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Parallel Plate Capacitor Help Please?
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
1 Answer
- Steve4PhysicsLv 71 decade agoFavorite Answer
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.
