Two identical droplets of water.... [Potential]

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SUMMARY

The discussion centers on calculating the electric potential of a merged spherical droplet of water, each initially carrying a charge of 7.92 pC and a surface potential of 144 Volts. When the two droplets merge, the total charge remains 15.84 pC, but the potential is not simply the sum of the individual potentials. The potential of the new droplet must be recalculated using the formula V = kq/r, where 'k' is Coulomb's constant, 'q' is the total charge, and 'r' is the radius of the new droplet.

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  • Understanding of electric potential and charge conservation
  • Familiarity with Coulomb's law and the formula V = kq/r
  • Basic knowledge of spherical geometry and volume calculations
  • Concept of merging charges and resultant effects on potential
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  • Study the derivation and applications of Coulomb's law in electrostatics
  • Learn how to calculate the radius of a merged spherical droplet from volume and charge
  • Explore the implications of charge distribution in conductive materials
  • Investigate the relationship between electric potential and electric field strength
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Homework Statement


Suppose you have two identical droplets of water, each carrying charge 7.92 pC spread uniformly through their volume. The potential on the surface of each is 144 Volts.

Now, you merge the two drops, forming one spherical droplet of water. If no charge is lost, find the potential at the surface of this new large water droplet. in V.

Homework Equations


I don't have any equations since it's "water droplet" with no specified shape.

General formula for potential due to a single point charge is V = kq/r

The Attempt at a Solution


Is this question easy or tricky? Shouldn't the answer be 144 + 144 = 288 V? Due to conservation of charges?
 
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Antonius said:
I don't have any equations since it's "water droplet" with no specified shape.
Which shape do you expect for a droplet without contact to anything else?
What is the potential from such an object?
Antonius said:
Is this question easy or tricky?
It is not tricky, but it is not as easy as adding two numbers.
Antonius said:
Due to conservation of charges?
Charge is conserved, but that does not mean you can just add potentials.
 
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The new droplet of water has what radius, assuming sphericity? What total charge? Ergo, what potential?
 
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