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Asked by dibyajitbhattacharyya2 | 02 Jul, 2021, 03:10: PM
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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
Answered by Thiyagarajan K | 02 Jul, 2021, 08:29: PM
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