Question 428: Electrostatics & Capacitance

Question

A charged particle (mass m and charge q)
moves along X-axis with velocity V
0
. When it
passes through the origin it enters a region
having uniform electric field
$\overrightarrow E  =  - E\widehat j$ which
extends upto x = d.
Equation of path of electron
in the region x > d is
![](https://imagex.cdn.examgoal.net/jgy2xukeuun9zd/7767b511-eaf9-4331-908d-d4a98a55fb61/384ce390-f250-11ea-8293-3b6cc591631c/file-jgy2xukeuun9ze-203w.png)

Accepted Answer

Let particle have charge q and mass 'm
We know, F = ma
y
= qE
$ 	herefore $ ${a_y} = {{qE} \over m}$
As electric field is along vertical direction so horizontal component of acceleration is zero.
$ 	herefore $ a
x
= 0
Applying, $S = {u_y}t + {1 \over 2}{a_y}{t^2}$
$ - y = 0	imes t + {1 \over 2}\left( {{{qE} \over m}} ight){t^2}$ ....(1)
Also, $x = {V_0}t$
$ \Rightarrow t = {x \over {{V_0}}}$
Putting in (1) we get,
$y =  - {1 \over 2}\left( {{{qE} \over m}} ight){\left( {{x \over {{V_0}}}} ight)^2}$
Differentiating with respect to x,
${{dy} \over {dx}} =  - {{qE} \over m}\left( {{x \over {V_0^2}}} ight)$
$ 	herefore $ Slope = $m = 	an 	heta  =  - {{qE} \over m}\left( {{x \over {V_0^2}}} ight)$
when x = d, then $m =  - {{qE} \over m}\left( {{d \over {V_0^2}}} ight)$
We know, equation of straight line whose slope is m,
y = mx + c
Here m = $\left( { - {{qEd} \over {mV_0^2}}} ight)$,
$ 	herefore $ $ - {1 \over 2}\left( {{{qE} \over m}} ight){\left( {{d \over {{V_0}}}} ight)^2} = \left( { - {{qEd} \over {mV_0^2}}} ight)d + c$
$ \Rightarrow $ c = ${{{qEd} \over {2mV_0^2}}}$
Putting the value of c, we get
$y =  - \left( {{{qEd} \over {mV_0^2}}} ight)x + {{qE{d^2}} \over {2mV_0^2}}$
$ = {{qEd} \over {mV_0^2}}\left( {{d \over 2} - x} ight)$
![](https://imagex.cdn.examgoal.net/1kjzee8vp/6eac8556-2e80-47e7-8995-3b89d7dfedbd/ab3e2160-57cd-11eb-89ab-e9eae596edd7/file-1kjzee8vq-242w.png)

Answer

y = ${{qEd} \over {mV_0^2}}\left( {x - d} \right)$

Answer

y = ${{qEd} \over {mV_0^2}}\left( {{d \over 2} - x} \right)$

Answer

y = ${{qEd} \over {mV_0^2}}x$

Answer

y = ${{qE{d^2}} \over {mV_0^2}}x$

Electrostatics & Capacitance

A charged particle (mass m and charge q) moves along X-axis with velocity V 0 . When it passes through the origin it enters a region having uniform electric field \overrightarrow E = - E\widehat j which extends upto x = d. Equation of path of electron in the region x > d is

Choose the correct answer: