AL phy EM 唔識做 =________= 2

10. (a) V increases linearly from 0 v to +6v, then to +9v and then drop to 0 v
(b) since E = = -dV/dr
E is uniform from 0 to +6v, +6 to +9v and +9v to 0 v.
The strength of E is strongest between +9v and 0v, the weakest between +6v to +9v

11. Q = CV
where Q is the charges on the plates, C is the capacitance and V is the voltage across the plates.
(a) Separation unchanged ==> C remains constant
Increase of V leads to increase of Q. Hence Q is doubled.
(b) Plate separation is halved ==> C is doubled
Hence, Q is halved.

12. (a) Since the electrostatic force is acting upward in order to balance the weight of the particle, the charges are -ve so that it is attracted by the +ve upper plate.

Electric field intensity = 385/0.005 v/m = 77 000 v/m
hence, (3.8x10^-15)g = 77000Q
where g is the acceleration due to gravity
Q is the charge on the particle
solve for Q gives Q = 4.94 x 10^-19 C

(b)(i) Increase of plate separation ==> decrease of electric field intensity
This leads to a decrease of the upward electrostatic force, hence the particle would fall down.
(ii) Increase of p.d. ==> increase of electric field intensity
Hence, the upward electrostatic force increases. The particle rises.

(c) Electrostatic force = 4.94x10^-19 x (100/0.005) N = 9.88 x10^-15 N
Acceleration = 9.88x10^-15/3.8x10^-15 m/s2 = 2.6 m/s2
Hence using equation of motion: v^2 = u^2 + 2as
velocity before reaching the lower plate = square-root[2x2.6x0.005] m/s = 0.16 m/s

13. Time of flight = 0.1/2x10^7 s = 5x10^-9 s
Consider the vertical motion of the electron
acceleration = (1.6x10^-19) x (250/0.05) /9.11x10^-31 m/s2 = 8.78 x 10^14 m/s2
Apply equation of motion: s = ut + (1/2)at^2 to the vertical motion of the electron
s = (1/2).(8.78x10^14).(5x10^-9)^2 m = 0.011 m
The max vertical deflection = 0.011 m

10)
圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/Feb11/Crazyelectrostatic1.jpg
11a) The capacitance of the parallel plate capacitor remains unchanged. So by Q = CV, the charge on each plate will be doubled if the voltage is doubled.

b) With the plate separation halved, the capacitance is doubled. By Q = CV again, the charge will be doubled if the voltage remains unchanged.

12a) E-field strength between the plates = 385/0.005 = 77000 V/m

Suppose that the charge is q, then

77000q = 3.8 x 10-15 x 10

q = 4.94 x 10-19 C

b i) With the separation increased, the E-field strength decreases (by E = V/d) and so the electric force on the particle decreases. Hence the particle will gradually fall down.

ii) With the p.d. increased, the E-field strength increases (by E = V/d) and so the electric force on the particle increases. Hence the particle will gradually go up.

c) Now the E-field strength is 100/0.005 = 20000 V/m

Hence the electric force on the particle = 4.94 x 10-19 x 10000 = 4.94 x 10-15 N

So net force on the particle = 3.8 x 10-14 - 4.94 x 10-15 = 3.31 x 10-14 N downward

So acc. = (3.31 x 10-14)/(3.8 x 10-15) = 8.70 m/s2

Using v2 - u2 = 2as with u = 0, a = 8.7 and s = 0.005, we have v = 0.29 m/s

13) Time of flight of the electron within the midway between the plates = 0.1/(2 x 107) = 5 x 10-9 s

E-field strength between the plates = 250/0.05 = 5000 V/m

So vertical acc. of electron when between the plates = 5000 x 1.6 x 10-19/(9.11 x 10-19) = 8.78 x 1014 m/s2 upward (neglecting gravity)

Applying s = ut + at2/2 in vertical direction with u = 0, a = 8.78 x 1014 and t = 5 x 10-9, we have s = 0.0110 m

which is the max. vertical deflection.

2011-02-18 09:12:12 補充：
12a) The charge is negative since the E-field is downward. To keep it in equilibrium, an upward force is needed.