# Suppose that the Particle in Exercise in 1.33 is an Electron Projected with Velocity Vx= 2.0 × 10^6 ms^−1. If E Between the Plates Separated by 0.5 cm is 9.1 × 10^2 N/C, - Physics

Numerical

Suppose that the particle 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, m= 9.1 × 10−31 kg)

#### Solution

Velocity of the particle, vx = 2.0 × 106 m/s

Separation of the two plates, d = 0.5 cm = 0.005 m

Electric field between the two plates, E = 9.1 × 102 N/C

Charge on an electron, q = 1.6 × 10−19 C

Mass of an electron, m= 9.1 × 10−31 kg

Let the electron strike the upper plate at the end of plate L, when deflection is s.

Therefore,

"s" = ("qEL"^2)/(2"mv"_"x"^2)

"L" = sqrt((2"dmv"_"x"^2)/("qE"))

= sqrt((2 xx 0.005 xx 9.1 xx 10^-31 xx (2.0 xx 10^6)^2)/(1.6 xx 10^-10 xx 9.1 xx 10^2))

= sqrt(0.025 xx 10^-2)

= sqrt(2.5 xx 10^-4)

= 1.6xx10^-2 m

= 1.6 cm

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

Concept: Coulomb’s Law - Force Between Two Point Charges
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 1 Electric Charges and Fields
Exercise | Q 1.34 | Page 50
NCERT Class 12 Physics Textbook
Chapter 1 Electric Charge and Fields
Exercise | Q 34 | Page 50
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