# Finding the Electric Potential

1. Nov 6, 2014

### Abel I Daniel

1. The problem statement, all variables and given/known data
8 Charged drops of water each of radius 1 mm and having a charge of 10^(-10) Coloumbs are combined to form a bigger drop.Determine the potential of the bigger drop.

2. Relevant equations
Does the radius becomes 4*10^(-3)m,when it combines?
Does the charge sum up to 9*10^(-10)?

3. The attempt at a solution
Used the formula -V=9*10^(9)*Q/r
Got the ans as 1800 V.

Last edited by a moderator: Nov 6, 2014
2. Nov 6, 2014

### Staff: Mentor

I don't think so. It's probably easier to work with the mm units that were given.
What's the volume of one rain drop?
If you combine 8 rain drops into one bigger drop, what is the volume of this larger drop? What's its radius?

3. Nov 7, 2014

### Abel I Daniel

I assume the volume to be (4/3)*(pi)*r^3,where r is 1 mm.
But what potential has to do with Volume.i am quit weak in this subject

4. Nov 7, 2014

### Staff: Mentor

Or more simply, 4/3 $\pi$, the volume of one rain drop,
From the formula you wrote in post #1, the potential V is a function of the radius r.
If 8 rain drops combine into one larger drop, what's the volume of the larger rain drop? What's the radius of the larger rain drop?

5. Nov 7, 2014

### Abel I Daniel

i hope it adds up linearliy,that means ,radius of larger frop =8* 1mm (radius of each smaller drop)

6. Nov 7, 2014

### Staff: Mentor

No, it doesn't.
The volumes add like you think they would, but the radius of the combined raindrop is NOT 8mm.

7. Nov 8, 2014

8. Nov 8, 2014

### Staff: Mentor

2 mm is more like it. I'm pretty sure the charge is additive.

9. Nov 13, 2014