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Two spherical cavities of radii `a` and `b` are hollowed out from the interior of a neutral conducting sphere of radius `R`. At the center of each cavity , a point charge is placed . Call these charges `q_(a)` and `q_(b)`.
The electric field inside the cavity of radius `a` at a distance `r` from the center of cavity is
image
A. `(1)/( 4 pi epsilon_(0)) (q_(a))/(r^(2))`
B. `(1)/( 4 pi epsilon_(0)) (q_(a))/(r^(2))`
C. `(1)/( 4 pi epsilon_(0)) ( q_(a) + q_(b))/( r^(2))`
D. zero

1 Answer

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by (92.8k points)
 
Best answer
Correct Answer - A
i. `sigma_(A) = ( - q_(a))/(4 pi a^(2)) , sigma_(B) = (-q_(b))/( 4 pi b^(2)) ,sigma_(R) = ( q_(a) + q_(b))/( 4 pi R^(2))`
ii. `E = (1)/( 4 pi epsilon_(0)) ( q_(a) + q_(b))/(r^(2))` ( Fig. S2.56)`
iii. `E_(a) = (1)/( 4 pi epsilon_(0)) (q_(a))/( r^(2))`
image

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