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In this article we would find the electric field (https://physicscatalyst.com/elec/electric-field.php) due to a line charge. By line charge we mean that charge is distributed along the one dimensional curve or line in space. Here in this article we would find electric field due to finite line charge derivation for two cases
- electric field due to finite line charge at equatorial point
- electric field due to a line of charge on axis
We would be doing all the derivations without Gauss’s Law (https://physicscatalyst.com/elec/gauss- law.php). At the same time we must be aware of the concept of charge density (https://physicscatalyst.com/article/charge-density-formulas-and-solved-example/). Here since the charge is distributed over the line we will deal with linear charge density given by formula
Electric field due to finite line charge at perpendicular distance
Positive charge is distributed uniformly along y-axis between and. We have to find electric field due to line charge at point on the x-axis at a distance from the origin.
Electric field due to Line Charge
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l
λ = N/m
q l
Q y = −a y = +a P x
Figure below shows the situation and we have to find the electric field at point due to continuous distribution of charge.
We can solve the problem by mentally dividing the line segment of length into differential parts of length , with each of these lengths carrying a differential amount of charge. If is the linear charge density then,
Where
P
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2 a dy dq λ
dq = λdy
λ = 2 Qa
From the symmetry of the configuration, the component of parallel to the line charge distribution is zero. When a positive test charge is placed at P, the upper half of the charge line apply force on it along downward direction, while the lower half applies a force of equal magnitude in upward direction. This, the top and lower parts of the segment contribute equally to the total field at P, according to symmetry.
So, from this symmetry argument
d E⃗x =
4 πϵ 0
Q
2 a
xdy (x^2 + y^2 )3/
d E⃗y = (^4) πϵ^1 0
Q
2 a
ydy (x^2 + y^2 )3/
d E⃗
The net electric field at point due to this line segment is perpendicular to the line segment and can be found if we calculate the sum of the fields from all segments along the line. For this, we must integrate from to. SO,
Or,
Integral Evaluation We will now evaluate the integral
We will now solve for this integral
d E⃗y = 0
P
y = −a y = +a
Ex = ∫
+a −a
dEx
Ex = ∫
+a −a
4 πϵ 0
Q
2 a
xdy (x^2 + y^2 )3/
Ex = ∫
+a −a
4 πϵ 0
Q
2 a
xdy (x^2 + y^2 )3/
xdy (x^2 + y^2 )3/
y = x tan θ
dy = xsec^2 θdθ
x^2 4 πϵ 0
Q
2 a
sec^2 θdθ (x^2 + x^2 tan^2 θ)3/
I = ∫ sec
(^2) θdθ (x^2 + x^2 tan^2 θ)3/
I = ∫ sec = ∫
(^2) θdθ (x^2 + x^2 tan^2 θ)3/
sec^2 θdθ [x^2 (1 + tan^2 θ)]3/
I = ∫
sec^2 θdθ x^3 (sec^2 θ)3/
∵ 1 + tan^2 θ = sec^2 θ
I = 1 ∫
x^3
sec^2 θ sec^3 θ
In vector form above electric field is written as
points away from the line charge if charge on rod is positive and towards the line charge if charge on it is negative.
From the above equation,
If then would be equal to as on denominator can be ignored and the field becomes that of an electric charge (https://physicscatalyst.com/elec/electrostatics-electric- charges.php)
So, if point is far away from the line charge such that is negligible in comparison to , then field at this point would be same as that of a point charge. In this case when our line segment is very long or the field point is very close to it so that , then equation
can be rewritten as,
If then. In this case field is given by equation
This is the electric field due to an infinite long line of charge. AT any point at a perpendicular distance from the line in any direction, the magnitude of field is given by relation
Ex = (^4) πϵ^1 [( − )] 0
Q
2 a
x
a √x^2 + a^2
(−a) √x^2 + a^2
Ex = (^4) πϵ^1 [ ] 0
Q
2 a
x
2 a √x^2 + a^2
Ex = (^4) πϵ^1 0
Q
x√x^2 + a^2
E⃗ = 1 ^i 4 πϵ 0
Q
x√x^2 + a^2
E⃗
x >> a x^2 + a^2 r^2 a
E⃗ = 1 ^i 4 πϵ 0
Q
x^2 P a x P a >> x
E⃗ = 1 ^i 4 πϵ 0
Q
x√x^2 + a^2
E⃗ = 1 ^i 2 πϵ 0
λ
x√( x a 22 ) + 1
a >> x x → 0
2 a^2
E⃗ = λ ^i 2 πϵ 0 x
P r
It is important to note here that this field is proportional to instead of as it was in case of point charge.
Electric Field due to a Line of Charge at axial point
In this case our problem is to find the electric field due to uniformly charged rod of length at a point at distance from one end of the rod assuming that is the total positive charge on the rod as shown below in the figure.
Let's start by defining a coordinate system for our problem. We'll use an x-coordinate system because we're dealing with a one-dimensional case. Let's suppose the left side of the rod is at the origin. Our infinitesimal charge element will be distance away from the origin in this coordinate system, and the length of this infinitesimal charge element will be as shown in the figure.
Charge on the infinitesimal length element is
This can be regarded as a point charge, hence electric field due to this element at point is given by equation,
E =
λ 2 πϵ 0 r 1/r 1/r^2
L P
r Q
x dx
dq dx
dq = QLdx
dq dE P
