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n small drops of same size are charged to V volts each. If they coalesce to form a signal large drop, then its potential will be?

n small drops of same size are charged to V volts each. If they coalesce to form a signal large drop, then its potential will be?

Grade:12

5 Answers

Avneet Kaur
16 Points
10 years ago

Let potential f 1 drop = Kq/r =V 

Volume of n drops = volume of bigger drop 

n  4/3 pi (r)^3  = 4/3 pi (R)^3

Therefore ,  R = ( (n)^1/3) r 

also charge remains conserved 

charge on 1 drop = q 

charge on big drop containing n drops = (nq) 

Therefore potential on big drop = K (nq)/ ((n)^1/3)r

=(( n)^2/3) Kq/r

= (( n)^2/3) V

Abhishek Parihar
16 Points
10 years ago

Thank you sooo much!!!Laughing

Ishaan Shekhar
11 Points
7 years ago
Charge on 1 drop =q
Charge on big drop=nq
n4/3πr^3=4/3πR^3
R=n^1/3r
Potential on big drop=k(nq)/R^3
V1=knq/n^1/3r
V1=n^2/3V. 
Panitji
13 Points
4 years ago
Answer is V N power 2/3 
Let potential of 1 drop is =  q
Volume of n drops  is equal to the volume of bigger drops 
Charge on big drop containing n drops  is equal to 
= NQ
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Hello students,
The solution of the above problem is attached.
I hope the solution will solve all your doubts.
Thank You,
All the Best for the Exams.
645-299_WhatsApp Image 2020-06-04 at 8.11.49 PM.jpeg

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