Question

If an electron enters into a space between the plates of a parallel plate capacitor at an angle $\alpha$ with the plates and leaves at an angle $\beta$ to the plates. The ratio of its kinetic energy while entering the capacitor to that while leaving 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 B ${\left(\frac{\cos \beta }{\cos \alpha }\right)}^2$

Let $u$ be the velocity of electron while entering the field and $v$ the velocity when it leaves the plates. Component of velocity parallel to plates will remain unchanged.
Hence,
$u\cos \alpha =v\cos \beta$
or $\frac{u}{v}=\frac{\cos \beta }{\cos \alpha }$
The desired ratio is
$\frac{\frac{1}{2}m{u}^2}{\frac{1}{2}m{v}^2}={\left(\frac{u}{v}\right)}^2={\left(\frac{\cos \beta }{\cos \alpha }\right)}^2$