Electrostatic potential energy

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Question

27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ________times that of a smaller drop.

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Solution

$Using conservation of volume,$
$Initial Vol = Final Vol$
$\begin{array}{l} \Rightarrow 27 (\frac{4}{3} π r^{3}) = (\frac{4}{3} π R^{3}) \\ \Rightarrow R = 3 r \end{array}$
${Potential of small drop, V}_{s} = k \frac{q}{r} = 10 V (g i v e n)$
${Self- potential energy of a small drop=U}_{S} = (\frac{3}{5} k \frac{q^{2}}{r})$
${Self- potential energy of big drop=U}_{B} = \frac{3}{5} k \frac{Q^{2}}{R}$
$Thus, \frac{U_{B}}{U_{S}} = {(\frac{Q}{q})}^{2} \times \frac{r}{R} = {(\frac{27 q}{q})}^{2} \times \frac{r}{3 r} = 243$
$\Rightarrow U_{B} = 243 U_{S}$

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Electrostatic potential energy

Remember concepts with our Masterclasses.

27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ________times that of a smaller drop.

Using conservation of volume,Initial Vol = Final Vol⇒2743πr3=43πR3⇒R=3rPotential of small drop, Vs=kqr=10 V (given)Self- potential energy of a small drop=US=35kq2rSelf- potential energy of big drop=UB=35kQ2RThus, UBUS=Qq2×rR=27qq2×r3r=243⇒UB=243US

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