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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
Solution
Charge on a particle of mass $m =- q$
Velocity of the particle $= v _{ x }$
Length of the plates $= L$
Magnitude of the uniform electric field between the plates $= E$
Mechanical force, $F =$ Mass $( m ) \times$ Acceleration (a) $\Rightarrow a=\frac{F}{m}$
$\Rightarrow a=\frac{q E}{m} \ldots \therefore(1)$$\text { [as electric force, } F=q E]$
Time taken by the particle to cross the field of length $L$ is given by,
$t=\frac{\text { Length of the plate }}{\text { Velocity of the particle }}=\frac{L}{V_{x}} \ldots(2)$
In the vertical direction, initial velocity, $u=0$ According to the third equation of motion, vertical deflection s of the particle can be obtained as,
$s=ut+\frac 12 at^2$
$\Rightarrow s=0+\frac{1}{2}\left(\frac{q E}{m}\right)\left(\frac{L}{V_{x}}\right)^{2}$ $[\text { From }(1) \text { and }(2)]$
$\Rightarrow s=\frac{q E L^{2}}{2 m V_{x}^{2}}$
Hence, vertical deflection of the particle at the far edge of the plate is $\frac{q E L^{2}}{2 m V_{x}^{2}} .$ This is similar to the motion of horizontal projectiles under gravity.
