Question

a thin circular plate of radius R is held closing the opening ona. thin hemispherical shell of same radius R.both bodies are made of insulating materials and have surface charge denisty sigma. plate is released keeping shell fixed. find max KE acquired

Answer 1

$K_{\rm max}=\frac{\pi\sigma^2 R^3}{2\epsilon_0}$

Answer 2

We note that the repulsive force on the plate comes solely from the mutual interaction between the charge distributions on the plate and on the hemispherical shell. When the plate (of area $A_p=\pi R^2$ and charge $Q_p=\sigma\pi R^2$) is in contact with the hemisphere (whose curved surface has area $2\pi R^2$ and charge $Q_h=2\pi R^2\sigma$), the mutual interaction energy (per unit $\epsilon_0$) is stored in the configuration. When the plate is released, all of this interaction energy is converted into kinetic energy of the plate as it moves far away (where the mutual interaction vanishes).

A useful shortcut is to note that had the two parts (plate + hemisphere) been put together they would form a full spherical shell of radius $R$ with uniform surface charge density $\sigma$. The potential (on the surface) of a full charged spherical shell is known to be

$$V_{\rm sphere}=\frac{\sigma R}{\epsilon_0}\,.$$

Now, by symmetry one may argue that for the hemisphere the potential on the flat face (the base) is exactly one–half of that, i.e.

$$V_{\rm hemi}=\frac{\sigma R}{2\epsilon_0}\,.$$

(One can show by direct integration over the hemisphere that the potential is uniform over its base.)

Thus the interaction energy between the plate and the hemisphere is given by

$$U_{\rm int}=Q_p\, V_{\rm hemi}=\sigma\pi R^2\,\biggl(\frac{\sigma R}{2\epsilon_0}\biggr)=\frac{\pi\sigma^2 R^3}{2\epsilon_0}\,.$$

Since when the plate is moved to infinity the interaction energy becomes zero, the maximum kinetic energy gained by the plate is equal to the decrease in potential energy:

$$K_{\rm max}=\frac{\pi\sigma^2 R^3}{2\epsilon_0}\,.$$

Below is a simple mermaid diagram summarizing the geometry:

Summary of Answers

• Explanation:
  The potential on the base of a uniformly charged hemisphere is $V=\sigma R/(2\epsilon_0)$. Multiplying by the charge on the plate $Q=\sigma\pi R^2$ gives the interaction energy $U_{\rm int}=\pi\sigma^2R^3/(2\epsilon_0)$. This energy becomes the kinetic energy when the plate is moved to infinity.