By using Gauss’s law:
Where; E is the electrostatic field
Q is the total charge enclosed by the surface
is the 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. Q=0
The electric flux through a cube of side 1 cm which encloses an electric dipole will be zero, as net charge enclosed by a cube is zero.
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