# Find vertical deflection

1. Feb 17, 2012

### Alouette

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

In an ink-jet printer, drops of ink are given a controlled amount of charge before they pass through a uniform electric field generated by two charged plates. If the electric field between the plates is 1.5x106N/C over a region 1.20cm long, and drops of mass 0.12nanograms enter the electric field region moving horizontally at 16m/s.

What is the vertical deflection between the point where it leaves the plates and where it strikes the paper, if the paper is 0.50cm from the end of the charged plates? Ignore other forces such as gravity and air resistance, and assume the field between the plates is vertical and constant, and drops abruptly to zero at the edge of the plates.

2. Relevant equations

a = F/m = qE/m

t = (.012)m/(16) m/s = 7.5x10^-4

Vertical velocity [Vv] = at

D = Do (starting displacement at end of plates) + (Vv*t')

3. The attempt at a solution

I found q to be 1.42x10^-16. and a turns out to be 1775. So Vv = a*t = 1.33

Then t' = (.50) cm/ (16) m/s^2 = 3.12x10^-4

Then multiplying Vv(t') and adding to Do, I got 9.16x10^-4 but this is wrong.

What have I done wrong? I followed the formulas didn't i?

Last edited: Feb 18, 2012
2. Feb 17, 2012

### ehild

How did you get the charge of the ink drops?

ehild

3. Feb 17, 2012

### Alouette

a = F/m = qE /m
a = q x 1.50x10^6N/C / 0.12^-12kg .. .. .. a = q (1.25^19) m/s² (1)

Time between plates .. t = 0.012m / 16m/s .. .. t = 7.5x10^-4 s (2)

Vertical displacement ..
d = ut + ½ at²
(5x10^-4)m = (0) + (½ x q (1.25x10^19) x (7.5x10^-4)²)

q = 2(5x10^-4) / (7.03x10^12) .. .. ►q = 1.42x10^-16 C

4. Feb 17, 2012

### ehild

What is the 5x10-4 m? I do not find it in the problem text.

ehild

5. Feb 17, 2012

### Alouette

Oh I see why you ask. I didn't include it, but it was part of the first part of this question:

"How much charge on a drop is necessary to produce a vertical deflection of 0.50mm at the end of the electric field?"

So the 0.50 mm = 5x10^-4 m.

And with this newfound q, I used it in:

a=q(1.25x10^19) >> (1.42x10^-16)(1.25x10^19) = 1775

6. Feb 17, 2012

### ehild

Why did you use 17 m/s horizontal velocity?
Are you sure you have to add the displacement outside the plates to Do?

ehild

7. Feb 18, 2012

### Alouette

Oh my goodness, it's a typo! :( supposed to be 16 m/s as the problem states. But I just ran the numbers again, wrong still..

I used 16 m/s because t' is the time to reach the paper. Is this wrong since it asks for the vertical deflection?

And Do, which = 0.50 mm from the first part, is the starting displacement of the plates. So someone suggested adding this to Vv * t' to get D, which is the displacement (deflection) i'm looking for.

I vaguely understand the problem... am I just using the wrong formulas as I stated in my attempt at solution?

Last edited: Feb 18, 2012
8. Feb 18, 2012

### ehild

It is all right.

Read the text again. It asks the deflection from the point where the drops leave the plates. Te deflection is simply 1.33*t'. Do not add Do.

ehild

9. Feb 18, 2012

### Alouette

You are quite amazing, thank you for your help once again, I got the right answer!

It's so confusing learning the formulas they give us in class, and then when it comes to problems it's like we have to derive a whole new set....very frustrating.

So the deflection is the vertical velocity of the drop multiplied by time to reach the paper? Basically d=vt?

10. Feb 18, 2012

### ehild

Yes, it is in this case. But you should add Do if they would ask the deflection from the place the ink would reach the paper without the electric field. Usually that is asked in problems like this one.

Always read the problem very carefully and make a drawing. You used the correct equations everywhere, but did not understand the question.

ehild

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