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.
Consider the arrangement shown in figure. The total energy stored is $U_1$ when key is closed. Now the key $K$ is made off (opened) and two dielectric slabs of relative permittivity ${ \in _r}$ are introduced between the plates of the two capacitors. The slab tightly fit in between the plates. The total energy stored is now $U_2$. Then the ratio of $U_1/U_2$ is
Initially the circuit is in steady state. Now one of the capacitor is filled with dielectric of dielectric constant $2$ . Find the heat loss in the circuit due to insertion of dielectric
A parallel plate capacitor with air between the plates has a capacitance of $8 \;pF \left(1 \;pF =10^{-12} \;F \right) .$ What will be the capacitance (in $pF$) if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6 ?$
A parallel plate capacitor has plate area $40\,cm ^2$ and plates separation $2\,mm$. The space between the plates is filled with a dielectric medium of a thickness $1\,mm$ and dielectric constant $5$ . The capacitance of the system is :
Three capacitors $A,B$ and $C$ are connected with battery $emf\, \varepsilon $. All capacitors are identical initially. If dielectric slab is inserted between plates of capacitor $A$ slowy with help of external force then