a) Do this by dividing the charged rod into 2 equal length segments and treating each segment as
a point change at the center of each segment.
b) How could you improve your estimate?
Solution
A charge Q is uniformly distributed over a line of length L, linear charge density, lemda=total
charge /total length
=Q/L
Q is equal to intergral of lemda dL from limit 0 to l. Then, we get
E = (1/4?*epsilon -not) (lemda)dl/r 2
Let the rod subtend angle 2(theta) at P. Consider element dl at distance l having charge dq=
lemda*dl
l = x tan?
dl = X sec 2
? d?
dE = k(lemda) dl/(x 2
+l 2
)
dE = k(lemda) X sec 2
? d? /(x 2
+(X tan?) 2
)
dE = k(lemda) X sec 2
? d?/(x 2
sec 2
?)
= k(lemda)d?/x
Here, dE Cos ? along x-axis
dE Sin? along y-axis which cancel each other above and below.
a) If we divide a rod 2 equal length so l=L/2,then

E =E1+E2
= electric field due to ntegral from 0 to L/2 +
electric field due to integral from L/2 to L
E= K *lemda *sin?/x
B) we get same electric field when integral from 0 to L.