Question

Two parallel plates having charges of equal magnitude but opposite sign are separated by 30.0 cm. Each plate has a surface charge density of 48.0 nC/m2 A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density. kN/C (b) Determine the potential difference between the plates. (c) Determine the kinetic energy of the proton when it reaches the negative plate. (d) Determine the speed of the proton just before it strikes the negative plate. 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 KN/C (h) How does your value of the electric field compare with that found in part (a)?

Answer 1

Given,

d = 30 cm ; sigma = 48 nC/m^2

a)we know that

E = sigma/e0

E = 48 x 10^-9/(8.85 x 10^-12) = 5423.73 N/C

Hence, E = 5.423 kN/C

b)E = V/d => V = Ed

V = 5423.73 x 0.30 = 1627.12 V

Hence, V = 1627.12 V

c)KE = U = qV

KE = 1.6 x 10^-19 x 1627.12 = 2.603 x 10^-16 J

Hence, KE = 2.603 x 10^-16 J

d)We know that

KE = 1/2 m v^2

v = sqrt (2KE/m)

v = sqrt (2 x 2.603 x 10^-16/1.67 x 10^-27) = 5.58 x 10^5 m/s

Hence, v = 5.58 x 10^5 m/s

e)F = ma

F = q E

q E = ma => a = q E/m

a = 1.6 x 10^-19 x 5.423 x 10^3/(1.67 x 10^-27) = 5.19 x 10^11 m/s^2

Hence, 5.19 x 10^11 m/s^2