An air filled parallel plate capacitor has capacity $C$. If distance between plates is doubled and it is immersed in a liquid then capacity becomes twice. Dielectric constant of the liquid is
$1$
$2$
$3$
$4$
A parallel plate capacitor $\mathrm{C}$ with plates of unit area and separation $\mathrm{d}$ is filled with a liquid of dielectric constant $\mathrm{K}=2$. The level of liquid is $\frac{\mathrm{d}}{3}$ initially. Suppose the liquid level decreases at a constant speed $V,$ the time constant as a function of time $t$ is Figure: $Image$
The capacity of a parallel plate condenser is $5\,\mu F$. When a glass plate is placed between the plates of the conductor, its potential becomes $1/8^{th}$ of the original value. The value of dielectric constant will be
A capacitor of capacitance $9 n F$ having dielectric slab of $\varepsilon_{ r }=2.4$ dielectric strength $20\, MV / m$ and $P.D. =20 \,V$ then area of plates is ....... $\times 10^{-4}\, m ^{2}$
A slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$, where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?
A parallel plate capacitor Air filled with a dielectric whose dielectric constant varies with applied voltage as $K = V$. An identical capacitor $B$ of capacitance $C_0$ with air as dielectric is connected to voltage source $V_0 = 30\,V$ and then connected to the first capacitor after disconnecting the voltage source. The charge and voltage on capacitor.