what is electric flux through a cube of sides 1 cm which encloses an electric dipole .
Answers
Answer:
What is the electric flux through a cube of side 1 cm whose encloses an electric dipole? Since net charge enclosed in the surfaces bound by the cube is zero(dipole consists of equal and opposite charges), according to the Gauss Law, it can be claimed that the net flux through the cube is zero.
Answer:
It will be zero. Since net charge of a dipole is zero. Therefore, flux through the cube will be zero.
Explanation:
Since, according to the Gauss’ law of electrostatics, electric flux through any closed surface is given by,
phiE = ointEcdotdS = q/varepsilon0 ... (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.
Since net charge enclosed in the surfaces bound by the cube is zero(dipole consists of equal and opposite charges), according to the Gauss Law, it can be claimed that the net flux through the cube is zero.