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Figure 22-35 shows the d eflecting plates of an ink-jet printer, with superimposed coordinate axes. An ink drop with a mass m of 1.3xx10^(-10)kg anda negative charge of magnitude Q=1.5xx10^(-13)C enters the region between the plates, initially moving along the x axis with speed v_(x)=18 m//s. The length L of each plate is 1.6 cm. The plates are charged and thus produce an elecric field at all points between them. Assume that field vecE is downward directed, is uniform, and has a magnitude of 1.4xx10^(6)N//C. What is the vertical deflection of the drop at the far edge of the plates? (The gravitational force on the drop is small relative to the electrostatic force acting on the drop and can be neglected.) |
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Answer» Solution :KEY IDEAS The drop is negatively charged and the electric field is directed downward. From Eq. 22-45, a constant electrostatic force of magnitude QE acts upward on the charged drop. Thus, as the drop travels parallel to the x axis at constant speed `v_(x)`, it ACCELERATES upward with some constant acceleration `a_(y)`. Calculations: Applying Newtons.s second law (F = ma) for components along the y axis, we FIND that `a_(y)=(F)/(m)=(QE)/(m). "" (22-47)` Let t represent the TIME required for the drop to pass through the REGION between the plates. During t the vertical and horizontal displacements of the drop are `y=(1)/(2) a_(y)t^(2) and L=v_(x)t,"" (22-48)` respectively. Eliminating t between these two equationsand substituting Eq. 22-47 for `a_(y)`, we find `y=(QEL^(2))/(2mv_(x)^(2))` `=((1.5xx10^(-13)C)(1.4xx10^(6)N//C)(1.6xx10^(-2)m)^(2))/((2)(1.3xx10^(-10)kg)(18m//s)^(2))` `=6.4xx10^(-4)m=0.64mm.` (Answer) |
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