# What charge must a spherical raindrop of diameter 1.0 mm carry if it remains suspend

1. Feb 19, 2012

### duggielanger

1. The problem statement, all variables and given/known data

Two parallel plates are placed 0.10 m apart with one vertically above the
other and their edges aligned. The potential difference of the upper plate is
100 kV relative to the lower plate. What charge must a spherical raindrop of
diameter 1.0 mm carry if it remains suspended between the plates? Assume
that the electric field between the plates is uniform and give your answer in
units of the charge on an electron. (The density of water is 1000 kg m
−3
.)

2. Relevant equations

F=mg
v=0.75r^3d
and my be Eel=qV

3. The attempt at a solution
I first found the mass of the raindrop. And first i need to find the radius
So 1.0mm/1000=0.001 m and I want the radius so 0.001m/2=0.0005 m
Next I need to find the volume of the raindrop

v=0.75x3.14x0.01m
=2.945243113 ×10-10 m3
=2.945243113 ×10-10 m3 x 1000kg m^-3= 2.9x10^-7 kg

So we now have a mass.
Now the raindrop will have work done by gravity

F=mg
F=2.9x10^-7kg x 9.81 m s^-2= 2.9 x10^-6 kg m s^-2 or N

So I have downward force of 2.9 x 10^-6 N
Now its the next step I cant figure , I think I need to find upward force on the raindrop by finding the charge . But cant get my head around it

2. Feb 19, 2012

### cepheid

Staff Emeritus
Re: What charge must a spherical raindrop of diameter 1.0 mm carry if it remains sus

I think some of your basic physics (and geometry) is a little off here:

The first equation is okay.

For the second equation: I'm not entirely sure what you have written, but the equation for the volume of a sphere is $\mathcal{V} = (4/3)\pi r^3$ where r is the radius of the sphere. Note that 4/3 ≠ 0.75.

For the third equation: also wrong. The definition of the electric field at a point in space is the force per unit charge that would be exerted at that point. In other words, the electric field at a point in space is the force that a "test charge" of +1 C would feel if it were placed at that point. Hence, the units of electric field are N/C, and the relationship between electric force and electric field is given by $F_E = qE$, where q is the charge that is being placed in the field and feeling its influence.

The relationship between the voltage (a.k.a. potential difference) and the electric field is given by $E = V/d$ where d is the distance between the plates. Note that this relationship only holds true for a uniform electric field. From this, you can immediately see that the units for E-field must also be V/m = J/(Cm). This is perfectly consistent with what I said above, since N/C = (Nm)/(Cm) = J/(Cm).

EDIT: I changed my notation, since capital V was playing double duty as both voltage and volume.

Last edited: Feb 19, 2012
3. Feb 19, 2012

### duggielanger

Re: What charge must a spherical raindrop of diameter 1.0 mm carry if it remains sus

Hi cepheid that's what I meant for equation two just couldn't think how to put it down.
Yes it is uniform electric field.
So is my workings out for F=mg ok and do I now need to find E=V/d for the electric field and then use this to find the charge on the raindrop at a point in space .

4. Feb 19, 2012

### cepheid

Staff Emeritus
Re: What charge must a spherical raindrop of diameter 1.0 mm carry if it remains sus

Yes, but let's step back further for a moment.

What is the key physics principle that is being tested in this problem?

Hint: if the droplet is suspended in mid-air, then what must be true about the NET force on the droplet?

How many forces are acting on the droplet and contribute to this net force?

5. Feb 19, 2012

### duggielanger

Re: What charge must a spherical raindrop of diameter 1.0 mm carry if it remains sus

Its similar to the millikan Experiment
I think the Net force will be zero
There will be two forces acting on the droplet and both contribute to the net Force

6. Feb 19, 2012

### cepheid

Staff Emeritus
Re: What charge must a spherical raindrop of diameter 1.0 mm carry if it remains sus

Yeah, essentially. Here are the specific answers I was looking for:

The key principle behind the problem is Newton's second law of motion, which says that the net force on a particle is equal to its mass times its acceleration.

If the particle is suspended and remains motionless, then its acceleration must be 0, otherwise it would begin to move. Therefore, Newton's second law says that the net force on the particle must be 0.

Two forces are acting: the electric force, and gravity. Since these two must cancel each other out, they must be equal in magnitude and opposite in direction. That means that the electric force points upward.

So now, write down the equation (Newton's second law) for this situation, and set the acceleration to zero. This will lead to the result that the magnitudes of the electric force and the gravitational force are equal to each other. So, once you have equated these two forces, you can solve for the only thing that is unknown, which is q.

7. Feb 19, 2012

### cepheid

Staff Emeritus
Re: What charge must a spherical raindrop of diameter 1.0 mm carry if it remains sus

One more thing: if the electric force must be upward, then what must be sign of the charge q on the droplet (positive or negative)? Hint: what is the direction of the electric FIELD, given that the upper plate is more positive than the negative plate?

8. Feb 19, 2012

### duggielanger

Re: What charge must a spherical raindrop of diameter 1.0 mm carry if it remains sus

Ah so is it q*E=m*g as these are the two forces acting on the drop and since I want q I make this the subject and now we have q=m*g/E

9. Feb 19, 2012

### duggielanger

Re: What charge must a spherical raindrop of diameter 1.0 mm carry if it remains sus

Will it be negative

10. Feb 21, 2012

### cepheid

Staff Emeritus
Re: What charge must a spherical raindrop of diameter 1.0 mm carry if it remains sus

Yeah it will be negative, because the force needs to be upward, in the opposite direction of the field. Negative charges travel in the opposite direction of the electric field, because the electric field is defined as the force that a positive charge would feel.

Another way to think about this is that the negative charge will be attracted upwards towards the positive plate, and repelled from the negative plate below.

If you're wondering why the negative sign doesn't just pop out of the math, it's because the equation should have been E = -V/d. That's where the negative sign came from. So if V is positive, meaning that the upper plate has a higher potential than the lower plate, then E will be negative, which is downward, as is the case here.

11. Feb 21, 2012

### duggielanger

Re: What charge must a spherical raindrop of diameter 1.0 mm carry if it remains sus

Thank cepheid just quickly revised my answer as it was positive , even though I knew it should be negative. Thats what tiredness can do to you .
THANKS FOR THE HELP