Electric flux through 2 face
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Explanation:
As along the faces ABCD, AEHD and DCGH, he electric field lines are parallel to the surface of these faces the electric flux passing through theses faces is zero, as there is no component of electric field perpendicular to these face.
Faces AEFB, EFGH, and BFGC ar symmetrically placed with respect to the the charge, the electric flux through these faces is equal.
Therefore,
Total electric flux across the cube = 3 × electric flux across face EFGH
=
ϵ
Q
enclosed
( by Gauss's Law)
8ϵ
q
Therefore,
flux through face EFGH =
24ϵ
q
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