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MCQ
Asked by dibyajitbhattacharyya2 | 02 Jul, 2021, 15:10: PM
Expert Answer
If charged water droplets are accumulated in a conducting spherical container and half filled ,
then all the charges will appear in outer surface of conducting spherical container.
Potential of spherical container , VS = K × (Q / R ) ......................(1)
where K = 1/(4πεo ) , is Coulomb's constant, Q is total charge on the outer sutface of spherical container
and R is radius of spherical container.
Total charge Q = N × q ........................... (2)
where N is number of droplets accumulated in half filled spherical container and q is charge of each droplet .
Number of droplets, N = volume of water / volume of droplet
Number of droplets, N = [ (2/3) π R3 ] / [ (4/3) π r3 ] = R3 / ( 2 r3 ) .............................(3)
where r is radius of droplet.
At the instant of falling from tap, droplet is at a potential V
Hence we have , V = K × ( q / r ) or q = ( V × r ) / K ........................(4)
Using Eqn.(3) and eqn.(4) , we get total charge Q by rewriting eqn.(2)
Q = [ R3 / ( 2 r3 ) ] × [ ( V × r ) / K ] = ( R3 V ) / ( 2 K r2 ) ...........................(5)
By substituting total charge Q using eqn.(5) , we get potential of spherical container VS from eqn.(1) as
VS = (1/2) ( R2 / r2 ) V
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