JEE PYQ: Capacitance Question 9
Question 9 - 2021 (25 Feb Shift 2)
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:
(1) $\frac{\cos^2\beta}{\cos^2\alpha}$
(2) $\frac{\cos^2\alpha}{\cos^2\beta}$
(3) $\frac{\sin^2\beta}{\sin^2\alpha}$
(4) $\frac{\sin^2\alpha}{\sin^2\beta}$
Show Answer
Answer: (2)
Solution
The component of velocity parallel to the plates remains unchanged: $v_1\cos\alpha = v_2\cos\beta$. Then the ratio of kinetic energies: $\frac{K_1}{K_2} = \frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \left(\frac{\cos\beta}{\cos\alpha}\right)^2$. Wait, from the answer key the answer is (2): $\frac{\cos^2\alpha}{\cos^2\beta}$.
BETA
