(a) As the e beam is deflected downward, the electric field must be acting in upward direction from the down plate to the upper one. So, lower plate must be positively charged.
(b) Initially, the beam has only horizontal velocity which remain constant through the length of the plate, l. So, using, s = vt, we have, time taken to travel the length of the plates,
t = s/v = l/Vx
Along vertical, initial velocity, u = 0
But the beam will experience a force due to which it is deflected downward, the acceleration, a ,for which can be found by using kinematic equation, s = ut + 1/2 at^2, where, s = vertical displacement of the beam, y = 5 mm = 0.005 m, u = 0, t = l/Vx = 0.1/1.5*10^7 = 6.7*10^-9 s, hence,
y = 0 + 1/2 at^2 => a = 2y/t^2
= 2*0.005/(6.7*10^-9)^2
Thus, acceleration, a = 1.49*10^15 m/s^2
(c) Again, a = eE/m, where, e = 1.67*10^-19 C,
m = mass of electron = 9.1*10^-31 kg, so,
Electric field, E = ma/e
= 9.1*10^-31*1.49*10^15/1.67*10^-19
= 8.12*10^3 N/C
(d) Now, potential between plates, V = Ed
Where, d = separation between plates = 0.005 m
and, capacitance, C = oA/d
Hence, charge on the capacitor plates
Q = CV = oA/d * Ed = oAE
Therefore, Q = 8.85*10^-12*0.01*8.12*10^3
= 7.186*10^-10 C
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