Consider the leaky capacitor shown in the figure above. Specifications are as fo
ID: 1444428 • Letter: C
Question
Consider the leaky capacitor shown in the figure above. Specifications are as follows: The plate area is A =5.0 cm^2, and the gap distance is L = 1 mu m. The density of electrons that are not bound, and have the ability to move through the insulator, is n_e = 1 times 10^23 particles/m^3.The density of holes that are not bound, and have the ability to move through the insulator, is n_h = 1 times 10^23 particles/m^3, (a) Suppose that the electrons are immobile. For a current I = 2.0 mA across the imaginary boundary, what is the rate (in particles per second) at which holes pass by? (b) What average drift velocity (in meters/sec) must the holes have to give rise to this 2 mA current? (c) How many orders of magnitude smaller is the average drift velocity of a hole than the RMS thermal velocity of an electron moving in free space? (Use the law of equipartition to find the RMS thermal velocity of a free electron.)Explanation / Answer
A) particles per second = i/q
= 0.002/ 1.6×10^-19
=1.25 × 10^16 particles per second
B) i = n A e Vd
0.002 = 10^23 × 5×10^-4 ×1.6×10^-19 × Vd
Vd= 0.002/8 = 0.00025 m/s =2.5 × 10^-4 m/s
C) Electrons in free space acts like gas at room temperature
So. vrms = (3kT/m)
Putting mass of electron=9.1×10^-31kg, temperature as 298K, k =1.38×10^-23
we get Vrms = 1.16 ×10^5 m/s
drift velocity of hole is 9 orders of magnitude smaller than the Vrms of electron.