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  • A charged particle (mass m and charge q ) moves along X axis with velocity Vo. When It passes through the origin it enters a region having uniform electric field 

A charged particle (mass m and charge q ) moves along X axis with velocity Vo. When It passes through the origin it enters a region having uniform electric field  \vec{E}= -E\hat{j} which extends upto x=d. Equation of path of electron in the region x > d is :
Option: 1 y=\frac{qEd}{mV_{0}^{2}}\left ( x-d \right )
 
Option: 2 y=\frac{qEd}{mV_{0}^{2}}\left ( \frac{d}{2}-x \right )
Option: 3 y=\frac{qEd}{mV_{0}^{2}}\;x  
Option: 4 y=\frac{qEd^{2}}{mV_{0}^{2}}\;x

Answers (1)

best_answer

V_{Y}

 For \ \ x \leq d

Along x-axis

\\x=V_0t \\ v_x=V_0\\ a_x=0

So for x=d 

t=\frac{d}{V_0}

Now for y-axis

F=qE=ma\\ \\ \Rightarrow a=\frac{qE}{m}

So

y=\frac{1}{2}a_yt^2 \\ \\ v_y=a_yt \\ \\ a_y=\frac{qE}{m}

So at point P

Along x

v_x_p=V_0

x_p=d

Along y

v_{yp}=a_yt=\frac{qE}{m}*\frac{d}{V_0}

y_p=\frac{1}{2}a_yt^2=\frac{1}{2}*\frac{qE}{m}*(\frac{d}{V_0})^2

 

So now for x \geq d

y=-y_p-(v_{yp}t) ....(1)

x=x_p+(v_{xp}t) \\ \Rightarrow t=\frac{x-x_p}{v_{xp}}......(2)

Put equation (2) in equation (1)

We get

\\ y=-y_p-(v_{yp}*(\frac{x-x_p}{v_{xp}}.))\\ y=-(\frac{1}{2}*\frac{qE}{m}*(\frac{d}{V_0})^2) -((\frac{qE}{m}*\frac{d}{V_0})*(\frac{x-d}{V_0})) \\y=(\frac{qE}{m}*\frac{d}{(V_0)^2})*(-\frac{d}{2}-(x-d))\\ y=(\frac{qE}{m}*\frac{d}{(V_0)^2})*(\frac{d}{2}-x)

 

 

 

Posted by

avinash.dongre

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