A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?
$\frac{3 K \varepsilon_0 A}{4 d}$
$\frac{4 K \varepsilon_0 A}{3 d}$
$\frac{(K+1) \varepsilon_0 A}{2 d}$
$\frac{K(K+3) \varepsilon_0 A}{2(K+1) d}$
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
A parallel plate condenser with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed, then capacitance of the capacitor becomes
A parallel plate capacitor with plate separation $5$ $\mathrm{mm}$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness $2 \mathrm{~mm}$, while keeping the battery connections intact, the capacitor draws $25 \%$ more charge from the battery than before. The dielectric constant of the sheet is_____.
A parallel plate capacitor with plate area $'A'$ and distance of separation $'d'$ is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :
$\varepsilon(x)=\varepsilon_{0}+k x, \text { for }\left(0\,<\,x \leq \frac{d}{2}\right)$
$\varepsilon(x)=\varepsilon_{0}+k(d-x)$, for $\left(\frac{d}{2} \leq x \leq d\right)$
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 plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ......$ J.$