Consider a conducting solid sphere having a spherical cavity. Centre of the cavity and centre of the sphere coincide. A point sized dipole [negligible separation and cavity is large] is placed at centre of the cavity.

Now I know that Flux inside the cavity and outside the sphere is zero because the total charge enclosed $Q_{enclosed}$ is zero. $$Gauss's~Law:~~~~~ Electric~Flux = \oint E.∂s = \frac{Q_{enclosed}}{\epsilon_o}$$

But as we know Electric field for dipole has radial and tangential components as shown here - enter image description here

Now consider a spherical Gaussian Surface just below the inner surface of the cavity. So the radial component of Electric field is $||$ to the area vector and but tangential component is ⟂ to the area vector of the Gaussian surface, which suggests Flux is non-zero. This means there are some more charges other than dipole. This is contradictory to the earlier statement [that flux was Zero]. But how is this possible ?

1. What are charges on the inner and outer surfaces of the sphere ?

For outer surface, I think charges should be induced[where net charges would be zero]. But don't know the truth and confused due to the contradiction.

2. Does the Gauss's Law not apply as the way I have thought here ?

Now here I know that we may not be able to take spherical Gaussian surface because the Electric field due to dipole is not constant in magnitude and not even perpendicular everywhere for the surface, but then what should be the correct way to go about when I have limited knowledge of high school math and physics? If I need to know some new concept, then mention it too.

  • $\begingroup$ The sentence "point-sized dipole" is flawed since the very definition a dipole requires separation distance between the two charges with specific orientation, realistically if you really really stick two opposite charges in the same point, they would annihilate leaving zero electric field, please clarify further your question. $\endgroup$
    – AGawish
    Commented Jul 2, 2021 at 12:31
  • $\begingroup$ Assume the separation negligible and cavity to be large. $\endgroup$ Commented Jul 2, 2021 at 12:32
  • $\begingroup$ Okay assuming just really small distance, by Gauss's law the flux is still zero, apply a spherical Gaussian surface on each charge separately then sum: $D_1=q/4\pi r^2$ and $D_2=-q/4\pi r^2$ thus: $D_t=D1+D2=0$, please correct me if I misunderstood your question. $\endgroup$
    – AGawish
    Commented Jul 2, 2021 at 13:27
  • $\begingroup$ yes, but I have mentioned in the question about the electric field components of the dipole, what about them ? $\endgroup$ Commented Jul 2, 2021 at 13:34
  • $\begingroup$ I have got some contradiction while considering the electric field components and that is creating issue $\endgroup$ Commented Jul 2, 2021 at 13:41

1 Answer 1


Sorry, I do not understand the problem. We are using gauss law, on the spherical surface just inside the cavity, yes? So flux is Q(enclosed)/epsilon using the Gauss law, which gives 0 because the charge configuration is a dipole.

Then you are trying to find flux in another way, using Integral of (E.dA) over the whole sphere, right? So make the diagram of the sphere and visualize the directions of the small area vectors on all the points, pointing radially outward according to the accepted convention. Then draw the electric field lines of the dipole, using the formula you have. You will see that yes, there is flux at points, but there is +ve flux at some and -ve flux at some and they cancel out. Remember, flux is a scalar but it is dot product of dA and E at all the points, so you have to consider the sign of it while summing it over the sphere.

The flux is thus zero through both methods.

  • $\begingroup$ Okay the field lines do give a bigger picture now. So flux confusion is cleared. Do write about the induced charges. $\endgroup$ Commented Jul 2, 2021 at 14:06

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