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Q) A conducting sphere of radius R floats half submerged in a liquid dielectric medium of permittivity $\epsilon_1$. The region above the liquid is a gas of permittivity $\epsilon_2$. The total free charge on the sphere is $ Q$. Find a radial inverse-square electric field satisfying all boundary conditions and determine the free, bound, and total charge densities at all points on the surface of the sphere. Formulate an argument to show that this electric field is the actual one.

By the condition that potential should be same at the interface between dielectric $1$ and dielectric $2$, we get that potential as a function of $r$ is same in both media. But my question is that, when we use the second boundary condition, i.e., the normal components of Electric displacements have a difference of the free charge density on the interface, we get something weird.

$D$ inside the conducting sphere is $0$ and the potential is of the form $\frac{A}{r}$ due to spherical symmetry and boundary conditions.

$D1n$(Normal component of Electric displacement in dielectric 1) at $r=R= -\frac{\epsilon_1 A}{r^2}$ $D2n$(Normal component of Electric displacement in dielectric 2) at $r=R = -\frac{\epsilon_2 A}{r^2}$

So, using the conditions, we get:

$D1n - 0 =\sigma$

$D2n - 0 = \sigma$

Which is obviously wrong. So where did I make a mistake?

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  • $\begingroup$ I could help you with finding the electric field intensity. You could construct a gaussian surface of radius $r$ around the sphere, and calculate the magnitude of the electric field intensity (which is radial) to be - $E = \dfrac{Q}{2\pi\epsilon_0(\epsilon_1+ \epsilon_2)r^2}$, for $r > a$. For $r < a$ , it is obvious that $E = 0$. $\endgroup$ – vs_292 Jun 26 '17 at 7:51
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You have made an incorrect assumption about the surface charge density on the conducting sphere.
If you have a parallel plate capacitor with air between the plates there is a uniform charge density on the plates.
What does introducing a dielectric half way into the capacitor do to the surface charge density on the plates?

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  • $\begingroup$ Oh right! How could I not know! $\endgroup$ – Razor Jun 27 '17 at 14:02

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