An electron is made to enter symmetrically be | vedclass

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

Question

(C) The time taken by the electron to cross the plates of length $L = 10 \ cm = 0.1 \ m$ with a horizontal velocity $V_x = 10^6 \ m/s$ is:$t = \frac{L

Vedclass · Accepted Answer

$1.6 	imes 10^6 \ m/s$

Vedclass · Answer

(C) The time taken by the electron to cross the plates of length $L = 10 \ cm = 0.1 \ m$ with a horizontal velocity $V_x = 10^6 \ m/s$ is:$t = \frac{L}{V_x} = \frac{0.1}{10^6} = 10^{-7} \ s$The electric field $E = 9.1 \ V/cm = 9.1 	imes 10^2 \ V/m = 910 \ V/m$.The acceleration of the electron in the vertical direction is:$a_y = \frac{eE}{m} = \frac{1.6 	imes 10^{-19} 	imes 910}{9.1 	imes 10^{-31}} = \frac{1.6 	imes 10^{-19} 	imes 9.1 	imes 10^2}{9.1 	imes 10^{-31}} = 1.6 	imes 10^{14} \ m/s^2$The vertical component of velocity $V_y$ at the exit is given by $V_y = u_y + a_y t$,where $u_y = 0$:$V_y = 0 + (1.6 	imes 10^{14}) 	imes 10^{-7} = 1.6 	imes 10^7 \ m/s$. Wait,re-calculating: $E = 9.1 \ V/cm = 910 \ V/m$. $a_y = (1.6 	imes 10^{-19} 	imes 910) / (9.1 	imes 10^{-31}) = 1.6 	imes 10^{14} \ m/s^2$. $V_y = 1.6 	imes 10^{14} 	imes 10^{-7} = 1.6 	imes 10^7 \ m/s$. Given the options,there might be a typo in the question's provided options or the field value. If $E = 9.1 \ V/m$,then $V_y = 1.6 	imes 10^5 \ m/s$. If $E = 910 \ V/m$,$V_y = 1.6 	imes 10^7 \ m/s$. Given the options,$1.6 	imes 10^6 \ m/s$ is the closest intended answer if $E = 91 \ V/m$.

Similar Questions

An electron of mass $m$,charge $e$ falls through a distance $h$ meters in a uniform electric field $E$. Then the time of fall is:

Two metal plates having a potential difference of $800 \; V$ are $0.02 \; m$ apart horizontally. $A$ particle of mass $1.96 \times 10^{-15} \; kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is:

$A$ uniform electric field of $10^3 \, V/m$ is directed along the $y$-axis. $A$ particle of mass $1 \, g$ and charge $10^{-6} \, C$ is projected from the origin into the field with a velocity of $10 \, m/s$ along the positive $x$-axis. Its speed after $10 \, s$ is (in $m/s$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module