Question 764: Electrostats & Capacitance

Question

Assuming the drops to be spherical, 27 identical drops of mercury are charged simultaneously to the same potential of 20 volt. If all the charged drops are made to combine to form one big drop, then potential of big drop will be

Accepted Answer

Potential at surface of a sphere is,
$$ V=K \frac{q}{r}=20 \mathrm{~V} $$
where, $q$ and $r$ are the charge and radius of the small drop respectively.
As the volume of 27 small drops equals volume of the large drop,
$$ 27\left(\frac{4}{3} \pi r^3ight)=\frac{4}{3} \pi R^3 $$
where, $R$ is radius of large drop.
$$ 	herefore \quad R=3 r $$
The total charge on large drop is $\mathrm{Q}=27 \mathrm{q}$ Hence, potential at surface of this drop is
$$ \begin{aligned} & V^{\prime}=\frac{K Q}{R}=K \frac{(27 q)}{(3 r)}=9 \frac{K q}{r}=9 	imes 20 \ 	herefore & V^{\prime}=180 V \end{aligned} $$

Answer

90 V

Answer

180 V

Answer

270 V

Answer

360 V

Electrostats & Capacitance

Assuming the drops to be spherical, 27 identical drops of mercury are charged simultaneously to the same potential of 20 volt. If all the charged drops are made to combine to form one big drop, then potential of big drop will be

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