Question 296: Electrostatics & Capacitance

Question

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 :

Accepted Answer

$$ \because $$ $${v_1}\cos \alpha  = {v_2}\cos \beta $$
$${{{v_1}} \over {{v_2}}} = {{\cos \beta } \over {\cos \alpha }}$$
Then the ratio of kinetic energies
$${{{k_1}} \over {{k_2}}} = {{{1 \over 2}m{v_1}^2} \over {{1 \over 2}m{v_2}^2}} = {\left( {{{{v_1}} \over {{v_2}}}} \right)^2} = {\left( {{{\cos \beta } \over {\cos \alpha }}} \right)^2}$$
$$ \Rightarrow $$ $${{{k_1}} \over {{k_2}}} = {{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}$$

Answer

$${{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}$$

Answer

$${{\cos \beta } \over {\cos \alpha }}$$

Answer

$${{{{\sin }^2}\beta } \over {{{\cos }^2}\alpha }}$$

Answer

$${{\cos \beta } \over {\sin \alpha }}$$

Electrostatics & 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 :

Choose the correct answer: