Question

Eight dipoles of charges of magnitude q are placed inside a cube. Then, the total electric flux coming out of the cube will be
\[\begin{align}
  & A.\,\dfrac{8q}{\varepsilon } \\
 & B.\,\dfrac{16q}{\varepsilon } \\
 & C.\,\dfrac{q}{\varepsilon } \\
 & D.\,0 \\
\end{align}\]

Answer 1

Hint: The electric flux is the rate of flow of the electric field through a given area/surface. The dipole is a pair of positive and negative charges. The total electric flux coming out of the cube will be the sum of charges by epsilon.

Complete answer:
According to Gauss law, closed surfaces of various shapes can surround the charge. The net flux through closed surface equals \[\dfrac{1}{{{\varepsilon }_{0}}}\]times the net charge enclosed by that surface. The product of the closed integral of the electric field and a small area is also the net electric flux. The mathematical representation of the same is given as follows.
\[\Phi =\oint{E.dA}=\dfrac{{{q}_{net}}}{{{\varepsilon }_{0}}}\]
From the statement, we can notice that the charges enclosed by the cube are 8 pairs of +q and –q. So, while computing the flux of the electric field through the surface S, we should consider only these charges.
Now, we will compute the total electric flux coming out of the cube.
The charge present on one dipole is, \[\begin{align}
  & q=(+e)+(-e) \\
 & \therefore q=0 \\
\end{align}\]
So, the charge present on all the dipoles will also be zero.
\[\begin{align}
  & {{q}_{net}}=8\times q \\
 & \Rightarrow {{q}_{net}}=8\times (+q)+(-q) \\
 & \therefore {{q}_{net}}=0 \\
\end{align}\]
Consider the formula for computing the electric flux.
 \[\Phi =\dfrac{{{q}_{net}}}{{{\varepsilon }_{0}}}\]
Substitute the value of the net charge present in the cube obtained in the above equation.
\[\begin{align}
  & \Phi =\dfrac{0}{{{\varepsilon }_{0}}} \\
 & \therefore \Phi =0 \\
\end{align}\]
\[\therefore \] The total electric flux coming out of the cube will be zero.

Thus, option (D) is correct.

Note:
Gauss law connects the electric field with its source. The net flux will be independent of the shape and size of the closed surface (Gaussian surface) surrounding the charge. As the electric field due to many changes is the vector sum of the electric fields produced by the individual charges, the flux through any closed surface can be expressed as the sum of charges by \[{{\varepsilon }_{0}}\].