What is the electric flux through a cube of side 1 cm which encloses an electric dipole
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according to Gauss theorem electric flux through a closed surface is 1/€ times the change enclosed by the closed surface
i.e. integration of E.ds=q/€
where q is the charge enclosed by the closed surface .
now coming to the question ,
dipole is enclosed in the cube but we know dipole is the system of two same magnitude charge which opposite sign
and take small distance .
it means inside cube charge=0
so electric flux =q/€=0/€=0
i.e. integration of E.ds=q/€
where q is the charge enclosed by the closed surface .
now coming to the question ,
dipole is enclosed in the cube but we know dipole is the system of two same magnitude charge which opposite sign
and take small distance .
it means inside cube charge=0
so electric flux =q/€=0/€=0
abhi178:
I hope this is helpful
Answered by
1
Hey !!
Zero ( as net charge enclosed by the surface is zero ).
DETAILED EXPLANATION
In a cubic surface, the net electric charge will be zero since dipole carries equal and opposite charges. It is observed that the net electric flux through closed surface will be
= Charge enclosed / ε₀
and because the charge enclosed is zero.
electric flux is also zero.
Good luck !!
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