Answer to Question #139329 in Electricity and Magnetism for Diksha

Question

A very thing charged rod is placed along x axis with its midpoint at the origin.The charge per unit length of rod is Lemda c/m,the electric field vector at y=-a on u axis is

Answer 1

The Gauss's theorem is given by a equation:

$div(E) = -4\pi\lambda$

$\int _V div (E) = \int_S E ds$ - Gauss's theorem

$\int_V div (E) = -4\pi\lambda \int_L dl = -4\pi \lambda L$

$\int_S Eds = 2\pi |y|LE(x,y)$

If we've integrated it and evaluated E(x,y) then :

$E(x,a) = -2\frac{\lambda}{|a|}$