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