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A parallel-plate capacitor has 2.8cm � 2.8cm electrodes with surface charge densities �1.0�10?6C/m2. A proton traveling parallel to the electrodes at 1.0�106m/s enters the center of the gap between them.

By what distance has the proton been deflected sideways when it reaches the far edge of the capacitor? Assume the field is uniform inside the capacitor and zero outside the capacitor.

Express your answer to two significant figures and include the appropriate units.  Edit Comment  Solutions
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From Gauss's Law:

Electric field in capacitor = (charge density) / (epsilon_nought)

= 1x10^-6 / (8.85x10-12) = 1.12x10^5 N/C

Force on proton = qE = 1.6x10^-19 x 1.12x10^5 = 1.807x10^-13 N

Acceleration of proton = F/m = 1.807x10^-13 / 1.67x10^-27 = 1.082x10^13m/s^2

Time spent passing between plate distance / speed = 0.028 / 1x10^6 = 2.8x10^-8s

Displacement (sideways deflection) = ut + at^2/2 = 0 + (1.082x10^13)x(2.8x10^-8)^2 / 2 = 4.24x10^-3m (= 4.24mm)  Edit you are a lifesaver Member 338 Tue, 26 Sep '17
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