512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is ________ V.
Answer (integer)
128
Solution
Let charge on each drop = q<br><br>radius = r<br><br>$v = {{kq} \over r}$<br><br>$\Rightarrow$ $2 = {{kq} \over r}$<br><br>radius of bigger<br><br>${4 \over 3}\pi {R^3} = 512 \times {4 \over 3}\pi {r^3}$<br><br>$R = 8r$<br><br>$\therefore$ $$v = {{k(512)q} \over R} = {{512} \over 8}{{kq} \over r} = {{512} \over 8} \times 2$$<br><br>$= 128V$
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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