A uniform electric field $\vec E$ exists between the plates of a charged condenser. A charged particle enters the space between the plates and perpendicular to $\vec E$ . The path of the particle between the plates is a
A uniform electric field $\vec E$ exists between the plates of a charged condenser. A charged particle enters the space between the plates and perpendicular to $\vec E$ . The path of the particle between the plates is a
- [JEE MAIN 2013]
- A
straight line
- B
hyperbola
- C
parabola
- D
circle
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
A positive charge particle of $100 \,mg$ is thrown in opposite direction to a uniform electric field of strength $1 \times 10^{5} \,NC ^{-1}$. If the charge on the particle is $40 \,\mu C$ and the initial velocity is $200 \,ms ^{-1}$, how much distance (in $m$) it will travel before coming to the rest momentarily
A positive charge particle of $100 \,mg$ is thrown in opposite direction to a uniform electric field of strength $1 \times 10^{5} \,NC ^{-1}$. If the charge on the particle is $40 \,\mu C$ and the initial velocity is $200 \,ms ^{-1}$, how much distance (in $m$) it will travel before coming to the rest momentarily
- [JEE MAIN 2022]
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)$
An electron and a proton are in a uniform electric field, the ratio of their accelerations will be
An electron and a proton are in a uniform electric field, the ratio of their accelerations 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........
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]