Okay, here's the question:
In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The pattern on the paper is controlled by an electrostatic valve that determines at each nozzle position whether ink is squirted onto the paper or not. The ink drops have a radius of 20.0 um (micro) and leave the nozzle and travel toward the paper at a velocity of 24.0 m/s. The drops pass through a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length 2.50 cm where there is a uniform vertical electric field with a magnitude of 7.95×10^4 N/C .
If a drop is to be deflected a distance of 0.340 mm by the time it reaches the end of the deflection plate, what magnitude of charge must be given to the drop? (Assume that the density of the ink drop is 1000 kg/m^3
Alright, so I'm getting killed in this question and dont know what to do.
Me:
I'm keeping this and mind and overall, this is what my equation look like when I plug and chug.
## y= 1/2 (-qE/m) (L/v)^2 ##
>>> ## q= - (2ymv^2) / EL^2 ##
Now mine:
q= [-2(.34*10^-11)(1.0*10^-11)(23^2)]/[(7.9…
From this I got -2.081 *10^-16 C.... and it was wrong.
In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The pattern on the paper is controlled by an electrostatic valve that determines at each nozzle position whether ink is squirted onto the paper or not. The ink drops have a radius of 20.0 um (micro) and leave the nozzle and travel toward the paper at a velocity of 24.0 m/s. The drops pass through a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length 2.50 cm where there is a uniform vertical electric field with a magnitude of 7.95×10^4 N/C .
If a drop is to be deflected a distance of 0.340 mm by the time it reaches the end of the deflection plate, what magnitude of charge must be given to the drop? (Assume that the density of the ink drop is 1000 kg/m^3
Alright, so I'm getting killed in this question and dont know what to do.
Me:
I'm keeping this and mind and overall, this is what my equation look like when I plug and chug.
## y= 1/2 (-qE/m) (L/v)^2 ##
>>> ## q= - (2ymv^2) / EL^2 ##
Now mine:
q= [-2(.34*10^-11)(1.0*10^-11)(23^2)]/[(7.9…
From this I got -2.081 *10^-16 C.... and it was wrong.
-
can you say if the drop is falling vertically or moving horizontally ..
you say in the question that the drops go through a uniform vertical electric field this would suggest a drop moving horizontally between the plates ..
and I am assuming the field is down so that it pushes the charge down
I also assume that the gravity of the drop is negligible
this is a basic because the gravity is many thousand times lower than electrostatic force
==================
first we need the time in between the plates.. since velocity horizontal is fixed at 24.0 m/s we can find the time
d = 2.50 cm = 0.025 m
d = v*t
solve for t
t = d/v
= 2.5*10^-2 m/24.0 m/s
= 1.04 * 10^-3 seconds
y = 0.340*10^-3 m
y = 0.5*a*t^2
solve for a
a = 2y/t^2
= (2*3.40^10^-4)/(1.04*10^-3)^2
= 628.7 m/s^2
so we have the force thats needed we need to calculate the charge from this
F = Eq
F= ma
m = p*4*pi*r^3/3
where p is the density..
ma = Eq
solving for q
q = m*a/E
= p*4*pi*r^3*a/(3*E)
putting in the values
q = 1000*4*pi*(20*10^-6)^3*628.7/(3*7.95*10^…
= 2.65*10^-13 coulombs
hope that helps ...
you say in the question that the drops go through a uniform vertical electric field this would suggest a drop moving horizontally between the plates ..
and I am assuming the field is down so that it pushes the charge down
I also assume that the gravity of the drop is negligible
this is a basic because the gravity is many thousand times lower than electrostatic force
==================
first we need the time in between the plates.. since velocity horizontal is fixed at 24.0 m/s we can find the time
d = 2.50 cm = 0.025 m
d = v*t
solve for t
t = d/v
= 2.5*10^-2 m/24.0 m/s
= 1.04 * 10^-3 seconds
y = 0.340*10^-3 m
y = 0.5*a*t^2
solve for a
a = 2y/t^2
= (2*3.40^10^-4)/(1.04*10^-3)^2
= 628.7 m/s^2
so we have the force thats needed we need to calculate the charge from this
F = Eq
F= ma
m = p*4*pi*r^3/3
where p is the density..
ma = Eq
solving for q
q = m*a/E
= p*4*pi*r^3*a/(3*E)
putting in the values
q = 1000*4*pi*(20*10^-6)^3*628.7/(3*7.95*10^…
= 2.65*10^-13 coulombs
hope that helps ...
-
y = 1/2 ( q E/m) ( L/V)^2
0.34 x 10^(-3) = (1/2 ) (q x 7.95 x 10^4/ (4/3 pi r^3 x density)) ( 2.5 x 10^-2/ 24)^2
q = 8.408 x 10^-14
0.34 x 10^(-3) = (1/2 ) (q x 7.95 x 10^4/ (4/3 pi r^3 x density)) ( 2.5 x 10^-2/ 24)^2
q = 8.408 x 10^-14