Class-12th Physics, CBSE and UP Board
Electric Charges and Fields,
Chapter-1, Exercise -1, Additional Exercise, Q-1.33
NCERT Solutions for Class 12th Physics
A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vₓ (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL² /(2m vₓ² ). Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.
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Charge on a particle of mass m = — q
Velocity of the particle = vx
Length of the plates = L
Magnitude of the uniform electric field between the plates = E
Mechanical force, F = Mass (m) x Acceleration (a)
=>a = F/m
a = qE/m ……………… (1) [as electric force, F = qE]
Time taken by the particle to cross the field of length L is given by,
t=(Length of the plate)/(Velocity of the particle)
t=L/Vx——————–(2)
In the vertical direction, initial velocity, u = 0
According to the third equation of motion, vertical deflection s of the particle can be obtained as.
s = ut + at2/2
=> 0 + 1/2 (qE/m) (L/Vx)2 [From (1) and (2)]
=> s= (qEL2)/(2mV2x)
Hence, vertical deflection of the particle at the far edge of the plate is (qEL2)/(2mV2x).
This is similar to the motion of horizontal projectiles under gravity.