Question
If the proton would not hit one of the plates, what would be the magnitude of its vertical displacement as it exits the region between the plates?

K 2.00 cm 0 1.00 cm
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Answer #1

As the proton is positively charged. the force will be in the downwards direction. We use the kinematics equations with the mass of the proton  

w =EeL²/ 2mvₓ² =

= 364 x1.6 x10⁻¹⁹ (.02)² /2 x 1.67x 10⁻²⁷ x(1.6 x10⁶)²

w=2.72 x10⁻⁶ m

the proton will displaced 2.72μm downwords

all the data has been taken from standard question

E can be calculated as follows 2M -t -2x 9小X/03k o.vosXI.(Xlof. E-304 N/

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