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# Eight oil drops of same size are charged to a potential of 50 V each.These oil drops are merged into one single large drop.What will be the potential of the large drop?Sir explain it with answer.

Ravneet
104 Points
2 years ago
Volume of 8 drop of radius ‘r’ = Volume of big drop of radius ‘R’
8 X  4/3pi(r3 )= 4/3pi(R3)
R = 2r
potental of small drop  = kq/r = 50
potential of big drop = kq/2r = 50/2 = 25V
Ravneet
104 Points
2 years ago
potential of bigger drop = k(8q)/2r = 4kq/r   Becouse R = 2r
and kq/r = 50 V
so potential of bigger drop = 4 X 50  =  200 V
Chaitanya Prakhar Sharma
15 Points
2 years ago
Use
Potential =(n^2/3) kq/R=(n^2/3)v
n=no of drops
v=voltage given
put n=8 and v=50
Let's check its working
According to our problem
Potential of big drop = (8^2/3)*50
=  4*50
=200
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