Potential difference of spherical rain drops

In summary, the question is asking about calculating the potential of a raindrop, and it's not clear what concept the question is trying to test or how to go about answering it.
  • #1
chanella35
11
0
I absolutely have no idea how to answer this question. If someone could please explain to me what concept this question is trying to test and how to go about answering this question it would be immensely appreciated :)

THE QUESTION:

Two equal spherical rain-drops are both at a potential V. If they
coalesce into a single spherical drops, what will be its potential ?
 
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  • #2
Hint: calculate the capacitance of the drops and the charge on each, given V. Then put them together and go backwards from charge to find V'.
 
  • #3
marcusl said:
Hint: calculate the capacitance of the drops and the charge on each, given V. Then put them together and go backwards from charge to find V'.

but how do i calculate anything when i wasnt given any values of variables in the question?
 
  • #4
I think this question is quite tricky. It's hard to say it's gravitational potential or electric potential. We need a bit of analysis:

Rain is usually not pure water, i.e. it's electrically conducting. From the time the rain drop is formed to the time it falls through a long distance, it accumulates an amount of charges due to "rubbing" with the surrounding. So, we can view the rain drop as a conducting charged sphere. Assume that V is electric potential and the rain drop can be treated as isolated (so that V is due to ONLY the rain drop itself, no external effect by other rain drops). Let Q and R denote the charge and the radius of the rain drop. You can calculate V in term of Q and R, can't you? :wink:

Then when 2 identical rain drops coalesce, it forms a bigger rain drop of charge 2Q. The radius of the new rain drop can be calculated (the total volume is conserved). Again, the electric potential of this new rain drop V' can be calculated in term of Q and R, as it's an isolated conducting spherical rain drop. Finally find the ratio of V' and V, and thus, express V' in term of V.

So what if V is gravitational potential? My best guess is, it's quite pointless or way too easy if so. For 2 identical rain drops to coalesce, they must be at the same height. The newly formed rain drop should be at that height as well, and thus, its potential is 2V (as mass is doubled). Too obvious! I have just only considered the gravitational potential due to the gravitational field of the Earth. But taking into account the gravitational potential due to gravitational field of the rain drop itself is again quite pointless, and more importantly, impossible to lead to a uniform potential at every point in the rain drop.
 
  • #5


The concept being tested here is the conservation of electric charge. When two objects with different potentials come into contact, the charge will redistribute itself until both objects have the same potential. In this case, the two rain drops have the same potential before they coalesce, and this potential will remain the same after they merge into a single drop.

To calculate the potential of the merged drop, we can use the equation V = kQ/r, where V is the potential, k is the Coulomb's constant, Q is the charge, and r is the radius of the spherical drop. Since the two drops have the same potential (V), we can set up the equation as follows:

V = k(Q1/r1) = k(Q2/r2)

Since the drops are equal in size and have the same potential, we can set Q1 = Q2 and r1 = r2. This simplifies the equation to:

V = k(Q/r) + k(Q/r)

The two charges (Q) cancel out, leaving us with:

V = 2k(Q/r)

This means that the potential of the merged drop will be twice the potential of the original drops. In other words, the potential of the merged drop will be 2V. This is because the charge is now concentrated in a single drop, making the potential stronger.

In conclusion, the potential of the merged spherical rain drop will be twice the potential of the individual drops before they coalesced. This concept can be applied to other situations involving the merging of objects with different potentials.
 

Related to Potential difference of spherical rain drops

1. What is potential difference of spherical rain drops?

The potential difference of spherical rain drops refers to the difference in electric potential energy between two points on the surface of a spherical rain drop. It is caused by the separation of positive and negative charges within the drop due to the Earth's electric field.

2. How is the potential difference of spherical rain drops measured?

The potential difference of spherical rain drops can be measured using an electrometer, which measures the electric potential difference between two points on the surface of the drop. It can also be calculated using the equation V = Ed, where V is the potential difference, E is the electric field strength, and d is the distance between the two points.

3. What factors affect the potential difference of spherical rain drops?

The potential difference of spherical rain drops can be affected by several factors, including the size and shape of the drop, the strength of the Earth's electric field, and the presence of other charged objects nearby. Additionally, the potential difference may increase as the drop falls through the atmosphere and experiences changes in air temperature and humidity.

4. Why is the potential difference of spherical rain drops important?

The potential difference of spherical rain drops is important because it plays a role in the formation of lightning. As rain drops fall through the atmosphere, they can become highly charged due to the Earth's electric field. When these charged drops collide with other particles, it can create a difference in potential and lead to the discharge of electricity in the form of lightning.

5. Can the potential difference of spherical rain drops be harnessed for energy?

While the potential difference of spherical rain drops can generate electricity through lightning, it is currently not a viable source of energy due to its unpredictable nature and the difficulty in harnessing it. However, scientists are continuing to research and develop ways to harness the energy of lightning for practical use in the future.

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