Question

(a) Use Gauss’ law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λ C/m.
(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
(c) Find the work done in bringing a charge q from perpendicular distance r1 to r2 (r2 > r1).

Answer 1

Electric field (E) due to a straight uniformly charged infinite line is 2Kλ/R

Drawn the required graph

work done = 2qKλ(r2 -r1)/R

•consider a lineear wire with uniform charge density λ C/m

•Consider a cylindrical Gaussian surface of length l & radius r to find

electric feild (E) at P

•By Gauss law ,

•The electric flux linked with an enclosed body of any shape is always equal to 1/E• times the total charge enclosed by the body

i.e,

∫E.ds = q(enclosed)/E•

∫E.ds + ∫E.ds + ∫E.ds = λL/E•

(top) (bottom) (curved)

At top & bottom E is perpendicular to ds => E.ds = 0

∫E.ds = λL/E•

(curved)

Since E is uniform

E∫ds = λL/E•

E(2πRL) = λL/E•

E = λ/2πE•R

E = 2λ/4πE•R

E = 2Kλ/R

•electric field (E) due to a straight uniformly charged infinite line is 2Kλ/R

c) work done = qv = qEd

= 2qKλ(r2 -r1)/R