Question

If N drops of same size each having the same charge, coalesce to form a bigger drop. How will the following vary with respect to single small drop? (3) (i) Total charge on bigger drop (ii) Potential on the bigger drop (iii) Capacitance​

Answer 1

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

Answer 2

r= radius

q= charge

v= potential  

total charge on the bigger drop= sum of all  the charge on small drops 

Q = Nq 

volume of N small drops= V = N x 4/3 πr³ 

volume of the bigger drop = 4/3 πR³  

so, 

4/3 πR³ = N x 4/3 π r³  

R = N¹/₃r 

 potential on bigger drop  = V = 1/ 4π Ɛo Q/R

= 1/4πƐ0 Nq/N¹/₃r 

 = N raise to power 2/3 x 1/ 4πƐo q/r 

V = N²/₃ x v 

capacitance  C = 4πƐoR 

C = 4πƐoN¹/₃r 

C= N¹/₃(4πƐor) 

C = N¹/₃C