A uniform electric field $E =(8\,m / e ) V / m$ is created between two parallel plates of length $1 m$ as shown in figure, (where $m =$ mass of electron and $e=$ charge of electron). An electron enters the field symmetrically between the plates with a speed of $2\,m / s$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be........
A uniform electric field $E =(8\,m / e ) V / m$ is created between two parallel plates of length $1 m$ as shown in figure, (where $m =$ mass of electron and $e=$ charge of electron). An electron enters the field symmetrically between the plates with a speed of $2\,m / s$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be........
- [JEE MAIN 2022]
- A
$\tan ^{-1} (4)$
- B
$\tan ^{-1}(2)$
- C
$\tan ^{-1}\left(\frac{1}{3}\right)$
- D
$\tan ^{-1} (3)$
Similar Questions
An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E.$ The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance $h.$ The time of fall of the electron, in comparison to the time of fall of the proton is
An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E.$ The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance $h.$ The time of fall of the electron, in comparison to the time of fall of the proton is
- [NEET 2018]
In Millikan's oil drop experiment, a charged drop falls with terminal velocity $V$. If an electric field $E$ is applied in vertically upward direction then it starts moving in upward direction with terminal velocity $2V$.If magnitude of electric field is decreased to $\frac{E}{2}$, then terminal velocity will become
In Millikan's oil drop experiment, a charged drop falls with terminal velocity $V$. If an electric field $E$ is applied in vertically upward direction then it starts moving in upward direction with terminal velocity $2V$.If magnitude of electric field is decreased to $\frac{E}{2}$, then terminal velocity will become
A uniform electric field, $\vec{E}=-400 \sqrt{3} \hat{y} NC ^{-1}$ is applied in a region. A charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $2 \sqrt{10} \times 10^6 ms ^{-1}$. This particle is aimed to hit a target $T$, which is $5 m$ away from its entry point into the field as shown schematically in the figure. Take $\frac{ q }{ m }=10^{10} Ckg ^{-1}$. Then-
$(A)$ the particle will hit $T$ if projected at an angle $45^{\circ}$ from the horizontal
$(B)$ the particle will hit $T$ if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal
$(C)$ time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu s$ as well as $\sqrt{\frac{5}{2}} \mu s$
$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu s$
A uniform electric field, $\vec{E}=-400 \sqrt{3} \hat{y} NC ^{-1}$ is applied in a region. A charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $2 \sqrt{10} \times 10^6 ms ^{-1}$. This particle is aimed to hit a target $T$, which is $5 m$ away from its entry point into the field as shown schematically in the figure. Take $\frac{ q }{ m }=10^{10} Ckg ^{-1}$. Then-
$(A)$ the particle will hit $T$ if projected at an angle $45^{\circ}$ from the horizontal
$(B)$ the particle will hit $T$ if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal
$(C)$ time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu s$ as well as $\sqrt{\frac{5}{2}} \mu s$
$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu s$
- [IIT 2020]
A wooden block performs $SHM$ on a frictionless surface with frequency, $v_0$. The block carries a charge $+Q$ on its surface. If now a uniform electric field $\vec{E}$ is switched-on as shown, then the $SHM$ of the block will be
A wooden block performs $SHM$ on a frictionless surface with frequency, $v_0$. The block carries a charge $+Q$ on its surface. If now a uniform electric field $\vec{E}$ is switched-on as shown, then the $SHM$ of the block will be
- [IIT 2011]
A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options then correctly describe the trajectory of the mass ? (Curves are drawn schematically and are not to scale).
A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options then correctly describe the trajectory of the mass ? (Curves are drawn schematically and are not to scale).
- [JEE MAIN 2020]