27 identical drops are charged at 22V each. They combine to form a bigger drop. The potential of the bigger drop will be _____________ V.
Answer (integer)
198
Solution
<p>Let the charge on one drop is q and its radius is r.</p>
<p>So for one drop $V = {{kq} \over r}$</p>
<p>For 27 drops merged new charge will be Q = 27 q and new radius is R = 3r</p>
<p>So new potential is</p>
<p>$V' = {{kQ} \over R} = 9{{kq} \over r} = 9 \times 22$ V</p>
<p>= 198 V</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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