Potential difference between two large plate changing with time

[AGNuke]

The potential difference between two large parallel plates is varied as V=at; a is a positive constant and t is time. An electron starts from rest at t=0 from the plate which is at lower potential. If the distance between the plates is L, mass of electron m and charge on electron -e then find the velocity of the electron when it reaches the other plate.


I attempted the question by equating the relation between time & velocity and time & distance.

$$A=\frac{Vq}{x}=\frac{eat}{mL}$$
$$\int_{0}^{v}dv=\int_{0}^{t}{Adt} \Rightarrow v=\frac{eat^{2}}{2mL}$$
$$\int_{0}^{L}dx=\int_{0}^{t}vdt+\int_{0}^{t}atdt \Rightarrow L=\frac{ea}{2mL}\int_{0}^{t}t^{2}dt+\frac{ea}{mL}\int_{0}^{t}t^{2}dt=\frac{eat^3}{2mL}$$

Now dividing the cube of first equation with square of second equation, to eliminate the time, I got $$v=\left ( \frac{eaL}{2m} \right )^{1/3}$$

Now my problem is that the answer stated is $$v=\left ( \frac{9eaL}{2m} \right )^{1/3}$$

Any problems as for what I have done?

 

[tiny-tim]

Hi AGNuke!

I don't understand why you're integrating adt …

$$\int_{0}^{L}dx=\int_{0}^{t}vdt+\int_{0}^{t}atdt$$

why not just use x = ∫vdt ?

 

[AGNuke]

Oops... my fault. I thought that since acceleration is also changing, I better write it too, but I forgot that I already made it up for it when writing v.

Thanks tim, I got my 9 in the place.