A parallel plate capacitor with air between the plate has a capacitance of $15 pF$. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant $3.5.$ Then the capacitance becomes $\frac{ x }{4}\,pF$.The value of $x$ is $............$
$10.5$
$1.05$
$105$
$108$
Two condensers of capacities $2C$ and $C$ are joined in parallel and charged upto potential $V$. The battery is removed and the condenser of capacity $C$ is filled completely with a medium of dielectric constant $K$. The $p.d.$ across the capacitors will now be
A capacitor of capacitance $15 \,nF$ having dielectric slab of $\varepsilon_{r}=2.5$ dielectric strength $30 \,MV / m$ and potential difference $=30\; volt$ then the area of plate is ....... $ \times 10^{-4}\; m ^{2}$
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$?$
A capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process
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 :