Imagine if we have an electric dipole inside a cube (a conductor). The net electric charge inside the cube is zero. Hence, using Gauss's law, the net electric flux coming out of the cube must be zero. This way the net electric field inside the cube should also be zero. Can somebody draw a diagram of the zero electric fields here because I am not able to imagine it?

  • $\begingroup$ Who said that a dipole produces zero field? $\endgroup$ Sep 10, 2021 at 6:41
  • $\begingroup$ I'm voting to close this question because imagination is opinion-based. $\endgroup$ Sep 10, 2021 at 6:46
  • 3
    $\begingroup$ Visualization of a fact can't be opinion-based. $\endgroup$
    – Swami
    Sep 10, 2021 at 7:10
  • $\begingroup$ What do you mean by the net electric field? Please give an expression of what needs to be plotted. $\endgroup$
    – my2cts
    Sep 10, 2021 at 21:05

2 Answers 2


The key word in your post is net. Only the net flux must be zero, which means as much of the field must be pointing in as is pointing out of the surface. Look at this visualisation from Wikipedia

dipole field lines

If you draw any box around this dipole, you will notice as many field lines going into the box as are coming out. That's all Gauss's law tells us.


The Gauss law in electrostatics is stated as: $$\int_{\text{Gauss surface}}\mathbf{E} \cdot d\mathbf s = \frac{q}{\epsilon_0}$$

where $\mathbf{E}$ is the electric field, $q$ is the charge enclosed by the surface over which the integral is evaluated and $\epsilon_0$ is the permittivity of free space.

The integral $\int \mathbf{E} \cdot d\mathbf{s}$ is equal to the electric flux through the enclosing surface. In the case you describe, the cube encloses a dipole. So the net electric flux through the surface of the cube evaluates to zero according to Gauss law. But this does not imply that the electric field is zero due to the dipole. In fact, a dipole does have an electric field associated with it, which is given by $${\mathbf E}_{\text{dipole}} = \frac{\vec{p}}{r^3}. \qquad \square$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.