Find the position of the electron at any time t

[PhysicsTest]

Homework Statement

The plates of a parallel - plate capacitor are d m apart. At t=0 an electron is released at the bottom plate with a velocity v0 (m/s)
normal to the plates. The potential of top plate with respect to the bottom is -Vm sin(wt).
a. Find the position of the electron at any time t
b. Find the value of the electric field intensity at the instant when the velocity of electron is 0.

Relevant Equations

F=ma; E = -dV/dx

At present i am only attempting the part a, i want to use the equation
$F=ma ; qE = ma;$ ---> eq1
The electric field is given by the formula
$E = -\frac {dV} {dx}$
$v = \frac {dx} {dt} => dx = v_0 {dt}$ ---> eq2 (?)
$E = -\frac{dV} {v_0 dt}$
Here $V = -V_m \sin(\omega t)$ hence
$E = \frac {-V_m\omega\cos(\omega t)} {v_0}$ substituting this in eq1
$m*\frac{d^2x} {dt^2} = q* \frac {-V_m\omega\cos(\omega t)} {v_0}$
$md^2x=\frac {-qV_m\omega} {v_0} \cos(\omega t) dt^2$
integrating once
$mdx = \frac {-qV_m} {v_0}\sin(\omega t) dt$
integrating one more time
$mx = \frac{qV_m} {v_0\omega} \cos(\omega t)$
$x = \frac{qV_m} {m v_0\omega} \cos(\omega t)$ ----> Answer
But i am not very convinced with eq2. Is my solution correct?

 

[TSny]

At present i am only attempting the part a, i want to use the equation
$F=ma ; qE = ma;$ ---> eq1

OK

The electric field is given by the formula
$E = -\frac {dV} {dx}$
$v = \frac {dx} {dt} => dx = v_0 {dt}$ ---> eq2 (?)
$E = -\frac{dV} {v_0 dt}$

eq (2) only holds for immediately after t = 0. So, the last equation above is not valid.

If the capacitor had a constant voltage $V$, what would be the electric field between the plates? You can assume that this relation between $E$ and $V$ still holds at each instant of time even when V is changing with time. This would not be true if $V$ is changing very rapidly with time (that is, if $\omega$ is very large). But I think you can safely assume that they do not want you to worry about the high frequency effects.

 

[Delta2]

Something doesn't look right with $$E=-\frac{dV}{v_0dt}$$, it is like saying that the electric field depends on the initial velocity of the electron, which doesn't seem right to me (unless we were looking for the electric field produced by the electron, but I am sure we can take this as negligible for this problem).

 

[PhysicsTest]

If i remove the equation 2, i have few confusions with signs but hope i have done correctly.
$F=ma ; qE = ma;$ ---> eq1
The electric field is given by the formula
$E = \frac {V_m\sin(\omega t)} d$
$-q \frac {V_m\sin(\omega t)} d = m \frac{d^2x} {dt^2}$
$-dm {d^2x} = qV_m\sin(\omega t) dt^2$
Integrating twice
$dmdx = \frac {qV_m\cos(\omega t)} {\omega} dt$
$dmx = \frac{qV_m\sin(\omega t)} {\omega^2}$
$x = \frac{qV_m\sin(\omega t)} {\omega^2 md}$

 

[PhysicsTest]

For the 2nd part b, using the constant energy
$\frac {mv_0^2} 2 = qV_{req}$
$E_{req} =\frac {V_{req}} d$
$E_{req} = \frac {mv_0^2} {2qd}$

 

[TSny]

$F=ma ; qE = ma;$ ---> eq1
The electric field is given by the formula
$E = \frac {V_m\sin(\omega t)} d$

OK. Just to check, if the time $t$ is such that the right hand side of the last equation above is positive, what is the direction of the field?

$-q \frac {V_m\sin(\omega t)} d = m \frac{d^2x} {dt^2}$

Are you taking $q$ to be a positive number so that the charge on the electron is $-q$?

$-dm {d^2x} = qV_m\sin(\omega t) dt^2$
Integrating twice

This is a very strange way to split up a second derivative. But the idea of integrating twice is good. If you do indefinite integrals, then there will be constants of integration that you must determine.

Here's a suggestion. Let $v$ represent the instantaneous velocity of the electron. Then the F = ma equation can be written as $m \large \frac{dv}{dt} = -q \frac {V_m\sin(\omega t)} d$. The first derivative can be "split up" to give

$\large m dv = -q \frac {V_m\sin(\omega t)} d dt$.

Now you can integrate to find $v$ as a function of time. If you do indefinite integration, then don't forget the constant of integration.

 

[PhysicsTest]

OK. Just to check, if the time is such that the right hand side of the last equation above is positive, what is the direction of the field?

E is negative of Voltage, top plate is -ve compared to bottom plate which is assumed "0". Hence the V is decreasing from bottom to top plate. The E field will then be -ve of this V, hence it will be from Top to Bottom plate, but since it is -ve charge, it will be from bottom to top plate. Sorry for the confusion.

Are you taking q to be a positive number so that the charge on the electron is -q?

Yes

$mv = q\frac {V_m\cos(\omega t)} {d\omega} + C$ when t=0; v=v0
hence $C=mv_0 - \frac {qV_m} {\omega d}$
$m\frac {dx} {dt} = q\frac {V_m\cos(\omega t)} {d\omega} + (mv_0 - \frac {qV_m} {\omega d})$
$mx = q\frac{V_m\sin(\omega t)} {d{\omega}^2} + (mv_0 - \frac {qV_m} {\omega d})$
$x = q\frac{V_m\sin(\omega t)} {md{\omega}^2} + (v_0 - \frac {qV_m} {md\omega })$

 

[TSny]

E is negative of Voltage, top plate is -ve compared to bottom plate which is assumed "0". Hence the V is decreasing from bottom to top plate.

The potential of the top plate relative to the bottom plate is $V = -V_m \sin(\omega t)$. So, at $t = 0$, there is no potential difference between the plates and $E = 0$ at this instant. During the first half-cyle of the oscillating voltage, $V$ will be negative and, as you say, the potential decreases as you go from the bottom plate to the top plate.

The E field will then be -ve of this V, hence it will be from Top to Bottom plate

No. If the top plate is at a lower potential than the bottom plate, then $E$ will point from the bottom plate to the top plate. (V decreases as you move in the direction of E.) So, during the first half-cylce, E points upward.

but since it is -ve charge, it will be from bottom to top plate.

Since the charge on the electron is negative, the direction of the force on the electron will be opposite to the direction of the field. So, during the first half-cycle, the force on the electron will be downward.

$mv = q\frac {V_m\cos(\omega t)} {d\omega} + C$ when t=0; v=v0
hence $C=mv_0 - \frac {qV_m} {\omega d}$
$m\frac {dx} {dt} = q\frac {V_m\cos(\omega t)} {d\omega} + (mv_0 - \frac {qV_m} {\omega d})$

OK

$mx = q\frac{V_m\sin(\omega t)} {d{\omega}^2} + (mv_0 - \frac {qV_m} {\omega d})$

Are you missing an overall factor of $t$ in the second term of the right hand side? Make sure that your equation agrees with $x = 0$ at $t = 0$.