Between the plates of a parallel plate condenser, a plate of thickness ${t_1}$ and dielectric constant ${k_1}$ is placed. In the rest of the space, there is another plate of thickness ${t_2}$ and dielectric constant ${k_2}$. The potential difference across the condenser will be
$\frac{Q}{{A{\varepsilon _0}}}\left( {\frac{{{t_1}}}{{{k_1}}} + \frac{{{t_2}}}{{{k_2}}}} \right)$
$\frac{{{\varepsilon _0}Q}}{A}\left( {\frac{{{t_1}}}{{{k_1}}} + \frac{{{t_2}}}{{{k_2}}}} \right)$
$\frac{Q}{{A{\varepsilon _0}}}\left( {\frac{{{k_1}}}{{{t_1}}} + \frac{{{k_2}}}{{{t_2}}}} \right)$
$\frac{{{\varepsilon _0}Q}}{A}({k_1}{t_1} + {k_2}{t_2})$
In a medium of dielectric constant $K$, the electric field is $\vec E$ . If ${ \varepsilon _0}$ is permittivity of the free space, the electric displacement vector is
A capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process
An air capacitor of capacity $C = 10\,\mu F$ is connected to a constant voltage battery of $12\,V$. Now the space between the plates is filled with a liquid of dielectric constant $5$. The charge that flows now from battery to the capacitor is......$\mu C$
Two parallel plate capacitors of capacity $C$ and $3\,C$ are connected in parallel combination and charged to a potential difference $18\,V$. The battery is then disconnected and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $9$. The final potential difference across the combination of capacitors will be $V$
A parallel plate capacitor having capacitance $12\, pF$ is charged by a battery to a potential difference of $10\, V$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates. The work done by the capacitor on the slab is.......$pJ$