Gauss LawEasy
Question
-E = (PE)final - (PE)initial = −K/3R + K/2RA 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 BCFG is
Options
A.q/4π∈0L
B.zero
C.q/2π∈0L
D.q/3π∈0L
Solution
The flux for both the charges exactly cancels the effect of each other
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