64 identical drops each charged upto potential of $10 ~\mathrm{mV}$ are combined to form a bigger drop. The potential of the bigger drop will be __________ $\mathrm{mV}$.
Answer (integer)
160
Solution
The problem involves 64 identical drops, each charged up to a potential of $10 \mathrm{~mV}$, that are combined to form a bigger drop. We are asked to find the potential of the bigger drop.
<br/><br/>
The potential of each drop is given by:<br/><br/>
$V = \frac{Kq}{r}$<br/><br/>
where $K$ is the Coulomb constant, $q$ is the charge on each drop, and $r$ is the radius of each drop.
<br/><br/>
The radius of the bigger drop is:<br/><br/>
$R = 4r$<br/><br/>
since the 64 identical drops combine to form a bigger drop.
<br/><br/>
The total charge on the 64 identical drops is:<br/><br/>
$Q = 64q$
<br/><br/>
The potential of the bigger drop is:<br/><br/>
$$V_{bigger} = \frac{KQ}{R} = \frac{K(64q)}{4r} = 16 \frac{Kq}{r} = 16V = 16 \times 10 \mathrm{~mV} = 160 \mathrm{~mV}$$
<br/><br/>
Therefore, the potential of the bigger drop is $160 \mathrm{~mV}$.
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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