Two infinite line charges each having uniform charges density `lamdba` .pass through the mid points of two pairs of opposite faces of a cube of edge `L` as shown in figure .The modulue of the total electric flux due to the both the line charges through the face

Options

A. `CDHG `is `(lambdal)/(2epsilon_(0))`
B. `AEHD`is `(lambdal)/(3epsilon_(0))`
C. `ABCD` is `(lambdal)/(2epsilon_(0))`
D. `CDHG` is `(lambdal)/(3epsilon_(0))`

Answer

C

Explanation

the charge inside the cube due to each line is `lambdaL`.Flux due to horizontal wire through the faces `AEDH`,`BCGF` is zero as the electric field produced will be parallel to these faces.
But the total flux due to this line is `(lambdaL)/(epsilon_(0))` as it is equally distributed among the remaining four faces .Similarly flux due to vertical coire through the faces `DHGC,AEFB` is zero.
,.Flux through `CDHG=O+(lambdaL)/(4epsilon_(0))=(lambdaL)/(4epsilon_(0))`
:.Flux through `AEHD=O+(lambdaL)/(4epsilon_(0))=(lambdaL)/(4epsilon_(0))`
Flux through `ABCD=(lambdaL)/(4epsilon_(0))=(lambdaL)/(4epsilon_(0))=(lambdaL)/(2epsilon_(0))`