# Tricky exercise (electricity)

1. Mar 11, 2004

### broegger

i'm having trouble with this:

A spherical raindrop has a charge Q = 3*10^-11. The potential at the surface is V = 500 V (V = 0 infinitely far away).

a) Determine the radius of the drop.

My first thought was to consider the raindrop as a point charge and then integrate the electric field from this point charge from a distance that equals the unknown radius R to infinity and, finally, equate this to 500 V.. but I'm not sure how to do this or if it's even right...

Another, identical, raindrop is brought in touch with the first one.

b) Determine the magnitude of the repelling force between the raindrops just before they touch each other.

Here, again, I would consider the drops as point charges and then calculate the Coulomb force when the distance between them equals 2R. Can you do that even though the charge distribution is in reality homogenous?

The two raindrops is now brought together and forms a single, spherical raindrop.

c) What is the potential at the surface of this new raindrop?

Here I would integrate the electric field from the 'old' radius to the radius of the new drop and then subtract this from the 500 V.. Again, I have no idea if you can do that...

2. Mar 11, 2004

### HallsofIvy

Staff Emeritus
I would not assume a point charge. Treat it, rather, as a uniform charge density. That is, since the volume of the sphere is (4/3)&pi;R3, the charge density is (4Q)/(3&pi;R3) and the electric force at distance r from the center is (4Q)/(3&pi;R3r2) Integrate that over the sphere of radius R:
$$\frac{4Q}{3\piR^3}\int_{r=0}^R\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi}(\frac{1}{r^2})(r^2 sin(\phi)d\phi d\theta dr$$
(If you have a formula for the field due to a homogeneous sphere of charge at a point insidethe sphere, you can use that instead)

Set that equal to 500V and solve for R.

Yes. As long as you are outside a homogeneous sphere, you can treat
the sphere as a point charge

That's not the way I would do it. You can assume that the new drop has twice the volume and twice the charge of the old sphere.
Use V= (4/3)&pi;R3 to determine the radius of the new sphere (it's not twice the radius!) and integrate the force function to find the field. (or, if you have it, use the formula.)

3. Mar 11, 2004

### broegger

a) You mean solving this equation

$$\int_{R}^{\infty} \mathbf{\vec{E}} \bullet d\mathbf{\vec{r}} = \int_{R}^{\infty} \frac{1}{4\pi{}\epsilon_{0}} \frac{Q}{r^2} dr = 500 V$$

with respect to R is wrong. If so, why?

4. Mar 12, 2004

### Palpatine

Use Gauss Law to find C = 4 * pi * e0 * R

and of course q = CV