Class-12th Physics, CBSE and UP Board
Electric Charges and Fields,
Chapter-1, Exercise -1, Additional Exercise, Q-1.34
NCERT Solutions for Class 12th Physics
Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vₓ = 2.0 × 10⁶ m s⁻¹. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate? (|e|=1.6 × 10⁻¹⁹ C, mₑ = 9.1 × 10 ⁻³¹kg.)
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Velocity of the particle, Vx = 2.0 x 106 m/s
Separation of the two plates, d = 0.5 cm = 0.005 m
Electric field between the two plates, E = 9.1 x 102 N/C
Charge on an electron, q = 1.6 x 10–19 C
Mass of an electron, me= 9.1 x 10-31 kg
Let the electron strike the upper plate at the end of plate L, when deflection is s. Therefore,
s= (qEL2)/(2mV2x)
=>L = √ (2smV2x)/qE
= √ 2 x 0.005 x 9.1 x 10-31)/(1.6 x 10–19 x 9.1 x 102
= √ 0.00025 = 0.016 m=1.6 cm
Therefore ,the electron will strike the upper plate after travelling 1.6cm.