Moving Plates Capacitor!

Nice Problem , however pretty simple as well ! I see you are contunisly posting beautiful physics problems , I appreciate you for this . Keep Posting more !

Let at $t=t$ , $x=vt$ distance is moved by both plates , hence sepration b/w plates is increasing , and other has decresing , But potential diff is same on both capacitor , So charge will flow from higher capcitance plate( 2 - plate ) to lower capicatance plate (1-plate) .

So we just equate pot. diff of both plates :

$\displaystyle{{ V }_{ 1 }={ V }_{ 2 }\\ { { q }_{ 1 } }/{ { C }_{ 1 } }=\sfrac { { q }_{ 2 } }{ { C }_{ 2 } } \\ (2{ Q }_{ o }-q)(d+x)=(2{ Q }_{ o }+q)(d-x)\quad ...\quad \quad (:{ Q }_{ o }=1\mu C\quad )\\ \cfrac { 2{ Q }_{ o }-q }{ 2{ Q }_{ o }+q } =\cfrac { d-x }{ d+x } \\ Apply\quad comp.\quad and\quad divdendo:\\ \cfrac { -q }{ 2{ Q }_{ o } } =\cfrac { -x }{ d } \\ q=\left( \cfrac { 2{ Q }_{ o } }{ d } \right) x\\ I=\cfrac { dq }{ dt } =\left( \cfrac { 2{ Q }_{ o } }{ d } { u }_{ o } \right) =6mA\\ Ans.}$

$q=CU$ so $\frac{dq}{dt}=Uo \frac{dC}{dt}$ we all know that $C=\epsilon\frac{A}{d}$ that yields $i=-\epsilon\frac{A}{d^2}Uo$ then $i=-\epsilon\frac{A}{d}\frac{Uo}{d}$ then $i=-C\frac{Uo}{d}$ that yields $i=-\frac{Qo}{d}$ because $Qo=CUo$ please note that this is half of the charge in the circuit ,finaly we need the absolute value of the current then $|i|=|-\frac{3*2\mu}{10^-3}|$ $|i|=|-\frac{6\mu}{10^-3}| = 6mA$