Two paralilel plates having charges of equal magnitude but opposite sign are sep

Question

Two paralilel plates having charges of equal magnitude but opposite sign are separated by 30.0 cm. released from rest at the positive plate. (o) Determine the magnitude of the electric field between the plates from the charge density 4.85 KN/C (b) Determine the potential difference between the plates 145103V c) Determine the kinetic eneray of the proton when it reaches the negative plate 2332 10-16 (d) Determine the speed of the proton just before it strikes the negative plate. 279310"2 Your response differs from the correct answer by more than 10%. Double check your calculations. km/s e) Determine the acceleration of the proton m/s2 towards the negative plate (f) Determine the force on the proton. N towards the negative plate (9) From the force, find the magnitude of the electric field. (h) How does your value of the electric field compare with that found in part (a)?

Explanation / Answer

given
d = 30 cm = 0.3 m
sigma = 43 nC/m^2 = 43*10^-9 C/m^2

a) E = sigma/epsilon

= 43*10^-9/(8.854*10^-12)

= 4856 N/C

= 4.86 kN/C

b) delta_V = E*d

= 4.86*10^3*0.3

= 1.46*10^3 V

c) KE = workdone on proton

= q*delta_V

= 1.6*10^-19*1.46*10^3

= 2.24*10^-16 J

d) use, KE = (1/2)*m*v^2

==> v = sqrt(2*KE/m)

= sqrt(2*2.24*10^-16/(1.67*10^-27))

= 5.18*10^5 m/s

e) a = F/m

= q*E/m

= 1.6*10^-19*4.86*10^3/(1.67*10^-27)

= 4.66*10^11 m/s^2

f) F = q*E

= 1.6*10^-19*4.86*10^3

= 7.78*10^-16 N

g) E = F/q

= 7.78*10^-16/(1.6*10^-19)

= 4.86*10^3 N/C

h) both are same.

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