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What is the electric flux through a cube of side 1 cm which encloses an electric dipole?

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#### Solution

As per the Gauss's law of electrostatics, electric flux through a closed surface is given by

`phi_E=ointvecE.vec(ds)=Q/epsilon_0 `

where

*E =* Electrostatic field

*Q* = Total charge enclosed by the surface

∮E=∮E→.ds→=0ε0=0" role="presentation" style="position: relative;">∮E=∮E→.ds→=0ε0=0

∈_{0} = Absolute electric permittivity of free space

In the given case, cube encloses an electric dipole. Therefore, the total charge enclosed by the cube is zero, i.e. *Q* = 0.

Therefore, from (i), we have

`phi_E=ointvecE.vec(ds)=0/epsilon_0=0`

Concept: Electric Flux

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