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, V

_{x}= 2.0 x 10^{6}m/sSeparation of the two plates, d = 0.5 cm = 0.005 m

Electric field between the two plates, E = 9.1 x 10

^{2}N/CCharge on an electron, q = 1.6 x 10

^{–}^{19}CMass of an electron, m

_{e}= 9.1 x 10^{-31}kgLet the electron strike the upper plate at the end of plate L, when deflection is s. Therefore,

s= (qEL

^{2})/(2mV^{2}_{x})=>L = √ (2smV

^{2}_{x})/qE= √ 2 x 0.005 x 9.1 x 10

^{-31})/(1.6 x 10^{–}^{19}x 9.1 x 10^{2}= √ 0.00025 = 0.016 m=1.6 cm

Therefore ,the electron will strike the upper plate after travelling 1.6cm.