Charged dielectric/conductor in capacitor It is a standard problem to consider a dielectric or a conductor between the parallel plates of a capacitor. But what happens to capacity, voltage, charge, inserting between the plates of an ideal capacitor a charged dielectric or a charged conductor (without contact with the plates)? 
 A: If you insert a conductor without touching either plates you end up with two capacitors in series with the widths less than that of the original capacitor you had before you inserted the conductor.
A: There are two case it depends on the battery is connected with constant voltage then There are two main cases:-
1)When Battery is connected  $V$=constant
   a)Capacity:-

$$C_{0}=\frac{Q}{V_{0}}$$.For a medium permitivity constant will be written as In general,
$$\epsilon_{m}=\epsilon_{0}K$$($\epsilon_{m}$ the permitivity of medium)
So, Potential can be written as $$V_{0}=\frac{q}{4\pi\epsilon_{m}r}$$ 
I  just defined the $\epsilon_{m}$ so 
$$V=\frac{q}{4\pi\epsilon_{0}Kr}$$
$$V=\frac{V_{0}}{K}$$
So $$C=\frac{Q}{\frac{V_{0}}{K}}$$
So $$C=C_{0}K$$.
Hence here

$$C =KC_{0}$$

(Here $C_{0}$ is defined for the air )
b)Charge:-
 $Q$ is directly propotional to $C$ Hence 
$$Q=Q_{0}K$$
2) When Battery is not connected Q=constant
a) Potential 
Again $$C=KC_{0}$$
So V is inversely proportional to the the capacitance so,
$$V=\frac{V_{0}}{K}$$
This is the case when the dielectric is added.when it will be remove the cases will change.
