591Conserving the volume of both the spheres..we have..
4/3Ï€r3 *27 = 4/3Ï€R3
Solving this eqn we get
R=3r = 3 mm
Now the total charge on 27 drops= 27*10-12
Therefore potential on the surface of the big drop = kq/r
= 9*109 * 27*10-123*10-3
=81 V
27 charged water droplets each with a diameter of 2 mm and a charge of 10-12 C coalesce to form a single drop. We have to calculate the potential of the bigger drop. How do I proceed? The answer is 81V.
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2 Answers
591
1161For this you need a concept of self energy. You have to find the self energy of all the 27 droplets and the bigger drop.
Self energy can be calculated by the formula 1/2ξE2 where E is the electric field in that reason. You have to go by integration.
- Shreya Sharma Sir, I'm still not getting the answer. Do we have to integrate and find out the E for the small drops and then integrate for the large sphere?Upvote·0· Reply ·2013-05-22 07:24:45