Question

An electron of mass me and charge q=−e is injected horizontally midway between two very large oppositely charged plates. The upper plate has a uniform positive charge per unit area +σ and the lower plate has a uniform negative charge per unit area −σ . You may ignore all edge effects. The particle has an initial velocity v⃗ 0=v0xˆ . What is the y-component of the position of the particle when the particle reaches the plane defined by x=L?
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Answer 1

The velocity in the x direction remains constant as the acceleration is only in the Y direction. This can be checked for consistency by finding the direction of the field inside the capacitor. So now obtain the time taken by the x component of velocity to travel a distance of L.

This is the same time taken by the y component starting initially from rest to travel a distance of L. So use V = U + at, you have acceleration and time, simple substitution will land you at the answer :). U_y = 0

Answer 2

find the time taken by electron to reach dist l i.e (l/vo)....
now use this time in the equation V = u + at where u is zero, a is Ee/m, and t is l/vo...
and u will get Eel/mvo as the ans

Answer 3

Hello
Do you have a movement like the launch of a projectile, only the acceleration is upward since the charge is negative and the electric field is directed downward.
The equations of motion will
\[x(t)=x_{0}+v_{0x}t\] and
\[y(t)=y_{0}+v_{0y}+\frac{1}{2}at^{2}\] well
\[x_{0}=y_{0}=0\] and \[v_{0x}=v_{0}\] so
\[x(t)=v_{0}t\] and
\[y(t)=\frac{1}{2}at^{2}\]
When the electron travels a distance L and a height y rises, it travels a distance L at time t equal to:
\[x=L=v_{0}*t\] or \[t=L/v_{0}\]
And the acceleration?
May it is obtained by electric force that acts on the electron during the time t given above.
The resultant force will be:
\[F=qE=eE=2e\sigma/\epsilon_{0}=ma\] this is
\[a=2e\sigma/(m\epsilon_{0})\], now with a and t into y(t) we get
\[y(t=L/v_{0})=\frac{e\sigma}{m\epsilon_{0}}(\frac{L}{v_{0}})^{2}\]
Hope this helps