An electron enters a parallel plate capacitor | vedclass

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Question

(A) The electron is subjected to an electric force due to the electric field $E$ between the capacitor plates.The time taken by the electron to cross

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(A) The electron is subjected to an electric force due to the electric field $E$ between the capacitor plates.The time taken by the electron to cross the region of length $x$ between the plates is $t = \frac{x}{u}$.In this time, the acceleration of the electron in the $y$-direction is $a_y = \frac{F}{m} = \frac{eE}{m}$.The vertical velocity component $v_y$ acquired by the electron upon leaving the capacitor is $v_y = a_y t = \left(\frac{eE}{m}ight) \left(\frac{x}{u}ight) = \frac{eEx}{mu}$.The horizontal velocity component remains constant at $v_x = u$.The angle of deflection $	heta$ is given by $	an 	heta = \frac{v_y}{v_x} = \frac{eEx/mu}{u} = \frac{eEx}{mu^2}$.From this expression, we see that $	an 	heta \propto \frac{1}{u^2}$.Therefore, $\frac{	an 	heta_2}{	an 	heta_1} = \frac{u_1^2}{u_2^2}$.Given $	an 	heta_1 = 0.4$ and $u_2 = 2u_1$, we have:$	an 	heta_2 = 	an 	heta_1 \left(\frac{u_1}{u_2}ight)^2 = 0.4 \left(\frac{u_1}{2u_1}ight)^2 = 0.4 \left(\frac{1}{4}ight) = 0.1$.

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