An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 N C–1 [Fig. 1.10(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. 1.10(b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’.
The electric force on a charged _particle in an electric _field is :
F = qE
where F is the force on the particle, q is the charge on the particle, and E is the electric field strength.
For an electron with charge q = -1.6 × 10^-19 C, falling through a distance of 1.5 cm in an electric field of magnitude E = 2.0 × 10^4 N C^-1, the force on the electron is:
F = qE = -1.6 × 10^-19 C × 2.0 × 10^4 N C^-1 = -3.2 × 10^-15 N
Since the force is in the opposite direction to the motion of the electron, it slows down the electron's fall. The acceleration of the electron is:
a = F/m
where m is the mass of the electron, which is 9.11 × 10^-31 kg.
a = -3.2 × 10^-15 N / 9.11 × 10^-31 kg = -3.52 × 10^15 m s^-2
The negative sign(-ve) indicates that the acceleration is in the opposite direction to the electric field. Using the equation of motion for uniformly accelerated motion, the time taken by the electron to fall through a distance of 1.5 cm is:
s = (1/2)at^2
where s =distance fallen & t = time taken.
1.5 × 10^-2 m = (1/2) × (-3.52 × 10^15 m s^-2) t^2
t = √[(1.5 × 10^-2 m) / (1/2) × (-3.52 × 10^15 m s^-2)] = 6.49 × 10^-8 s
For a proton with charge q = 1.6 × 10^-19 C falling through the same distance in the same electric field, the force on the proton is:
F = qE = 1.6 × 10^-19 C × 2.0 × 10^4 N C^-1 = 3.2 × 10^-15 N
Since the force is in the same direction as the motion of the proton, it accelerates the proton's fall. The acceleration of the proton is:
a = F/m
where m = mass of the proton, which is 1.67 × 10^-27 kg.
a = 3.2 × 10^-15 N / 1.67 × 10^-27 kg = 1.91 × 10^12 m s^-2
Using the equation of motion for uniformly accelerated motion, the time taken by the proton to fall through a distance of 1.5 cm is:
s = (1/2)at^2
where s =distance fallen & t = time taken.
1.5 × 10^-2 m = (1/2) × (1.91 × 10^12 m s^-2) t^2
t = √[(1.5 × 10^-2 m) / (1/2) × (1.91 × 10^12 m s^-2)] = 1.32 × 10^-5 s
Comparing the time taken for the electron and proton to fall through the same distance, we see that the proton falls faster than the electron due to the opposite directions of the electric force and the charges of the particles. This is analogous to free fall under gravity, where objects with different masses fall at the same rate in a vacuum. However, in the presence
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