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In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of 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 mass m = 1.00�10?11kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 18.0m/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 D0 = 2.00cm , where there is a uniform vertical electric field with magnitude E = 7.75�104N/C
1) If a drop is to be deflected a distance d = 0.310mm by the time it reaches the end of the deflection plate, what magnitude of charge q must be given to the drop? Assume that the density of the ink drop is 1000kg/m3 , and ignore the effects of gravity.
q = _______ C

Respuesta :

Answer:

[tex]q = 6.48 \times 10^{-14} C[/tex]

Explanation:

Deflection in the drop is due to electric field force

so we will have

[tex]F = qE[/tex]

acceleration of the drop is given as

[tex]a = \frac{qE}{m}[/tex]

[tex]a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}[/tex]

[tex]a = 7.75 \times 10^{15} q[/tex]

now we know that time to cross the plates is given as

[tex]t = \frac{D}{v}[/tex]

[tex]t = \frac{0.02}{18}[/tex]

[tex]t = 1.11 \times 10^{-3} s[/tex]

now the deflection is given as

[tex]d = \frac{1}{2}at^2[/tex]

[tex]0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2[/tex]

[tex]0.310 \times 10^{-3} = 4.78 \times 10^9 q[/tex]

[tex]q = 6.48 \times 10^{-14} C[/tex]

The magnitude of charge q must be given to the drop should be [tex]q = 6.48 \times 10^{-14}C[/tex]

Calculation of magnitude of charge q:

The magnitude of the electric field refers to the force per charge on the test charge. Since the electric field represents the force per charge, its units should be force units divided by charge units.

Deflection in the drop is because of electric field force

So,

[tex]F = qE[/tex]

Now acceleration of the drop should be

[tex]a = qE \div m\\\\= q(7.75 \times 10^{4}) \div (1.00 \times 10^{-11})\\\\= 7.75 \times 10^{15}q[/tex]

Now

time to cross the plates should be

[tex]t = D \div v\\\\t = 0.02 \div 18\\\\= 1.11 \times 10^{-3}s[/tex]

Now deflection should be

[tex]d = \frac{1}{2}at^2\\\\ 0.310 \times 10^{-3}= \frac{1}{2}(7.75 \times 10^{15}q) (1.11 \times 10^{-3})^2\\\\ 0.310 \times 10^{-3}=4.78 \times 10^{9}q\\\\q = 6.48 \times 10^{-14}C[/tex]

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