Question

Suppose two plates lie in parallel horizontal planes, one plate in the xy plane at z = 0 and the other plate in the plane that is parallel to the xy plane at z = 10 mm. Between the plates is a constant electric field directed vertically upward (that is, in the positive z direction). A proton and an electron are launched in the positive x direction with the same initial velocity from position (0, 0, 5.0 mm).

Part A :If the proton strikes a plate at (200 mm , 0, 10 mm), where does the electron strike?

Answer 1

The expression for the electric force acting on the charge q is, And the expession for the force by using Newton's second law is, F ma By equating the above two equation of forces, and rearrange for the acceleration as follows: xa Now, the expression for the speed along z-direction using kinematic equation is, 2mz GE

Using this equation, the distance in x-direction is expressed as follows: r-v(t) 2mz GE 2z wa constant for both proton and electron m, m (9.1x10- kg) 1.67 x10" kg) - (200 mm) -27 = 4.7 mm Therefore, the electron strikes at (4.7 mm, 0, 0)