Lets consider a system of two opposite charges separated by a certain distance (dipole), if we ask what is the net charge for this system? the answer would be zero. The net charge (what I have come to know until now) is said to be addition of sign of charges irrespective of the distance between the charges.

Lets consider the following case:
According to Gauss's law the total electric flux through a closed surface enclosing a charge is equal to $\frac{1}{\epsilon_0}$ times the magnitude of charge enclosed.

In the below figure we have a $+q$ charge enclosed in the guassian surface, so there is a flux linked with the guassian surface, which will be equal to $\frac{1}{\epsilon_0}$ times the $+q$ charge.

enter image description here

Now, replace the $+q$ charge with the dipole in which charges are separated by a certain distance.

Magnitude of electric field ($E$) due to an electric dipole at a distance $r$ from its centre in a direction making an angle $\theta$ with the dipole is given by the equation,$$E=\frac{1}{4\pi\epsilon}.\frac{p\sqrt{3\cos^2\theta+1}}{r^3}$$
where, $p=2aq$ ($2a$ is the distance of separation of the charges $q$).

Electric field around the dipole will be zero if and only if distance of separation of the dipole charges is zero.

Although we can't expect field lines to be perpendicular to the surface, we can expect some contribution due to the electric field components (perpendicular to the surface) of the enclosed dipole (with charges separated by a distance). From the dipole equation it follows that there exists electric field around dipole when charges are separated by distance, so there is flux through the surface which implies that there is charge enclosed in the surface (from gauss law). But net charge of dipole is said to be zero. By gauss law, net charge is not zero. Then what is net charge? From gauss law we can directly find the $q$. Can we say this as net charge?

Then, is it that net charge can't be said directly just by adding the signs and is it a function of electric field?

If I have misunderstood, pardon me and explain.

Sometimes the problem might not be understood well due to communication error, if any correction required, comment on the part.


Every field line from the dipole must begin on one charge and end on the other. That means that if a field line passes out of your surface it must pass back in through it again. The surface as a whole will have the same number of field lines going in as out, so the net flux through the surface will be zero.

| cite | improve this answer | |
  • $\begingroup$ @JohnRennie: Thank you for the answer. Gauss's law is appreciable and it seems that charge is the function of flux. $\endgroup$ – Immortal Player Feb 25 '14 at 17:32

A non-zero electric field on your Gaussian surface does not mean a non-zero flux. There are positive and negative contributions to the flux due to the electric field pointing in & out in different places. Your cosine term confirms this.

Apparently the contributions must cancel in this case since the net enclosed charge is zero.

| cite | improve this answer | |

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.