Finding charge on suspended water drop in electric field

[thursdaytbs]

So the question reads:

"A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8480 N/C. The mass of the water drop is 3.50 x 10^-9 kg. (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?"

The correct answers are:
(a) Positive, so that the electrostatic force points upward.
(b) 2.53 x 10^7 protons

I really wasn't sure where to begin, so I wrote some equations down.
I tried using F=Eq, or E = F/q where the E = 8480, but I wasn't sure about F or q. There was an equation that said something = q/A but I don't have A or q.

I think the main part I'm missing is converting the 3.50 x 10^-9kg of H2O into how much charge it has. Can anyone give me some direction as to where to go for this problem? Any help is greatly appreciated, Thanks.

 

[Andrew Mason]

So the question reads:

"A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8480 N/C. The mass of the water drop is 3.50 x 10^-9 kg. (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?"

The correct answers are:
(a) Positive, so that the electrostatic force points upward.
(b) 2.53 x 10^7 protons:

You are correct that F=qE gives the upward force. What is the downward force (hint: think mechanics not electricity)?

AM

 

[wais]

Firstly, great job on writing down relevant equations and identifying what information is missing. This is the first step in solving any physics problem.

To solve this question, we need to use the equation F=Eq, where F is the electrostatic force, E is the electric field, and q is the charge. We also know the mass of the water drop, which we can use to find the number of excess electrons or protons residing on the drop.

To start, let's rearrange the equation to solve for q: q = F/E. We have the value for E (8480 N/C), but we still need to find the value for F. To do this, we can use the equation F=ma, where m is the mass and a is the acceleration. In this case, we can use the acceleration due to gravity, which is 9.8 m/s^2. Plugging in the values, we get F = (3.50 x 10^-9 kg)(9.8 m/s^2) = 3.43 x 10^-8 N.

Now, we can plug this value for F into our original equation, q = F/E, to find the charge on the water drop. q = (3.43 x 10^-8 N)/(8480 N/C) = 4.05 x 10^-12 C.

Since we know that the electric field is directed upward, the excess charge on the water drop must also be positive in order for the electrostatic force to point in the same direction.

To find the number of excess electrons or protons, we can use the fact that the charge on one electron is 1.6 x 10^-19 C. So, dividing our charge on the water drop (4.05 x 10^-12 C) by the charge on one electron, we get 2.53 x 10^7 electrons/protons.

Therefore, the excess charge on the water drop is positive and there are 2.53 x 10^7 protons or electrons residing on the drop.

I hope this helps guide you in solving this question. Remember to always start by identifying what information you have and what you need to find, and then using the relevant equations to solve for the missing values. Good luck!