An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity 10⁶ m s⁻¹. If the magnitude of the electric field between the plates i…

Question

An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity 10⁶ m s⁻¹. If the magnitude of the electric field between the plates is 9.1 V cm⁻¹, then the vertical component of velocity of the electron is (mass of electron =9.1×10⁻³¹ kg and charge of electron =1.6×10⁻¹⁹ C)

Accepted Answer

(1) 16×10⁶ m s⁻¹. The field is transverse, so the horizontal motion is uniform: time inside the plates t=L/(vₓ)=(0.10)/(10⁶)=10⁻⁷ s. Convert the field: E=9.1 V cm⁻¹=910 V m⁻¹. Transverse acceleration a=(eE)/m=(1.6×10⁻¹⁹×910)/(9.1×10⁻³¹)=1.6×10¹⁴ m s⁻². v_y=at=1.6×10¹⁴×10⁻⁷=1.6×10⁷=16×10⁶ m s⁻¹.

Step-by-step solution

Why the other options are wrong