Example 21.

11: Field of a line of charge

Positive electric charge Q is distributed uniformly along a line with length 2a, lying along the
y-axis, between y = −a and y = +a. Find the electric field at point P on the x-axis at a distance
x from the origin.

Solution
As in Example 21.10, we are interested in calculating the electric field due to a continuous distribution of
charge. Figure 1 illustrates geometry of the problem:

y
dyˇ -> dQ
r r=
a x2 + 2
y y
α P dEx
α
x
O x
a dEy dE

total charge Q

Fig. 1. Calculating the electric field of a line of charge of total length 2a at point P localized
on the axis that is perpendicular to the charge distribution and goes through the charged
line center. The total charge Q on the line is assumed to be positive.

By analyzing the situation, we see that the distribution of charge is symmetric with respect to the x-
axis. Thus, we can use similar arguments we used with the charged ring to deduce that the y-component
 is zero on the x-axis because the contributions to Ey from the upper and lower
of the total electric field E
halves of the charged line exactly cancel each other. For illustration purposes, let us first do the full, general
calculation, without assuming the symmetry of the problem.
As with the ring in Example 21.10, we can divide the line into infinitesimal segments of length dy. Each
segment carries charge dQ and acts as a point-charge source of electric field. Charge dQ can be expressed
using uniform linear charge density λ on the line:
Q
λ= = constant
2a
as
Q
dQ = λdy = dy
2a
 at point P that corresponds to the electric field of a point
Each segment dy then creates electric field dE
charge dQ:
 
dE = 1 dQ r̂ = 1 Q dy
r̂,
4π0 r2 4π0 2a (x2 + y 2 )

1
  at point P is then given by
where r̂ is a unit vector pointing from segment dy to point P . The total field E
the integral of all elementary fields dE  over the whole line:
  +a  
 =  = 1 Q dy
E dE 2 + y2)
r̂. (1)
line −a 4π 0 2a (x

In Eq. (1), direction of unit vector r̂ changes as we move to different locations on the line.
Because we are dealing with a two-dimensional problem, we can describe the electric field dE of a typical
line segment dy using its components dEx and dEy along the x- and y-axis, respectively:
 
 
dEx = dE  cos(α)
 
 
dEy = − dE  sin(α).

From Fig. 1, it follows that

x
cos(α) =
(x2
+ y 2 )1/2
y
sin(α) = 2 ,
(x + y 2 )1/2
and, therefore,
 
1 Q xdy
dEx =
4π0 2a (x2 + y 2 )3/2
 
1 Q ydy
dEy = − .
4π0 2a (x2 + y 2 )3/2
Components Ex and Ey of the total field of the whole charged line are then given by integrals
 +a      +a
1 Q xdy 1 Qx dy
Ex = 2 2 3/2
= 2 2 3/2
(2)
−a 4π 0 2a (x + y ) 4π 0 2a −a (x + y )
 +a      +a
1 Q ydy 1 Q ydy
Ey = − 2 2 3/2
=− 2 2 3/2
, (3)
−a 4π0 2a (x + y ) 4π0 2a −a (x + y )

as x is constant along the whole line. We can evaluate the integral in Eq. (2) using a substitution

y = x tan(α)
x
dy = dα
cos2 (α)
limits of α : −A → A
 
a
A = arcsin
(a2 + x2 )1/2
After this substitution, Eq. (2) is transformed to
   +A
1 Qx xdα 1 1
Ex = 2 (α) x3 3/2
=
4π0 2a −A cos [1 + tan2 (α)]
   +A
1 Q dα 1
= 2 (α)
=
4π0 2ax −A cos [1 + tan2 (α)]3/2
   +A
1 Q dα (cos2 (α))3/2
= 2 3/2
=
4π0 2ax −A cos (α) [cos2 (α) + sin2 (α)]
   +A  
1 Q 1 Q +A
= cos(α)dα = [sin(α)]−A =
4π0 2ax −A 4π0 2ax

2
    
1 Q a a
= − − =
4π0 2ax (a2 + x2 )1/2 (a2 + x2 )1/2
    
1 Q 2a 1 Q
= = . (4)
4π0 2ax (a2 + x2 )1/2 4π0 x(a2 + x2 )1/2

Similarly, we can evaluate the integral in Eq. (3) using a substitution

z = x2 + y 2
dz = 2ydy
limits of z : (x2 + a2 ) → (x2 + a2 )

which gives
  (x2 +a2 )
1 Q dz
Ey = − = 0, (5)
4π0 2a (x2 +a2 ) 2z 3/2

 of a charged line at point P shown

as we integrate over a region of zero width. Thus, the total electric field E
in Fig. 1 has only x-component and can be written as
 
 1 Q
E = Ex ı̂ = ı̂. (6)
4π0 x(a2 + x2 )1/2

where ı̂ is the unit vector in the direction of the positive x-axis. Equation (6) holds for a positive charge
Q distributed uniformly around the line; if we instead assume a negative charge −Q, the symmetry of the
problem remains but the x-component of the total electric field points in the direction of the negative x-axis.
As we have already mentioned, the fact that the y-component of the total electric field E  is zero could
be deduced from the symmetry of the problem. Thus, we see that using symmetry, we can often simplify
our calculations. We have to be cautious, however, to use the symmetry only when it is justified by the
geometry of the problem. In the case of a uniformly charged line, if the point P is moved away from the
axis of symmetry, both Ex and Ey are generally non-zero and we have to evaluate integrals (2) and (3) with
appropriate limits of integration!

Note
Let us evaluate the total electric field of a charged line given by Eq. (6) for distances x which are either
much larger or much smaller than the line length 2a.

(1) x  2a
In this limit, we can neglect a2 in the denominator of Eq. (6) relative to x2 and obtain

 (x2a) = Ex,(x2a) ı̂ = 1 Q
E ı̂. (7)
4π0 x2
which is equal to the electric field of a point charge Q situated at the line center. This is the same result we
found for a uniformly charged ring in Example 21.10.
In Fig. 2, the ratio of the exact electric field of a charged line given by Eq. (6) to the approximate electric
field of Eq. (7) is shown as a function of the normalized distance from the line (x/a) (blue curve). As we can
see, for x/a > 10, the two solutions differ by less than 0.5%. Therefore, for distances x > 10a, we can use
the approximate expression (7) with less than 0.5% error.

3
(2) x  2a
In this limit, we can neglect x2 in the denominator of Eq. (6) relative to a2 and obtain

 (x2a) = Ex,(x2a) ı̂ = 1 Q λ
E ı̂ = ı̂. (8)
4π0 xa 2π0 x
Thus, the electric field decreases with distance as ∼ 1/x, slower than the electric field of a point charge. As
we will see in Example 22.6, this distance dependence also describes the electric field of an infinite charged
line with a uniform charge density λ.
In Fig. 2, the ratio of the exact electric field of a charged line given by Eq. (6) to the approximate electric
field of Eq. (8) is shown as a function of the normalized distance from the line (x/a) (red curve). As we can
see, for x/a < 0.1, the two solutions differ by less than 0.5%. Therefore, for distances x < 0.1a, we can use
the approximate expression (8) with less than 0.5% error.
Depending on the level of error we can tolerate in our calculations, we can use the above approximate
solutions Eq. (7) or Eq. (8) instead of the full solution of Eq. (6) to calculate the electric field of a charged
line. This approach (i.e. using a simpler approximate solution instead of a more complex exact solution)
is a common practice in physics which often allows analytical rather than numerical treatment of various
problems and gives more insight into the physical nature of the problem.

1.0

1 Q
0.8 Eapprox =
4πε0 x a
Eexact / Eapprox

0.6

0.4

0.2 1 Q
Eapprox =
4πε0 x2

0.01 0.1 1 10
x/a

Fig. 2. Comparison of the exact electric field Eexact given by Eq. (6) and approximate
electric fields Eapprox given by Eq. (7) (blue curve) or Eq. (8) (red curve) as a function of
the normalized distance x/a from a uniformly charged line of finite length 2a with the linear
charge density λ. The total charge of the line Q = 2aλ.