Need help with Quantinization of Charge concept and calculations

[AceInfinity]

Homework Statement

An oil drop whose mass is 3.30x10_-15kg moves at a constant speed of 1.70m/s upward between two horizontal parallel plates. The plates are 1.00cm apart and are connected to a 6.75x10₂V battery. If the oil drop is negative and the top plate is positive, calculate the number of excess electrons that the oil drop has.

Spoiler

Answer: 3

Homework Equations

Not sure, but I have a list of equations that we use, and i'd guess:
• delta V = (delta E)/(q)
• E = (delta V)/(delta d)
• E = Fe/q

The Attempt at a Solution

I don't know where to start.

 

[gneill]

If the velocity is constant, what does that say about the net force on the oil drop?
What forces are at play?

 

[AceInfinity]

constant velocity means no acceleration, and Fnet = 0? If you could provide me a short explanation of the foruma's you use and the steps to the final solution that would better help me I think, I need to figure this out by tomorrow, and I get the concepts somewhat, I just don't understand what some of the variables in the formulas mean and why you'd use them. I can figure things out better that way with my style of learning.

If you could tell me if I have the right idea by using... Fnet = Fe + Fg? And since Fnet = 0 then if it's just a magnitude then Fe = Fg?

 

[Matterwave]

Yes, now answer gneill's second question. What forces are present in this problem (hint: gravity is one), and then you can try to balance forces.

 

[AceInfinity]

did you read my post correctly?

"Fnet = Fe + Fg ... Fe = Fg" I need help beyond that point.

 

[gneill]

Presumably you know that the gravitational force will be the weight of the oil drop, or Fg = mass*g. It is interesting to note that g has units m/s² which are entirely equivalent to the units N/kg. There's a similar expression for the force due to an electric field, Fe = q*E. Here E is the electric field strength in N/C or equivalently -- and this is important -- V/m.

Now, with the information given, can you think of a way to arrive at the magnitude of the electric field? Do you know anything about electric fields between charged plates?

 

[AceInfinity]

Alright, so from here:

Fg = Fe
Fg = qE

but you need to solve for E first so you'd use: E = v/m
E = 675V / 0.05m
E = 1.35x10⁴

m x gravity = q x 1.35x10⁴
(m x gravity) / 1.35x10⁴ = q
q = 4.1555555...x10^-19

instead of gravity though you'd use the 1.7m/s

Then the charge of an electron we divide into that (1.6x10^-19)

4.155555.x10^-19 / 1.6x10^-19

= 2.5972...
Answer = 2

Btw, I forgot that the teacher changed the answer key, the answer is actually 2 because you can't round up since a decimal of an electron doesn't exist. I hope that was right.

 

[gneill]

Alright, so from here:

Fg = Fe
Fg = qE

but you need to solve for E first so you'd use: E = v/m
E = 675V / 0.05m
E = 1.35x10⁴

Where did the 0.05m distance come from? Your problem statement claimed that the plate separation was 1.0cm.

m x gravity = q x 1.35x10⁴
(m x gravity) / 1.35x10⁴ = q
q = 4.1555555...x10^-19

instead of gravity though you'd use the 1.7m/s

Huh? How can a velocity be interchanged with gravity?

Then the charge of an electron we divide into that (1.6x10^-19)

4.155555.x10^-19 / 1.6x10^-19

= 2.5972...
Answer = 2

Btw, I forgot that the teacher changed the answer key, the answer is actually 2 because you can't round up since a decimal of an electron doesn't exist. I hope that was right.

Your result should be very close to an integer before rounding, any difference being attributable to experimental error, numerical truncation issues, or the use of inexact constants.

 

[AceInfinity]

I wasn't looking for criticism, I was looking for some help, or some steps in the right direction.

But from what you say... The electric field strengh would be equivilant to E = v/m, so then I would end up with (6.75x10^2V)/(3.30x10^-15Kg) = 2.0454545...x10^17 N/C...??

 

[gneill]

I wasn't looking for criticism, I was looking for some help, or some steps in the right direction.

But from what you say... The electric field strengh would be equivilant to E = v/m, so then I would end up with (6.75x10^2V)/(3.30x10^-15Kg) = 2.0454545...x10^17 N/C...??

The 'm' in E = V/m is the meter, a unit of length. The V is the volt. Electric field strength is measured in units of volts per meter, or Newtons per coulomb.

Once you have the electric field strength you can equate the electric force to the gravitational force (since there is no acceleration hence no net force on the oil drop).

Since electric charge comes in discrete lumps the size of the charge on the electron (let's call it q_e here), then let the total charge on the oil drop be n*q_e. Solve for n.

 

[AceInfinity]

E = V/m
= E = (6.75x10²V)/(0.10m)

Therefore E = 6750N/C

E = N/C

Therefore N = E/C

N = (6750N/C)/(1.60x10^-19C)
= 4.21875x10²²
= ANS / 1.60x10^-19 <--- to find out how many electrons there are by finding how many charges of one electron can fit into the oil drop's total charge.

= 2.63671... x10⁴²

lol the answers agree with the final answer, all except for the exponents I believe. I'm not sure.

 

[gneill]

E = V/m
= E = (6.75x10²V)/(0.10m)

Therefore E = 6750N/C

You're off by an order of magnitude. 1cm is not 0.1m

E = N/C

Therefore N = E/C

N = (6750N/C)/(1.60x10^-19C)
= 4.21875x10²²
= ANS / 1.60x10^-19 <--- to find out how many electrons there are by finding how many charges of one electron can fit into the oil drop's total charge.

= 2.63671... x10⁴²

lol the answers agree with the final answer, all except for the exponents I believe. I'm not sure.

No idea what you're calculating there. It looks as though you were going for a force (N), but why did you insert a charge? You don't know the charge yet.

You want to balance the electric force with the gravitational force. The unknown in that balance is the charge, which you want to solve for.

 

[AceInfinity]

E = V/m
E = (6.75x10²V)/(0.01m)
E = 67500 N/C

Fg = Fe
m x a_g = Eq
(3.30x10^-15kg)(9.81m/s²) = (67500 N/C)(q)
q = [(3.30x10^-15kg)(9.81m/s²)] / (67500 N/C)
q = 4.796x10^-19

*Doing this as I type it out and that -19 looks promising...

4.796x10^-19 / (1.60x10^-19) = 2.9975

and we would round up to 3...

Thankyou!

 

[gneill]

My pleasure. This last bit of figuring is short, crisp, and correct. Nicely done.