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(C) The electron enters the electric field perpendicular to the plates. To turn it by $90^{\circ}$,the electric field must exert a force that causes a

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$10$

Vedclass · Answer

(C) The electron enters the electric field perpendicular to the plates. To turn it by $90^{\circ}$,the electric field must exert a force that causes a parabolic trajectory such that the displacement along the field direction is $d$ and along the initial velocity direction is $R = 1.0 \ cm = 0.01 \ m$.Let the initial velocity be $u$. The time taken to travel distance $R$ is $t = R/u$.In this time,the displacement in the direction of the electric field is $d = \frac{1}{2} a t^2 = \frac{1}{2} (\frac{qE}{m}) (\frac{R}{u})^2$.For the electron to emerge at $90^{\circ}$,the path must be such that the electric field deflects it by $d$ while it covers $R$. Given the geometry,the effective deflection required is $d = R = 0.01 \ m$.Thus,$R = \frac{1}{2} (\frac{qE}{m}) (\frac{R^2}{u^2}) \implies E = \frac{2 m u^2}{q R^2} = \frac{4 (KE)}{q R^2}$.Given $KE = 8.0 	imes 10^{-17} \ J$,$q = 1.6 	imes 10^{-19} \ C$,and $R = 0.01 \ m$:$E = \frac{4 	imes 8.0 	imes 10^{-17}}{1.6 	imes 10^{-19} 	imes (0.01)^2} = \frac{32 	imes 10^{-17}}{1.6 	imes 10^{-19} 	imes 10^{-4}} = \frac{32}{1.6} 	imes 10^6 = 20 	imes 10^5 \ N/C$.Wait,re-evaluating the trajectory: The electron enters perpendicular to the field. It follows a parabolic path. To exit at $90^{\circ}$ at a distance $R$,the field must be $E = \frac{2(KE)}{qR} = \frac{2 	imes 8.0 	imes 10^{-17}}{1.6 	imes 10^{-19} 	imes 0.01} = \frac{16 	imes 10^{-17}}{1.6 	imes 10^{-21}} = 10 	imes 10^5 \ N/C$.Thus,$y = 10$.

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