Question

1. Consider a parallel plate capacitor. The magnitude of the electric field in the capacitor is E = 2 x 10³N/C, and the plate separation is 5 x 10^-3m. What is the V (voltage) between the plates?
   1. 1.77 V
   2. 3.00 V
   3. 8.85 V
   4. 12.4 V
   5. 17.7 V

answer: B

2. Continuing from the previous problem: A test charge q₀= -2 x 10-3 C is released from rest at a negative plate of the capacitor. What is the kinetic energy of the test charge just before it reaches a positive plate?
   1. 6 x 10^-3J
   2. 2 x 10^-3J
   3. 0 J
   4. -3 x 10^-3J
   5. -6 x 10^-3J

answer: A

I not sure how they got those answers

Answer 1

E = 2 E+3 N/C ; d = 5.0 E-3 m

we assume the field E across the capacitor is uniform

Potential across the plates V = Ed = 2.0E+3 * 5.0E-3 = 10 V - ideal

due to fringe effect the field is not completely uniform across. at the ends of the plates the field extends slightly outside the area of the plates. Thus Ed >10 and V

V = 12.24 - (d)

test charge q_o = -2E-3 C

Force on the charge F = Eq_o

work done W = Fd = Eq_od = KE of the charge

= 2.0E+3 * 2.0E-3 * 5.0E-3 = 20 E-3 J