1. By Squadron Leader Zahid Mir CS&IT Department , Superior University
PHY-AP Electric Field due to Line of Charge By
Squadron Leader Zahid Mir
CS&IT Department , Superior University

2. Line of Charge Consider a +ve charge Q is distributed
x-axis 2a P qo + r x
y-axis
z-axis Ѳ dy dQ y
Consider a +ve charge Q is distributed
uniformly along a line with length ‘2a’
lying along y-axis between y= -a and y=+a.
We want to calculate electric field at a point
‘P’ on the x-axis at a distance ‘x’ from the origin.
We divide the line into infinitesimal segments
each of which acts as a point charge ’dQ’.
Let the length of a typical segment at height
‘y’ be ‘dy’.
Linear Charge Density  = Q = Total Charge
2a Total Length

3. Line of Charge Charge density for the segment ‘dy’ is: dQ =  dy
dQ = (Q/2a) dy
The magnitude of electric field ‘dE’ at point ‘P’
dE = ¼o dQ/r2
dE = ¼o Q dy/ 2a(x2 + y2)
Electric field in terms of its x- and y-components
dEx = dE Cos & dEy = dE Sin
Here
Cos = x/r = x /(x2 + y2)1/2
Sin = y/r = y/(x2 + y2)1/2

4. Line of Charge (cont) Symmetry Argument
dEy2
dE2 Ѳ Ѳ O
X-axis Ѳ Ѳ
dE1 r
dEy1 y2
Symmetry Argument
For every charge element ‘dQ’ located at position +y1, there is another charge
element ‘dQ’ located at –y2.
When we add the electric fields due to the charge element at +y1 & -y2 we find the
y-components have equal magnitudes but opposite directions; So their sum is zero.
dEy = 0
Hence total contribution of electric field is along x-component.

5. Line of Charge (cont) dEx = dE Cos Here dE = ¼o Q dy/ 2a(x2 + y2)
Cos = x/r = x /(x2 + y2)1/2
dEx = ¼o Q dy/ 2a(x2 + y2) {x /(x2 + y2)1/2}
Ex =  x dy
4o (x2 + y2)3/2
here tan  = y/x
Ex = 
2o x(1+ x2/a2)1/2
Ex = 
2o x a -a

6. The field magnitude depends only on the distance of point ‘P’ from the line of charge.
2o r
E α /r
The direction of E is radially outward from the line if  is +ve and radially inward if  is –ve.