The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants \(K _1=2, K_2=3\), and \(K _3=5\) (as shown in the figure). If new capacitance is \(\frac{ n }{3} C\) then the value of \(n\) is ___________.

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance will become:

$A$ parallel plate capacitor of capacitance $200 \,\mu F$ is connected to a battery of $200 \,V$. $A$ dielectric slab of dielectric constant $2$ is now inserted into the space between the plates of the capacitor while the battery remains connected. The change in the electrostatic energy in the capacitor will be ...... $J$.

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is completely filled with a dielectric material of constant $K$,then the potential difference across the other capacitor will become:

$A$ parallel-plate capacitor of capacitance $40 \mu F$ is connected to a $100 V$ power supply. Now,the intermediate space between the plates is filled with a dielectric material of dielectric constant $K=2$. Due to the introduction of the dielectric material,the extra charge and the change in the electrostatic energy in the capacitor,respectively,are:

Two parallel plate capacitors of capacity $C$ and $3C$ are connected in parallel and charged to a potential difference of $18V$. The battery is then disconnected,and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $K=9$. The final potential difference across the combination of capacitors will be $V^{\prime}$. Find $V^{\prime}$. (in $V$)