Question

Hi really need help with this Physics question in the picture ASAP. Thank you.

Question 2 The positive and negative plates of a parallel-plate capacitor have an area of 3.60 cm by 3.60 cm. Their surface charge densities are +1.00x10-6 C/m2 and -1.00x10-6 C/m2, respectively. A proton moving parallel to the plates enters the middle of the space between them at a speed of 4.75x106 m/s. Assuming the field outside the capacitor is O and the field inside is unifornm how far to the side will the proton's path have deviated when it gets to the far end of the capacitor? Click "Submit Answer" below to submit all 2 of your answers. Submit Answer Tries 0/10

Answer 1

Charge of proton (q) = 1.6 times 10^-19 C Area of plates(A) = 3.60 times 3.60 cm^2 = 12.96 cm^2 = 12.96 times 10^-4 m^2 surface charge density of the plates is (sigma) = 1 times 10^-6 c/m^2 Initial velocity of proton (u) = 4.75 times 10^6 m/s Electric field b/w the plates E = sigma/E_0 mass of proton (m) = 1.67 times 10^-27 kg acceleration of proton (a) = Eq/m a = 1 times 10^-6 times 1.6 times 10^-19/8 854 times 10^-12 times 1.67 times 10^-27 a = 1.1 times 10^13 m/s^2 time taken by proton to reach the end of capacitor t = Distance/speed = 3.6 times 10^-2/4.79 times 10^6 = 7.6 times 10^-9 sec \r\n x = (4.75 times 10^-6)(7.6 times 10^-9) + 1/2(1.1 times 10^13)(7.6 times 10^-9)^2 x = 0.03631768 m = 3.6317 cm x = 3.63 cm Answer Therefore proton moves 363cm to the side of -ve place