Between two infinitely long wires having linear charge densities `lambda` and ` - lambda` , there are two points A and B as shown in the figure. The amount of work done by the electric field in moving a point charge `q_(0)` from A to B is equal to :

Options

A. `(lambda q_(0))/( 2pi epsi_(0))` In 2
B. `-(2lambda q_(0))/(pi espi_(0)) `In2
C. `( 2 lambda q_(0))/(pi epsi_(0))` IN 2
D. `(lambda q_(0))/(pi epsi_(0)) ` In 2

Answer

D

Explanation

Electric field at P is given by
`E= lamda/(2 pi e_0) (1/x+1/(3a-x))`
Work done is given by

`W= int_a^(2a) q0 Edx`
`=(lamda q_0)/(2 pi e_0) [ int_a ^(2a) (dx)/x+ int _a^(2a) (dx)/(3a-x)] (lamda q_0)/ (pi e_0) In 2`