Find charge q of particle passing through charged plates

[SchrodingersMu]

In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The ink drops, which have a mass of 1.3 10-8 g each, leave the nozzle and travel toward the paper at 21 m/s, passing through a charging unit that gives each drop a positive charge q by removing some electrons from it. The drops then pass between parallel deflecting plates 2.0 cm long where there is a uniform vertical electric field with magnitude 7.7 104 N/C. If a drop is to be deflected 0.35 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop?

So, I know that F= E*q=m*a
I can solve for t, . I still don't know a, the acceleration, or vfinal of y either. I was thinking of solving for vfy by using dy=vi+vf/2 * t.

Either way, I keep getting the wrong answers. They are supposed to be about ^-13.

Any help is appreciated!

 

[paisiello2]

Some of your numbers and formulae are hard to understand with typos, I think. But you seem to be on the right track.

You do know a in terms of q.

What other kinematic formulae do you know?

 

[amind]

Here is what I can deduce the data from OP
Mass of a drop ,m = 1.3e-8 g = 1.3e-11kg
Deflection , y = .35mm = 3.5e-4 m
Length of the plate ,l = 2cm = 2e-2m,
Velocity of the drop = 21m/s
E = 7.7e4.
Here's the solution,
a_{y} = \frac{qE}{m} ( Since F = qE )
Let t be the time required for the drop to pass the region and deflect,
y = \frac{1}{2}a_{y}t^{2} ... (1)
l = vt
∴ t = \frac{l}{v}
Substituting for t and a in (1) , we get,
y = \frac{qEl^{2} }{2mv^{2}}
Solving for q,
q = 1.30e-13 = 1.3\times10^-13
Note : the notation xey means x \times 10 ^y

 

[SchrodingersMu]

Sorry about the typos guys!
So, we can solve for t, even though the velocity is mostly in the x direction, because the time is the same for the x and y components of motion, right?
ALso, eq(1) would have originally been y= initial y position + velocity in y * time + (acceleration stuff), but the y velocity component is 0 initially, right?

Eventually, we have velocity in thr y direction, though, so why does eq(1) work? From what I see, it says the velocity in the y direction is always 0.

The answer is absolutely correct; I;m just trying to understand the answers both of you gave. I appreciate this very much. This is better feedback than my prof gives.

 

[SchrodingersMu]

I did this one by accident, so I'll make a cat to not waste the post

=^O_O^=

 

[amind]

Sorry about the typos guys!
So, we can solve for t, even though the velocity is mostly in the x direction, because the time is the same for the x and y components of motion, right?
ALso, eq(1) would have originally been y= initial y position + velocity in y * time + (acceleration stuff), but the y velocity component is 0 initially, right?

Eventually, we have velocity in thr y direction, though, so why does eq(1) work? From what I see, it says the velocity in the y direction is always 0.

The answer is absolutely correct; I;m just trying to understand the answers both of you gave. I appreciate this very much. This is better feedback than my prof gives.

The velocity is in the x direction , this is anologous to projectile motion , if you fire in the horizontal direction , the horizontal velocity is unaffected by gravity but the vertical component is affected by gravity so in this case the drop moving across the plates in the x-axis is deflected in the y-axis due to the vertical field.
The equation 1 works because acceleration is constant through out the motion.
If you still have doubts about the equation recall how it is derived.
ds = v.dt ,(now, since v - u= at and u = 0)
ds = at.dt , integrating both sides we get our equation.
Hope that clear things up.