Two metallic identical hemispheres of very large radius R are arranged in such a way that there is a very small gap between them. The hemispheres are charged with −Q and 3Q (Q>0). The flat surface of the hemisphere may be assumed as infinite surface. Then the electric field strength in the air gap between the hemispheres is

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$\frac{2 Q}{3 π ε_{o} R^{2}}$

$\frac{Q}{π ε_{o} R^{2}}$

$\frac{2 Q}{π ε_{o} R^{2}}$

$\frac{Q}{2 π ε_{o} R^{2}}$

answer is C.

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Detailed Solution

The charges will be distributed over the surface of the hemispheres so that the field inside them is equal to zero. To do this: (a) the same charges must be placed on spherical surfaces; b) charges of equal magnitude but opposite sign on flat surfaces of the same size
(see figure). From the law of conservation of charge, we get for charges $q_{1} a n d q_{2}$
Question Image

$q_{1} - q_{2} = - Q$ and $q_{1} + q_{2} = 3 Q$
Hence $q_{1} = Q a n d q_{2} = 2 Q$
Hence the electric field in the required region is $E = \frac{2 Q}{π ε_{o} R^{2}}$