If ${q}_{{f}}$ is the free charge on the capacitor plates and ${q}_{{b}}$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates, then bound charge $q_{b}$ can be expressed as
${q}_{{b}}={q}_{{f}}\left(1-\frac{1}{{k}}\right)$
${q}_{{b}}={q}_{{f}}\left(1-\frac{1}{\sqrt{{k}}}\right)$
${q}_{{b}}={q}_{{f}}\left(1+\frac{1}{\sqrt{{k}}}\right)$
${q}_{{b}}={q}_{{f}}\left(1+\frac{1}{{k}}\right)$
In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $=A$ )
Write the types of dielectric and explain them. Give some examples of dielectric substance.
A medium having dielectric constant $K>1$ fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is $d$. The capacitor is connected to a battery of voltage $V$. as shown in Figure ($a$). Now, both the plates are moved by a distance of $\frac{d}{2}$ from their original positions, as shown in Figure ($b$).
In the process of going from the configuration depicted in Figure ($a$) to that in Figure ($b$), which of the following statement($s$) is(are) correct?
When a dielectric material is introduced between the plates of a charges condenser, then electric field between the plates
The capacitance of a parallel plate capacitor with air as medium is $6\, \mu F$. With the introduction of a dielectric medium, the capacitance becomes $30\, \mu F$. The permittivity of the medium is..........$C ^{2} N ^{-1} m ^{-2}$
$\left(\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}\right)$