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

1.	What is the electric flux through a cube of side 1 cm which encloses an electric dipole?
Answer» Since, according to the Gauss’ law of electrostatics, electric flux through any closed surface is given by, \(\phi_E = \oint E \cdot S = q/ \varepsilon_0\) ....(i) Where, e = electrostatic field q = total charge enclosed by the surface ϵ₀ = absolute electric permittivity of free space so, in the given case, cube encloses an electric dipole. Therefore, the net charge enclosed within the cube is zero. i.e. q = 0. Therefore, from Eq. (i), we have ϕE = q/ϵ₀ = 0 i.e. electric flux is zero.

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

Answer»

Since, according to the Gauss’ law of electrostatics, electric flux through any closed surface is given by,

\(\phi_E = \oint E \cdot S = q/ \varepsilon_0\) ....(i)

Where, e = electrostatic field

q = total charge enclosed by the surface

ϵ₀ = absolute electric permittivity of free space so, in the given case, cube encloses an electric dipole. Therefore, the net charge enclosed within the cube is zero. i.e. q = 0.

Therefore, from Eq. (i), we have ϕE = q/ϵ₀ = 0 i.e. electric flux is zero.

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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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