Problem 3. -G Two charged parallel plates have charge densities o and -o, where

Question

Problem 3. -G Two charged parallel plates have charge densities o and -o, where o 40.0nC/m2. The plates are sepa by a distance 0.150m. An a particle is released at rest near the positive plate. Ignore gravity in this problem. (a) What is the magnitude and direction of the electric field between the plates? (b) What is the potential difference between the plates? (e) the energy of the a particle when itis released? (Take the potential enersy to be zero at the negative plate.) (d) How fast is the a particle moving when it reaches the negative plate? (e) By comparing the electrostatic force on the a particle with the gravitational force on it, show that itis appropriate to ignore gravity.

Explanation / Answer

a) E = sigma/eo = 40e-9/8.84e-12 = 4524.9 N/C

b) dV = e*d = 4524.9*0.15 = 678.735 V

c) U = dV*q = 678.735*2*1.6e-19 = 2.171952e-16 J

D)
U = 0.5*m*v^2

v = sqrt(2U/m) = 2.56*10^5 m/s

E) Fe = E*q = 4524.9*2*1.6e-19 = 1.447968e-15 N

Fg = m*g = 6.6e-27*9.8 = 6.468e-26 N

Fe / Fg = 2.23*10^10

Fe = 2.23*10^10* Fg

Fe is very much greater than Fg

Fg = gravational fore can be ignored

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