An electron enters horizontally between two parallel charged plates with an initial velocity of 4 10^6 ~m/s . The length of the plates is 9.1 ~cm . If the electron emerges from the electric field region at an angle of 45^ to its initial direction, the magnitude of the uniform electric field between the plates is (Given: mass of electron = 9.1 10⁻³¹ ~kg and charge of electron = 1.6 10⁻¹⁹ ~C )

Step-by-step solution

Let the initial horizontal velocity be v_x = 4 10^6 ~m/s . The time spent by the electron in the electric field is: t = L v_x = 9.1 10⁻² 4 10^6 ~s Since the electron emerges at an angle of 45^ , the final vertical component of velocity v_y is: (45^ ) = v_y v_x v_y = v_x = 4 10^6 ~m/s Using the first equation of motion for the vertical direction ( u_y = 0 ): v_y = a_y t = ( eE m ) t Substituting the expressions for v_y and t : v_x = ( eE m ) ( L v_x ) Rearranging to solve for E : E = m v_x^2 e L Substituting the giv