Parallel Plate Capacitor Help Please
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# Parallel Plate Capacitor Help Please

[From: ] [author: ] [Date: 11-05-02] [Hit: ]
07x10^13)x(5.13x10^-9)^2 / 2 = 1.19x10^-3m (= 1.Please check my arithmetic as its a very messy calculation.......
A proton traveling at a speed of 5.38 *10^6m/s enter the gap between the plates of a 2.76 cm wide parallel plate capacitor. The surface charge densities on the plate are +/- 8.38 *10^-6 C/m2. How far (in m) has the proton deflected sideways when it reaches the far edge of the capacitor? Assume that the electric field is uniform inside the capacitor and zero outside

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From Gauss's Law:
Electric field in capacitor = (charge density) / (epsilon_nought)
= 8.38x10^-6 / (8.85x10-12) = 9.47x10^5 N/C

Force on proton = qE = 1.6x10^-19 x 9.47x10^5 = 1.515x10^-13 N

Acceleration of proton = F/m = 1.515x10^-13 / 1.67x10^-27 = 9.07x10^13m/s^2

Time spent passing between plate distance / speed = 0.0276 / 5.38x10^6 = 5.13x10^-9s

Displacement (sideways deflection) = ut + at^2/2 = 0 + (9.07x10^13)x(5.13x10^-9)^2 / 2 = 1.19x10^-3m (= 1.19mm)

Please check my arithmetic as it's a very messy calculation.
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