A mass $m = 20\,g$ has a charge $q = 3.0\,mC$. It moves with a velocity of $20\,m/s$ and enters a region of electric field of $80\,N/C$ in the same direction as the velocity of the mass. The velocity of the mass after $3$ seconds in this region is.......$m/s$
A mass $m = 20\,g$ has a charge $q = 3.0\,mC$. It moves with a velocity of $20\,m/s$ and enters a region of electric field of $80\,N/C$ in the same direction as the velocity of the mass. The velocity of the mass after $3$ seconds in this region is.......$m/s$
- A
$80$
- B
$56$
- C
$44$
- D
$40$
Similar Questions
A particle of mass $m$ and charge $(-q)$ enters the region between the two charged plates initially moving along $x$ -axis with speed $v_{x}$ (like particle $1$ in Figure). The length of plate is $L$ and an uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $q E L^{2} /\left(2 m v_{x}^{2}\right)$
Compare this motion with motion of a projectile in gravitational field
A particle of mass $m$ and charge $(-q)$ enters the region between the two charged plates initially moving along $x$ -axis with speed $v_{x}$ (like particle $1$ in Figure). The length of plate is $L$ and an uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $q E L^{2} /\left(2 m v_{x}^{2}\right)$
Compare this motion with motion of a projectile in gravitational field
There is a uniform electric field of strength ${10^3}\,V/m$ along $y$-axis. A body of mass $1\,g$ and charge $10^{-6}\,C$ is projected into the field from origin along the positive $x$-axis with a velocity $10\,m/s$. Its speed in $m/s$ after $10\,s$ is (Neglect gravitation)
There is a uniform electric field of strength ${10^3}\,V/m$ along $y$-axis. A body of mass $1\,g$ and charge $10^{-6}\,C$ is projected into the field from origin along the positive $x$-axis with a velocity $10\,m/s$. Its speed in $m/s$ after $10\,s$ is (Neglect gravitation)
An electron is released from the bottom plate $A$ as shown in the figure $(E = 10^4\, N/C)$. The velocity of the electron when it reaches plate $B$ will be nearly equal to
An electron is released from the bottom plate $A$ as shown in the figure $(E = 10^4\, N/C)$. The velocity of the electron when it reaches plate $B$ will be nearly equal to
An electron moving with the speed $5 \times {10^6}$ per sec is shooted parallel to the electric field of intensity $1 \times {10^3}\,N/C$. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of $e = 9 \times {10^{ - 31}}\,Kg.$ charge $ = 1.6 \times {10^{ - 19}}\,C)$
An electron moving with the speed $5 \times {10^6}$ per sec is shooted parallel to the electric field of intensity $1 \times {10^3}\,N/C$. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of $e = 9 \times {10^{ - 31}}\,Kg.$ charge $ = 1.6 \times {10^{ - 19}}\,C)$
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]