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n identical droplets are charged to v volt each. If they coalesce to form a single drop, then its potential will be(A) n2/3v (B) n1/3v (C) nv (D) v/n

aniket anand , 11 Years ago
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anser 2 Answers
Arun

Last Activity: 4 Years ago

Dear Aniket
 
 

Let potential f 1 drop = Kq/r =V 

        Volume of n drops = volume of bigger drop 

            n  4/3 pi (r)^3  = 4/3 pi (R)^3

          Therefore ,  R = ( (n)^1/3) r 

also charge remains conserved 

charge on 1 drop = q 

charge on big drop containing n drops = (nq) 

Therefore potential on big drop = K (nq)/ ((n)^1/3)r

                                              =(( n)^2/3) Kq/r

                                             = (( n)^2/3) V

 

Hope it helps

Vikas TU

Last Activity: 4 Years ago

Let r, q and v be the radius, charge the potential of a small drop.  
The total charge on the bigger drop is the sum of all charge on small drop s 
Q = Nq 
The volume of N small drops = Nx 4/3 πr cube  
And for the bigger drop = 4/3 πR cube  
Hence,  
4/3 πR cube = N x 4/3 π r cube  
R = N raise to power 1/3 r  
So the potential on bigger drop  
V = 1/ 4π Ɛo Q/R = 1/4πƐ0 Nq/N raise to power 1/3r = N raise to power 2/3 x 1/ 4πƐo q/r 
V = N raise to power 2/3 x v 
And the capacitance  C = 4πƐoR 
C = 4πƐoN raise to power 1/3 r = N raise to power 1/3 (4πƐor) 
C = N raise to power 1/3c
 

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