Please answare this one too! The positive and negative plates of a parallel-plat

Question

Please answare this one too!

The positive and negative plates of a parallel-plate capacitor have an area of 1.30 cm by 1.30 cm. Their surface charge densities are +1.00 times 10^-6 C/m^2 and -1.00 times 10^-6 C/m^2, respectively. A proton moving parallel to the plates enters the middle of the space between them at a speed of 2.25 times 10^6 m/s. Assuming the field outside the capacitor is 0 and the field inside is uniform, how far to the side will the proton's path have deviated when it gets to the far end of the capacitor?

Explanation / Answer

given

L = 1.3 cm = 0.013 m

sigma = 1*10^-6 c/m^2

electric field between the plates, E = sigma/epsilon

= 1*10^-6/(8.854*10^-12)

= 1.13*10^5 N/c

now use, F = q*E

m*a = q*E

a = q*E/m

= 1.6*10^-19*1.13*10^5/(1.67*10^-27)

= 1.08*10^13 m/s^2

time taken to cross the the capacitor, t = L/v

= 0.013/(2.25*10^6)

= 5.78*10^-9 s

the dflection of proton, d = (1/2)*a*t^2

= (1/2)*1.08*10^13*(5.78*10^-9)^2

= 1.8*10^-4 m or 0.18 mm

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