Question

$1000$ identical drops of mercury are charged to the same potential at $10\text{V}$. All these are now combined to make one large drop. The potential of the large drop is

• A. $100\text{V}$
• B. $10\text{V}$
• C. $1000\text{V}$
• D. $90\text{V}$

Answer 1

The correct option is C $1000\text{V}$

Let the radius of smaller drop be $r$ & charge on each drop when charged to $10\text{V}$ be $q$.

Now, capacitance of each drop, $$C=4\pi {\epsilon }_{0}r$$As $q=CV=4\pi {\epsilon }_{0}r\times \left(10\right)$

Total amount of charge on $1000$ smaller drops will be,

$Q_{total}=1000q$

$\Rightarrow Q_{total}=1000\times 4\pi {\epsilon }_{0}r\left(10\right)$

$\Rightarrow Q_{total}=40000\pi {\epsilon }_{0}r.....\left(1\right)$

Now let's focus on the bigger drop,
assume its radius is $R$ & voltage be $V$.

If the bigger drop is made by merging all the smaller drops, then total charge & volume would remain the same.

So,

$\text{Volume}=\frac{4}{3}\pi R^3=1000\left(\frac{4}{3}\pi r^3\right)$

$\Rightarrow R=10r$

Now, capacitance of the bigger drop will be,

$C_{biggerdrop}=4\pi {\epsilon }_{0}R=40\pi {\epsilon }_{0}r$

Charge on the bigger drop:

$\Rightarrow Q=C_{biggerdrop}V=40\pi {\epsilon }_{0}rV.....\left(2\right)$

From the conservation of charge and using equation (1) and (2), we get

$40\pi {\epsilon }_{0}rV=40000\pi {\epsilon }_{0}r$

$\Rightarrow V=1000\text{volts}$

Hence, option (a) is the correct answer.

Key concepts: 1. Conservation of mass 2. Conservation of charge of isolated system 3. Capacitance of an isolated conductor 4. $Q=CV$