- #1
PhysicsTest
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- 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?
##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?