# What is the charge on the droplet?

## Homework Statement

A small droplet of oil with a mass of 2.00x10^-15 kg is held suspended in a region of uniform electric field directed upward with a magnitude E = 6,000 N/C.
a) What is the charge on the droplet?
b) Find the acceleration of the droplet if the electric field suddenly dropped to 0 N/C

## Homework Equations

E = F/q (for a uniform field)
F=ma

## The Attempt at a Solution

It sounds like a straight forward question, solve for q. But the droplet being suspended has me confused.

I assume I substitute F=ma into E=F/q to get E=ma/q, but if the droplet is suspended doesn't that mean a=0? And that would make E=0? There's obviously something I'm not understanding. Please help?

danago
Gold Member
If it is suspended, then the net force on it is zero, which means that the force due to the electric field completely balances the opposing force due to gravity. I think thats what its getting at Ahaaaa.. that would make sense. Then essentially part a is:
E=F/q = mg/q
q = E/mg = (6000 N/C) / (2.00x10^-15 kg)(9.8 N/kg)
q = 3.06x10^17 C

Correct?

And for part b:
if the electric field dropped to 0 N/C would that mean that the acceleration is just equal to gravitational acceleration (9.8 N/kg)?

danago
Gold Member
Id have a look at how you have calculated part (a) again. The charge you calculated is huge, and will result in an electric force in the order of 10^21 N, which is MUCH bigger than the gravitational force.

Are you saying that my equation is incorrect?

Because if it IS correct then I keep coming up with the same answer.. over and over. So am I to believe that my equation is at fault?

danago
Gold Member
Are you saying that my equation is incorrect?

Because if it IS correct then I keep coming up with the same answer.. over and over. So am I to believe that my equation is at fault?

Yea.

Try drawing a free body diagram of the drop and label all forces acting on it (electric and gravitational). For equilibrium, we should have Fgrav=Felec.

Perillux
Ahaaaa.. that would make sense. Then essentially part a is:
E=F/q = mg/q
q = E/mg = (6000 N/C) / (2.00x10^-15 kg)(9.8 N/kg)
q = 3.06x10^17 C

Correct?

And for part b:
if the electric field dropped to 0 N/C would that mean that the acceleration is just equal to gravitational acceleration (9.8 N/kg)?

You messed up when you tried solving for q. You had the equation E = mg/q correct but then when you solved for q you said:
q = E / mg
but it should be:
q = mg / E

If it is suspended then the only forces acting on it are the forces of gravity in the downward direction and the electric field force in the upward direction. They have to balance out, as you said above, to be suspended (in equilibrium).

since force is = qE then so is mg, so mg=qE
(2.00x10^-15 kg)(9.8 N/kg) = q(6000 N/C)
(1.96x10^-14 N) / (6000 N/C) = q
q = 3.27x10^-18 C

yes?

Perillux
I don't have a calculator right now, but that all looks correct yes.