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N small conducting liquid droplets, each of radius r, are charged to a potential V each. These droplets coalesce to form a single large drop without any charge leakage. Find the potential of the large drop. |
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Answer» SOLUTION :Let N small conducting liquid droplets each of radius r. CARRY a charge q so as to have a potential V) then `V=q/c=(q)/(4PI epsi_(0) r)` If the radius of single bigger drop formed due to coalescence of these droplets be r., then, `4/3 pi Nr^(3) =N xx 4/3 pi r^(3) rArr r=(N)^(1/3).r` and charge on this drop q.=Nq Potential of single drop `V.=(q.)/(C.)=(q.)/(4pi epsi_(0) r.) =(Nq)/(4pi epsi_(0) (N)^(1/3) .r) =(N)^(2/3) . (q)/(4pi epsi_(0) r)=N^(2/3).V` |
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