Two large conducting plates are placed 2 mm apart. The bottom plate is attached

Question

Two large conducting plates are placed 2 mm apart. The bottom plate is attached to ground so that its electrical potential is 0 V. The top plate is kept at a positive potential of 1000 V.
(a) Which plate is positively charged?
(b) What is the magnitude and direction of the electric field?
Magnitude V/m
direction
(c) A 4 g mass carries a negative charge of -4 µC and is places exactly at the center between the two plates. What is its potential energy? J
(d) If the mass in part (c) is initially at rest and then released, which plate does the mass hit?
(e) What is the velocity of the mass in part (d) just before it hits the plate? m/s

Explanation / Answer

(a) plate connected to earth is always at 0V so the other plate is positive charged. (b) Potential difference per unit legnth = (1000-0)/(0.002) = 5x10^5V/m. (c) Potential energy= electric field x distance. Electric field E = VQ where V we have found and Q is given. Thus, E = 5x10^5 * -4x10^-6 * 1mm = 5x10^5 * -4x10^-6 * 10^-3 = -20x10^-4 J. (d) the charge is negative so it will be attracted towards the positive plate. (e) force = eletcric field x charge. Thus, F = VQ^2 = 5x10^5 * (-4x10^-6)^2 = -8x10^-6. acceleration = force/mass = 0.002. Remember to take mass as 0.004Kg and not 4g. distance travelled s=0.001m. Thus, use formula v^2 = u^2 +2as where u=initial speed=0. Thus, v=sqrt(2*0.002*0.001) = 0.002m/s.

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