A charged particle q is placed at the centre O of cube of length L (ABCDEFGH). Another same charge q is placed at a distance L from Then the electrons flux through ABCD is:

Question

A charged particle q is placed at the centre O of cube of length L (ABCDEFGH). Another same charge q is placed at a distance L from Then the electrons flux through ABCD is: (A)  q/4π ε 0L  (B) zero (C)  q/2π ε 0L  (D)  q/3π ε 0L

Tardigrade Admin · Accepted Answer

Electric flux φ E= ointE.ds ds is the areal vector and E is the electric field vector. Therefore, E.ds=Eds cos θ θ is angle between E and ds But θ =90° Thus, φ E=0

Q. A charged particle q is placed at the centre $O$ of cube of length L (ABCDEFGH). Another same charge q is placed at a distance L from Then the electrons flux through ABCD is:

$q /4 π ε_{0} L$

zero

$q /2 π ε_{0} L$

$q /3 π ε_{0} L$

Electric flux $ϕ_{E} = \oint E . d s$
$d s$ is the areal vector and $E$ is the electric field vector. Therefore, $E . d s = E d s cos θ$ $θ$ is angle between $E$ and $d s$ But $θ = 90^{\circ}$ Thus, $ϕ_{E} = 0$

Q. A charged particle q is placed at the centre O of cube of length L (ABCDEFGH). Another same charge q is placed at a distance L from Then the electrons flux through ABCD is:

q/4πε0​L

zero

q/2πε0​L

q/3πε0​L

Electric flux ϕE​=∮E.ds ds is the areal vector and E is the electric field vector. Therefore, E.ds=Edscosθ θ is angle between E and ds But θ=90∘ Thus, ϕE​=0

Q. A charged particle q is placed at the centre $O$ of cube of length L (ABCDEFGH). Another same charge q is placed at a distance L from Then the electrons flux through ABCD is:

$q /4 π ε_{0} L$

$q /2 π ε_{0} L$

$q /3 π ε_{0} L$

Electric flux $ϕ_{E} = \oint E . d s$
$d s$ is the areal vector and $E$ is the electric field vector. Therefore, $E . d s = E d s cos θ$ $θ$ is angle between $E$ and $d s$ But $θ = 90^{\circ}$ Thus, $ϕ_{E} = 0$