An electron projected at an acute angle `theta_(1) (ne 0^(@), ne 90^(@`) with the plates between the plates of a charged parallel plate capacitor. It leaves at an acute angle `theta_(2)(ne 0^(@), ne 90^(@))` with the plates. What will the ratio of its initial and final (leaving the capacitor) kinetic energies ?

Options

A. `(cos^(2) theta_(2))/(cos^(2)theta_(1))`
B. `(cos^(2)theta_(1))/(cos^(2)theta_(2))`
C. `(sin^(2)theta_(2))/(sin^(2)theta_(1))`
D. `(sin^(2) theta_(1))/(sin^(2) theta_(2))`

Answer

A

Explanation

Electric field and acceleration is perpendicular to the plate. Hence, velocity along the plate will not change.
`v_(1) cos theta_(1) = v_(2)cos theta_(2)`
Ratio of kinetic energy = `(1/2 mv_(1)^(2))/(1/2 mv_(1)^(2)) = v_(1)^(2)/v_(1)^(2) (cos^(2)theta_(2))/(cos^(2)theta_(1))`