A problem of practical interest is to make a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$10$

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,$y = 10$.

Similar Questions

The kinetic energy of an electron which is accelerated through a potential of $100 \, V$ is:

An electron,initially at rest,is accelerated through a potential difference of $200 \ V$,so that it acquires a velocity of $8.4 \times 10^6 \ m/s$. The value of $e/m$ for the electron will be:

$A$ small sphere of mass $m$ carrying a charge $q$ is hanging between two parallel plates by a string of length $L$ as shown in the figure. The time period of the pendulum is $T_0$. When the parallel plates are charged as shown,the time period changes to $T$. The ratio $T / T_0$ is equal to:

An electron experiences a force equal to its weight when placed in an electric field. The intensity of the field will be

$A$ positively charged particle moving along the $x$-axis with a certain velocity enters a uniform electric field directed along the positive $y$-axis. Its

Vedclass Test Series

Exam Paper Generator

Online Exam Module