Question

An electron enters into a space between the plates of parallel plate capacitor at an angle of $\alpha$ with the plates and leaves at an angle of $\beta$ to the plates. The ratio of its $KE$ while leaving to entering the capacitor will be :

• A. ${\left(\frac{\cos \alpha }{\cos \beta }\right)}^2$
• B. ${\left(\frac{\cos \beta }{\cos \alpha }\right)}^2$
• C. ${\left(\frac{\sin \alpha }{\sin \beta }\right)}^2$
• D. ${\left(\frac{\sin \beta }{\sin \alpha }\right)}^2$

Answer 1

The correct option is A ${\left(\frac{\cos \alpha }{\cos \beta }\right)}^2$

We know that Electric field is always perpendicular to the plates of the capacitor.

Let the speed of the electron be $v_1$ when it enters the gap between the plates and $v_2$ when it leaves the plates.

As the electron only experiences force in the direction perpendicular to the plates, by the conservation of momentum in the direction parallel to the plates, we have

$v_1\text{cos}\left(\alpha \right)=v_2\text{cos}\left(\beta \right)\Rightarrow \frac{v_2}{v_1}=\frac{\text{cos}\alpha }{\text{cos}\beta }$

The ratio of kinetic energy is given by $(\frac{v_2}{v_1}{)}^2=(\frac{\text{cos}\alpha }{\text{cos}\beta }{)}^2$