In a question, the left plate of a capacitor carried a charge Q while the switch was off;
When the switch is closed and the current stops flowing again, you're supposed to find the final charge on the right plate.
It's apparently done this way(say the electric field $= E$ ) ;
$$E = E_1 + E_2 \ \ \ \ \ \ \ \ \ \ \ \ $$
$$E = \frac{Q+x}{2A\epsilon_0}+\frac{x}{2A\epsilon_0} $$
Here, $V$ 's the potential, $A$ 's the surface area of the plates, and $d$ 's the distance between capacitor plates;
$$ V = Ed \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
$$ V = \frac{(Q+2x)d}{2A\epsilon_0}$$ Now,$$ C = \frac{\epsilon_0A}{D} $$ Now solving this;
$$x = CE - \frac{Q}{2}$$
But the final potential across the capacitor is always equal to the cell potential $\epsilon$, so why can't it be done this way?
$$\frac{Q + x}{C} - \frac{-x}{C} = E \ \ \ \ \ \ \ \ $$
$$Q+x+x = CE$$
$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -x \ =\frac{Q-CE}{2}$$
