Question

21.If an electron enters in to a space between the plates of a parallel plates capacitor at an angle 'alpha' with the plates and leaves at an angle 'Beeta' to the plates. The ratio of its K.E while entering the capacitor to that while leaving will be

Answer 1

$$\begin{gathered} Theinitialkineticenergyofelectron=\frac{1}{2}m{u}^2. \\ whenelectronentereswithanangle\alpha andleavestheregionofcapacitorwith \\ anangle\beta then, \\ thekineticenergyatthetimeofleavingthecapacitor=\frac{1}{2}m{v}^2. \\ wherevisthevelocityatthetimeofleavingcapacitor. \\ \sin cenormaltothecapacitorplates,thereisnoforceactingontheelectron,so \\ u\sin \alpha =v\sin \beta . \\ orv=\frac{usin\alpha }{sin\beta } \\ Thereforeratioofinitialkineticenergytokineticenergyatthetimeofleavingthecapacitor \\ isgivenby, \\ {K}_1/K_2=\frac{{\displaystyle \frac{1}{2}}m{u}^2}{{\displaystyle \frac{1}{2}}m{v}^2}=\frac{u^2}{v^2}=\frac{u^2}{{\left(\frac{usin\alpha }{sin\beta }\right)}^2}=\frac{{\sin }^2\beta }{{\sin }^2\alpha }. \end{gathered}$$