Acceleration, energy, power??

Question

pLs heLp me solve these..i already know the answer in #1..i really dont know how to solve the rest...

1. the parallel cathode and plate of a vacuum tube are 2cm apart and have 120 V between them. what is the time required for an electron to travel from cathode to plate?

answer: 6.2 X 10^-9 sec

2. refer to question 1. what is the acceleration of the electron as it reaches the plate?

a. 3.71 X 10^14 m/sec^2

b. 1.04 X 10^15 m/sec^2

c. 2.07 X 10^15 m/sec^2

d. 3.57 X 10^15 m/sec^2

3. what is the energy dissipated as heat on the plate in question 1?

a. 1.89 X 10^-17 J

b. 3.21 X 10^-17 J

c. 1.11 X 10^-16 J

d. 4.00 X 10^-16 J

4. refer to question 1. if the plate current of the tube is 7.5 mA, how much power is dissipated as heat on the plate of the tube?

a. 0.9 W

b. 1.8 W

c. 5.2 W

d. 7.1 W

tHankz!!! 10 pointz for the verY best answer...^_^

Answer 1

#1  

In solving this, you probably used V = Ed, F = ma = Eq and

d = .5at^2.

So E= ma/q = V/d

substituting in a = 2d/t^2 gives...

V/d = m(2d/t^2) / q

V/d^2 = 2m/(qt^2)

1/t^2 = (qV)/(d^2 * 2m)

t= sqrt((d^2 * 2m)/(qV)) = sqrt((2E-2^2 * 2(9.11E-31)/(1.60E-19 * 120)) = 6.2 E -9 ...  You are correct!

#2

E = ma/q = V/d

a = (Vq)/(md) = (120 * 1.6E-19)/(9.11E-31 * 2E-2) = 1.04E15 m/sec^2

Answer B

#3

ΔKE = .5 * m * vf^2  (because there was no initial velocity)

vf = at = 1.04E15 * 6.2E-9 = 6448000

.5 * m * (6448000)^2 =

.5 * 9.11E-31 * (6448000)^2 = 1.89E-17 J

Answer A



#4

P = IV

where V is voltage and I is current

P = 7.5E-3 * 120 = .9 W

Answer A