Question 545: Electrostats & Capacitance

Question

A charged particle (mass m and charge q)
moves along X-axis with velocity V0. 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

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

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

Electrostats & Capacitance

A charged particle (mass m and charge q) moves along X-axis with velocity V0. 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: