Consider a charged water droplet of radius R and charge Q uniformly distributed over its surface. From energy considerations, obtain an expression for the minimum charge that permits the droplet to spontaneously split into (i) two droplets of equal size, (ii) three droplets of equal size. Show that both the processes are energetical ly allowed for a droplet of charge Q = 5 x 10^-10 C and radius R = 1 mm. Will these processes actually occur if the dielectric breakdown of the surrounding medium takes place at a field strength of 3 x 10⁶ Vm^-1 ?

 

could you please post the detailed soln

INPhO 2000 if i amnt mistaken......dekho..when i first saw the Q. this is all that i cud figure out.....

assume that there is a sphere and we are depositing charge over it uniformly on its surfare.....

assume there to be a charge q already and we are adding a charge of dq......work done to do so will be

dW = Kqdq/R..........integrating from 0 to Q

and W = KQ²/2R

this is stored as potential energy of the system

so when u splitt into 2 equal spheres.i assume that the total volume is unaltered......so

4/3pi R³ = 2 * 4/3pi r³

or r= R/2^1/3......

and the energy of  first sphere = E = KQ²/R + TA₁........(A1 and A2 are the surface areas of the first and the second 2 spheres)(TA= energy due to surface tension)

where T= surface tension of water...(given in the value list and not used in any other problem)

and energy of the smaller sphere = U = KQ²/4(R/2^1/3) + TA₂

and for this change to be feasible we need E = 2U

equating which and solving we get Q and for 3 bubbles i guess u shud do it using r= R/3^1/3

and E = 3U....

third part main mere palle nahi padh rahe(dielectric break down)

but i have given a very brief method for the first 2 parts.....

correct if wrong....

rate if useful......