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n identical droplets are charged to v volt each. If they coalesce to form a single drop, then its potential will be
  • a)
    n2/3v
  • b)
    n1/3v
  • c)
    nv
  • d)
    v/n
Correct answer is option 'A'. Can you explain this answer?
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n identical droplets are charged to v volt each. If they coalesce to f...
 

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n identical droplets are charged to v volt each. If they coalesce to f...
Solution:

When n identical droplets combine to form a bigger drop, then the total charge remains the same as the charge is conserved.

Let the radius of each droplet be r and the final radius of the bigger drop be R.

Then, the volume of each droplet is (4/3)πr^3 and the volume of the bigger drop is (4/3)πR^3.

As the number of droplets is n, the total volume of all droplets is n(4/3)πr^3.

As the droplets coalesce, their volumes add up to form the volume of the bigger drop. Therefore,

n(4/3)πr^3 = (4/3)πR^3

Simplifying, we get R = nr

The capacitance of a sphere of radius R is given by C = 4πεR, where ε is the permittivity of the medium.

The potential of a charged sphere is given by V = Q/C, where Q is the charge on the sphere.

Let Q be the charge on each droplet. Then, the total charge on all droplets is nQ.

When the droplets combine, the total charge remains the same. Therefore, the charge on the bigger drop is also nQ.

The capacitance of the bigger drop is C' = 4πε(nr)

The potential of the bigger drop is V' = nQ/C' = nQ/[4πε(nr)]

Simplifying, we get V' = (n^2/4πε)(1/r)

As R = nr, we have r = R/n

Substituting this in the above equation, we get

V' = (n^2/4πε)(n/R)

Simplifying, we get V' = n2/3v (where v is the potential of each droplet)

Therefore, the potential of the bigger drop is n2/3v. Hence, option (A) is the correct answer.
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