A charged particle q is placed at the centre O of cube of length L(A B C D E F G H). Another same charge q is placed at a distance L from O. Then the electric flux through ABCD is

Options

A. `(Q)/(epsilon_(0))`
B. `(5Q)/(6epsilon_(0))`
C. `(10Q)/(6epsilon_(0))`
D. zero

Answer

b

Explanation

A point charge is placed at the centre of acube, then net flux through six surfaces `=(Q)/(epsilon_(0))` Since all the surfaces are symmetrical , therefore net flux through one surface due to charge at `P,phi_(1)=0`
`therefor` Net flux due to charge at O is `phi_(2)=5(Q)/(6epsilon_(0))`
`therefore` The total electric flux through 5 faces of the cube other than ABCD is
`phi=phi_(1)+phi_(2)=0+(5Q)/(6epsilon_(0))=(5Q)/(6epsilon_(0))`