A parallel-plate capacitor has 1.5 cm times 1.5 cm electrodes with surface charg
ID: 1526945 • Letter: A
Question
A parallel-plate capacitor has 1.5 cm times 1.5 cm electrodes with surface charge densities plusminus 1.0 times 10^-6 C/m^2. A proton traveling parallel to the electrodes at 1.4 times 10^6 m/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.Explanation / Answer
E = sigma/ epsilon
= 1.4*10^-6/8.854*10^-12
= 158120.63 N/c
u = 1.4*10^6 m/s (inital velcoty of proton)
F = q*E
m*a =q*E
a = q*E/m
= 1.6*10^-19*158120.63/1.67*10^-27
= 1.514*10^13 m/s^2
time taken to cross the plate, t = d/u =0.015/ 1.4*10^6 = 0.00000001071 s
deflection = 0.5*a*t^2
= 0.5*1.514*10^13*(0.00000001071)^2
= 0.0008683 m