A uniform electric field is established between the plates of a parallel plate capacitor by holding one plate at ground and the other at a positive potential V as shown. A uniform magnetic field B is established perpendicular to the electric field (Fig. 13.2). A charge -q is released from rest from the lower plate. (i) Write down the equations of motion for the velocity components of the charge. (ii) Show that, at some time t later, the velocity of the electron in the x - direction is related to the distance y moved along the y-axis by vx = wy (iii) By applying the conservation of energy or otherwise to determine the square of the velocity in the x-y-plane, show that vy^2=(2qvy/md-w^2y^2)
Solution:
|