A capacitor is connected to a battery of voltage $V$. Now a di electric slab of dielectric constant $k$ is completely inserted between the plates, then the final charge on the capacitor will be
(If initial charge is $q_{0}$ )
$\frac{\varepsilon_{0} A}{d} V$
$\frac{k \varepsilon_{0} A}{d} V$
$\frac{\varepsilon_{0} A}{k d} V$
zero
A capacitor when filled with a dielectric $K = 3$ has charge ${Q_0}$, voltage ${V_0}$ and field ${E_0}$. If the dielectric is replaced with another one having $K = 9$ the new values of charge, voltage and field will be respectively
Assertion : In the absence of an external electric field, the dipole moment per unit volume of a polar dielectric is zero.
Reason : The dipoles of a polar dielectric are randomly oriented.
Explain the effect of dielectric on capacitance of parallel plate capacitor and obtain the formula of dielectric constant.
A capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process
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 $\left(\frac{d}{2} \leq x \leq d\right)$