# How to deal with Dipole in a cavity?

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 -

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.

• 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. Commented Jul 2, 2021 at 12:31
• Assume the separation negligible and cavity to be large. Commented Jul 2, 2021 at 12:32
• 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. Commented Jul 2, 2021 at 13:27
• yes, but I have mentioned in the question about the electric field components of the dipole, what about them ? Commented Jul 2, 2021 at 13:34
• I have got some contradiction while considering the electric field components and that is creating issue Commented Jul 2, 2021 at 13:41