The capacitance of an air-filled parallel plate capacitor is $9 \ pF$. If the space between the plates is filled with two dielectric slabs of thickness $d/3$ with dielectric constant $K_1 = 3$ and thickness $2d/3$ with dielectric constant $K_2 = 6$,the new capacitance will be ...... $pF$.

$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced is

$A$ parallel-plate capacitor of capacitance $40 \mu F$ is connected to a $100 V$ power supply. Now,the intermediate space between the plates is filled with a dielectric material of dielectric constant $K=2$. Due to the introduction of the dielectric material,the extra charge and the change in the electrostatic energy in the capacitor,respectively,are:

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is then inserted between the plates. What is the result?

When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates,then the energy stored:

$A$ voltmeter reads $4 \ V$ when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between the plates for the same configuration,the voltmeter reads $2 \ V$. What is the dielectric constant of the material?