A capacitor of capacitance C is charged up to a potential difference V. After removing the charging battery, the capacitor is connected  In series with an uncharged capacitor of the same capacitance. What will be the effect on the potential difference of the first capacitor in each case?

Two identical capacitors ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ are connected with a battery B, as shown. A dielectric slab is filled between the plates of ${\mathrm{C}}_2$, the battery remaining connected. What happens to the charge, the capacitance, the potential difference and the stored energy of each capacitor?

Two capacitors ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ Are given charges $Q_1$ and $Q_2$ And the charging battery is removed.  Then they are connected in parallel and a dielectric slab is inserted in ${\mathrm{C}}_2$. When will happen to ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ ?

Two capacitors ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ are given charges $Q_1$ and $Q_2$ and the charging battery is removed. They are then connected in parallel and a dielectric slab is inserted in ${\mathrm{C}}_2$. When will happen to ${\mathrm{Q}}_1$ and ${\mathrm{Q}}_2$ ?

Two capacitors ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ Are given charges $Q_1$ and $Q_2$ And the charging battery is removed. They are then connected in parallel and a dielectric slab is inserted in ${\mathrm{C}}_2$. What will happen to PD ?