A parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. A dielectric slab of dielectric constant $K,$ which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect $?$
The energy stored in the capacitor decreases $K$ times.
The potential difference between the plates decreases $K$ times.
The change in energy stored is $\frac{1}{2}C{V^2}\left( {\frac{1}{K} - 1} \right)$ છે.
The charge on the capacitor is not conserved.
A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?
A parallel plate capacitor with width $4\,cm$, length $8\,cm$ and separation between the plates of $4\,mm$ is connected to a battery of $20\,V$. A dielectric slab of dielectric constant $5$ having length $1\,cm$, width $4\,cm$ and thickness $4\,mm$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be......... $\in_{0}\,J$. (Where $\epsilon_{0}$ is the permittivity of free space)
How does the polarised dielectric modify the original external field inside it ?
A parallel plate capacitor having crosssectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ is inserted between the plates as shown in figure having dielectric constant $K (=4) .$ The ratio of new capacitance to its original capacitance will be,
A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with the capacitor are given by $Q_0, V_0, E_0$ and $U_0$ respectively. A dielectric slab is introduced between plates of capacitor but battery is still in connection. The corresponding quantities now given by $Q, V, E$ and $U$ related to previous ones are