# Solving Spherical Raindrop Homework

• stylez03
What is the difference between R_{new}^3} and R?The new radius is not the same as the respective radii of the original spheres.

## Homework Statement

If a spherical raindrop of radius 0.800 mm carries a charge of -1.50 pC uniformly distributed over its volume, what is the potential at its surface?

Two identical raindrops, each with radius and charge specified in part (A), collide and merge into one larger raindrop. What is the radius of this larger drop, if its charge is uniformly distributed over its volume?

## Homework Equations

$$V = \frac {K*q} {r}$$

## The Attempt at a Solution

I tried:

$$V = \frac {(\frac {1} {4*pi*(8.85*10^{-12}})*-1.50} {0.008}$$

The online program says I'm off by an additive constant, I'm not sure if I'm using pC value correctly, I've never seen that unit before in our lectures.

pC means picocoulombs which is 10-12 Coulombs

Note also that $0.800mm \neq 0.008m$

Hootenanny said:
pC means picocoulombs which is 10-12 Coulombs

Note also that $0.800mm \neq 0.008m$

Whoops, 0.800mm should be 0.0008 m

So for pC to C is just $$-1.50*10^{-12}$$

Last edited:
Two identical raindrops, each with radius and charge specified in part (A), collide and merge into one larger raindrop. What is the radius of this larger drop, if its charge is uniformly distributed over its volume?

Would it just be 2times the R? I don't think it would be, but I'm not sure how to figure this part out.

stylez03 said:
Two identical raindrops, each with radius and charge specified in part (A), collide and merge into one larger raindrop. What is the radius of this larger drop, if its charge is uniformly distributed over its volume?

Would it just be 2times the R? I don't think it would be, but I'm not sure how to figure this part out.
Volume is conserved.

$$\frac{8\pi R^3}{3} = \frac{4\pi R_{new}^3}{3}$$

Saketh said:
Volume is conserved.

$$\frac{8\pi R^3}{3} = \frac{4\pi R_{new}^3}{3}$$

So the radius is still the same?

stylez03 said:
So the radius is still the same?
No; it's not the same.

Think about it this way. You have two spheres of clay. You mash them together into a bigger sphere. Is the radius of the new sphere going to be the same as the respective radii of the original spheres?

The same thing is going on here.

Saketh said:
No; it's not the same.

Think about it this way. You have two spheres of clay. You mash them together into a bigger sphere. Is the radius of the new sphere going to be the same as the respective radii of the original spheres?

The same thing is going on here.

Okay, that was my original intuition that the radius will grow, and you said the volume is also conserved. Now that we know the radius will grow, how do I go about solving for the new R from knowing that each individual raindrop has a radius of r?

Once again, volume is conserved.

$$\frac{4\pi R^3}{3} + \frac{4\pi R^3}{3} = \frac{8\pi R^3}{3} = \frac{4\pi R_{new}^3}{3}$$

Where $$R_{new}$$ is the radius of the new raindrop.

Saketh said:
Once again, volume is conserved.

$$\frac{4\pi R^3}{3} + \frac{4\pi R^3}{3} = \frac{8\pi R^3}{3} = \frac{4\pi R_{new}^3}{3}$$

Where $$R_{new}$$ is the radius of the new raindrop.

I'm sorry I'm still not fully understanding that. I see from the equation above, you have the sum of the volume of the two raindrops, which gives the 3rd term, you said that the volume is conserved, equating that to 4th term. From what I can understand, $$R_{new}^3}$$ does not seem to be different from the original $$R$$.