JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 22)
Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b, and carries charge
Q. At its centre is a dipole $$\overrightarrow P $$
as shown. In this case :
_12th_April_Morning_Slot_en_22_1.png)
surface charge density on the inner surface is uniform and equal to $${{\left( {Q/2} \right)} \over {4\pi {a^2}}}$$
surface charge density on the inner surface of the shell is zero everywhere
surface charge density on the outer surface depends on $$\left| {\overrightarrow P } \right|$$
electric field outside the shell is the same as that of a point charge at the centre of the shell
Explanation
Total charge of dipole = 0, so charge induced on
outside surface = 0.
But due to non uniform electric field of dipole, the charge induced on inner surface is non zero and non uniform.
So, for any observer outside the shell, the resultant electric field is due to Q uniformly distributed on outer surface only and it is equal to.
$$E = {{KQ} \over {{r^2}}}$$
But due to non uniform electric field of dipole, the charge induced on inner surface is non zero and non uniform.
So, for any observer outside the shell, the resultant electric field is due to Q uniformly distributed on outer surface only and it is equal to.
$$E = {{KQ} \over {{r^2}}}$$
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