An electron of mass m and charge e enters symmetrically between two parallel plates of length L with an initial kinetic energy K . A uniform transverse electric field E is maintained between the plates. If the electron emerges from the region between the plates with a deviation angle , the magnitude of the electric field E is given by:

Step-by-step solution

The initial kinetic energy of the electron is K = 1 2 mv_x^2 , which gives mv_x^2 = 2K . The time taken by the electron to cross the plates of length L is t = L v_x . The transverse acceleration of the electron is a_y = eE m . The transverse velocity upon exiting the plates is v_y = a_y t = ( eE m ) ( L v_x ) . The angle of deviation is given by = v_y v_x = eEL mv_x^2 . Substituting mv_x^2 = 2K , we get = eEL 2K . Rearranging for E , we obtain E = 2K eL . Answer: 2K eL