The distance between plates of a parallel plate capacitor is $5d$. Let the positively charged plate is at $ x=0$ and negatively charged plate is at $x=5d$. Two slabs one of conductor and other of a dielectric of equal thickness $d$ are inserted between the plates as shown in figure. Potential versus distance graph will look like :

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel capacitor charged to a potential $V$. The two share the charge and the common potential is $V'$. The dielectric constant $K$ is

A capacitor when filled with a dielectric $K = 3$ has charge ${Q_0}$, voltage ${V_0}$ and field ${E_0}$. If the dielectric is replaced with another one having $K = 9$ the new values of charge, voltage and field will be respectively

A parallel plate capacitor has plate of length $'l',$ width $'w'$ and separation of plates is $'d'.$ It is connected to a battery of emf $V$. A dielectric slab of the same thickness '$d$' and of dielectric constant $k =4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored$?$

What is dielectric ?

Between the plates of a parallel plate condenser, a plate of thickness ${t_1}$ and dielectric constant ${k_1}$ is placed. In the rest of the space, there is another plate of thickness ${t_2}$ and dielectric constant ${k_2}$. The potential difference across the condenser will be