JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
An electron with kinetic energy \(K _{1}\) enters between parallel plates of a capacitor at an angle \('\alpha'\) with the plates. It leaves the plates at angle \(' \beta '\) with kinetic energy \(K _{2}\). Then the ratio of kinetic energies \(K _{1}: K _{2}\) will be ....... .
- A \(\frac{\sin ^{2} \beta}{\cos ^{2} \alpha}\)
- B \(\frac{\cos ^{2} \beta}{\cos ^{2} \alpha}\)
- C \(\frac{\cos \beta}{\cos \alpha}\)
- D \(\frac{\cos \beta}{\sin \alpha}\)
Answer & Solution
Correct Answer
(B) \(\frac{\cos ^{2} \beta}{\cos ^{2} \alpha}\)
Step-by-step Solution
Detailed explanation
velocity along the plate will not change. \(\therefore v _{1} \cos \alpha= v _{2} \cos \beta\) \(\frac{ K _{1}}{ K _{2}} \Rightarrow \frac{ v _{1}^{2}}{ v _{2}^{2}}=\frac{\cos ^{2} \beta}{\cos ^{2} \alpha}\)
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