www/802TEAL3D/visualization
s/electrostatics/index.htm
Integrating Along a Line of Charge
The Line of Charge
Electric Field Due to a
Circular Arc of Charge
Consider a plastic rod having a uniformly
distributed charge –Q that is bent into a
circular arc of radius r.
 The x-axis passes through the center of
the circular arc and the point P lies at the
center of curvature of the circular arc.
 We will determine the electric field E due
to the charged rod at point P.
 The equation for arc length is: s = r·q.
 Divide the circular arc into small, equal
pieces of length ds.

Illustrations

Each length ds will contain an equal
amount of charge dq.
Q
λ
L

Q
Ls λ
s
Uniform charge density allows us to set up
a proportional relationship between Q, s,
dq, and ds:
Q dq

s ds
dq
λ
ds

Each length ds containing charge dq
contributes to the net electric field at point
P and can be considered as a point charge:
E
k Q
r
2
becomes dE 
k  dq
r
2
The direction of E is towards the circular
arc because the charge dq is negative.
 For each symmetric length ds, the Ey
component of E are equal in magnitude
and opposite in direction and cancel out.


The net electric field
at point P is the sum
of the Ex components
for each length ds
from one end of the
circular arc to the
other end.
Ex
cos θ 
E
E x  E  cos θ
for each piece ds
dE x  dE  cos θ
dE x 
dE x 
k  dq
2
 cosθ
r
k  λ  ds
r
2
 cosθ 
dq  λ  ds
k  λ  cosθ
r
2
 ds
r is constant for every length ds along the length
of the circular arc.
 Each different length ds will have a different
angle q between the vector E and the vector Ex.
 The dEx equation has two variables that change,
q and s, therefore, we must express one variable
in terms of the other.


From the arc length equation: s = r·q
– Remember that r is constant, so as s changes
so does q.
s = r·q becomes ds = r·dq
– dq represents the angle at point P for a
particular length ds.
dE x 
k  λ  cos θ
2
 r  dθ
r
k  λ  cos θ
dE x 
 dθ
r
To determine the net
electric field at point P,
integrate from the
lower end of the rod –
q to the upper end of
the rod +q.

k  λ  cos θ
dEx 
 dθ
θ
θ
r

θ

θ

Left side of integral:

θ
θ

dE x  E x
Right side of integral: pull the constants
k, l, and r out in front of the integral
kλ

r

θ
θ

θ
θ
cos θ  dθ
cos θ  dθ 
θ
sin θ θ

Combining the constants and the result of the
integral:
kλ
kλ
θ
 sin θ θ 
  sin θ  sin  θ 
r
r
kλ
Ex 
  sin θ  sin  θ 
r
trig identity: sin  θ  sin θ
kλ
kλ
Ex 
  sin θ  sin θ  
  sin θ  sin θ 
r
r
kλ
Ex 
  2  sin θ 
r

Be sure to express the angle q in the correct
mode on your calculator (degrees or radians).
Keep in mind that the circular arc is going to
have a total length s that is some part of the
circumference of a circle (C = 2·p ·r).
Exactly how much of a circle this is will be
determined by the angles given.
 Should the circular arc begin at 0° and end
at 180° (or from 0 rad to p rad), substitution
into the sin function will give you an answer
of 0 N/C for the electric field.

– In that case, integrate from 0° to 90° (or 0 rad
to p/2 rad) and multiply this by 2 since each half
of the circular arc will contribute equally to the
net electric field at point P.

If given a problem in which you have a
two oppositely charged circular arcs (one
from 0° to 180° and the other from 180°
to 360°) arranged to form a ring of
charge, you can determine the electric
field of one-fourth of the circular arc and
multiply the answer by 4 since each
quarter will contribute equally to the net
electric field at the point P.
– Substituting the angles above into the sin
function results in an answer of 0 N/C for the
electric field.
Circular Arcs Within a Quadrant
When a circular arc lies entirely within one
quadrant, write equations for the Ex component
and the Ey component of the electric field
contribution for each element of charge dq along
the length of the circular arc.
 There is no
cancellation of
components for a
circular arc within
the single quadrant.

 Ex = E· cos q

p
2 dE
0
x


p
2
0
k l
Ex 

r
Ex 
Ex 
Ex 
Ex 
k l
r
k l
r
k l
r
k l
r
k l
 cos q  dq
r

p
2
0
cos q  dq
p
 sin q 02
p


 sin  sin 0 
2


 1  0
 Ey = E· sin q

p
2 dE
0
y


p
2
0
k l
Ey 

r
Ey 
Ey 
Ey 
Ey 
k l
r
k l
r
k l
r
k l
r
k l
 sin q  dq
r

p
2 sin q
0
 dq
p
   cos q 02
p


   cos   cos 0 
2


 0  1
Once you have the Ex component and the Ey
component, apply the Pythagorean theorem with
the two components to determine the
magnitude of the resultant electric field vector.
 Use a trig function (sin, cos, or tan) to
determine the direction of the resultant electric
field vector.
