Consider two large parallel plates charged with equal magnitude but opposite sig
ID: 585880 • Letter: C
Question
Consider two large parallel plates charged with equal magnitude but opposite signs as shown. The magnitude of the surface charge densities is sigma = 6.0 times 10^-22 C/m^2. The separation between the plates is d = 1.0 cm. Calculate the electric field to the left, in between, and to the right of the plates, respectively. A charge of magnitude q = 1.6 times 10^-19 C is placed just to the right of the negative plate and released from rest. It accelerates toward the positive plate. What is the sign of q? If the mass of the charge is m = 9.1 times 10^-31 kg, how long does it take to strike the positive plate? If the charge q is released with an initial velocity parallel to the plates, how all your answer in (b) change?Explanation / Answer
B)
Since it is moving towards the positive plate hence the charge on particle must be negative.
q = 1.6*10^-19 C
The force on the charge will be F = qE = 1.6*10^-19*6.78*10^-11 = 1.0848*10^-29 N
the mass of the charge m = 9.1*10^-31 kg
hence the acceleration a = F/m = 1.0848*10^-29/(9.1*10^-31) = 11.921 m/s^2
separation d = 1 cm = 0.01 m
hence time taken t = sqrt(2d/a) = sqrt(2*0.01/11.921) = 0.041 sec.