Question
Download Solution PDF512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ______V.
Answer (Detailed Solution Below) 128
Detailed Solution
Download Solution PDFExplanation:
The potential on a charged particle with charge Q and radius R is given as,
V = \(\frac{Q}{4π \epsilon_o R}\) -----(1)
Given,
No of mercury drops = 512, Potential of each drop = 2 V
Let us assume the charge on the smaller drop is q and the radius is r.
The potential on one charge particle = \(\frac{q}{4π \epsilon_o r}\)
⇒\(\frac{q}{4π \epsilon_o r}\) = 2 ⇒\(\frac{1}{4π \epsilon_o }\)= \(\frac{2r}{q }\) ----(2)
After joining the drops volume will be the same as the sum of the volume of individual drops. i.e.
\(\frac{4}{3 }\)πR3 = \(\frac{4}{3 }\)πr3 × 512
⇒R = 8r ----(3)
From equations (1), (2) & (3) we have-
∴ Potential on all 512 charge particles i.e single bigger drop = \(\frac{512q}{4π \epsilon_o R}\)= \(\frac{512q}{q \times 8r}\times 2r\) = 128 V.
Hence the correct value of potential is 128 V.
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