Eight identical spherical raindrops are charged such that each is at a potential V, relative to the potential infinitely far away. (Assume the water has enough impurities that the raindrops behave as conductors and hence the potential is the same throughout the raindrop.) They coalesce to make one spherical raindrop whose potential is:
A. V/8B. V/2C. 2VD. 4VE. 8V

Explanation:

Given that each of the rain drop sphere has a voltage of V

And their are 8 rain drops that want to coalesce

Electric potential is given as

U=Kq/r

If each has a radius r before coalesce then it volume is 4πr³/3

The volume of all the eight raindrops when they coalesce is 4πR³/3 which is equal to the 8 times each volume

Then,

4πR³/3=8× 4πr³/3

Then π, 4 and 3 cancel out

R³=8r³

Find cube root of both sides

R=2r

So the radius of the big raindrop is 2r

The electric potential of each of the sphere is

U=Kq/r= V

Now, for the eight raindrops the charges inside the big raindrops will be eight multiple each rain drops Q=8q

Therefore, electric potential of the big rain drops (Ub) is

Ub=KQ/R.

Where Q=8q and R=2r

Ub=K•8q/2r

Ub=4kq/r

Ub=4•kq/r. Since U=Kq/r= V

Then, Ub=4V

The answer is D.

power out to power in less than 100%

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right