Magnetism Exercise (06-08-2020)

Capacitance of the plates is

$C=\dfrac {\epsilon_0A}{d}$

Potential difference between the plates is

$V=\dfrac {i_0td}{\epsilon_0A}$

Force acting on the charge due to the electric field in the region between the plates is

$F=\dfrac {qV}{d}=\dfrac {i_0qt}{\epsilon_0A}$ acting vertically upwards.

The downward acceleration of the charge is

$\dfrac {dv}{dt}=g-\dfrac {i_0qt}{\epsilon_0mA}$

$\implies v=gt-\dfrac {i_0qt^2}{2\epsilon_0mA}$

The charge will just miss hitting the lower plate if at this point the vertical component of it's velocity is zero, which will happen at time

$t_0=\dfrac {2\epsilon_0mAg}{i_0q}$

Then,

$\dfrac d2=\displaystyle \int_0^{t_0} vdt =\dfrac 23 \dfrac {\epsilon_0^2m^2A^2g^3}{i_0^2q^2}$

$\implies i_0=\sqrt {\dfrac 43}\dfrac {\epsilon_0mAg^{\frac 32}}{q\sqrt d}$

Therefore $α=\sqrt {\frac 43}\approx \boxed {1.154}$ .

Consider a coordinate frame to be placed at the point of projection (Y-Axis upwards and X axis to the right)

Let the charge on the capacitor at time $t$ be $Q$ and since it is a constant current source:

$Q = i_o t$

The electric field between the plates is:

$E = \frac{Q}{A\epsilon_o}=\frac{i_ot}{A\epsilon_o}$

The X and Y components of acceleration of the particle are:

$\ddot{x}=0$ $\ddot{y} = -g + \frac{i_otq}{A\epsilon_om}$

Solving:

$\dot{x} = v_o$ $\dot{y} = -gt + \frac{i_ot^2q}{2A\epsilon_om}$

$x = v_ot$ $y = -\frac{gt^2}{2} + \frac{i_ot^3q}{6A\epsilon_om}$

Having solved the Y-motion, if the particle must just miss the bottom of the plate, it's minimum value must be $y_{min}= -d/2$ . This essentially means that that happens at the time when:

$\dot{y}=0$

Solving for time one gets:

$t_o = \frac{2A\epsilon_o mg}{i_o q}$

Finally, using the above expression:

$y(t) = -\frac{gt^2}{2} + \frac{i_ot^3q}{6A\epsilon_om}$

$-\frac{d}{2} = y \left(t_o\right)$

Finally, $i_o$ can be solved for and the resulting expression is (simplifications left out):

$i_o = \frac{2}{\sqrt{3}}\left(\frac{m g^{1.5}A \epsilon_o}{q\sqrt{d}}\right)$