A body having specific charge $8\,\mu {C} / {g}$ is resting on a frictionless plane at a distance $10\, {cm}$ from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of $100 \,{V} / {m}$ is applied horizontally toward the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be $....\, S.$
A body having specific charge $8\,\mu {C} / {g}$ is resting on a frictionless plane at a distance $10\, {cm}$ from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of $100 \,{V} / {m}$ is applied horizontally toward the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be $....\, S.$
- [JEE MAIN 2021]
- A
$1$
- B
$5$
- C
$7$
- D
$8$
Similar Questions
An electron falls through a distance of $1.5\; cm$ in a uniform electric field of magnitude $2.0 \times 10^{4} \;N C ^{-1} \text {[Figure (a)]} .$ The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Figure $(b)] .$ Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.
An electron falls through a distance of $1.5\; cm$ in a uniform electric field of magnitude $2.0 \times 10^{4} \;N C ^{-1} \text {[Figure (a)]} .$ The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Figure $(b)] .$ Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.
An electron falls through a distance of $1.5\, cm$ in a uniform electric field of magnitude $2.0\times10^4\, N/C$ as shown in the figure. The time taken by electron to fall through this distance is ($m_e = 9.1\times10^{-31}\,kg$, Neglect gravity)
An electron falls through a distance of $1.5\, cm$ in a uniform electric field of magnitude $2.0\times10^4\, N/C$ as shown in the figure. The time taken by electron to fall through this distance is ($m_e = 9.1\times10^{-31}\,kg$, Neglect gravity)
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 $\sqrt{\frac{5}{2}} \mu s$
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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]
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- [NEET 2018]