$A$ parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is

  • A
    Charge $Q$
  • B
    Potential $V$
  • C
    Capacity $C$
  • D
    Energy $U$

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When a dielectric material is introduced between the plates of a charged condenser,what happens to the electric field between the plates?

$A$ capacitor stores $60 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \mu C$ flows through the battery. The dielectric constant of the material inserted is:

$A$ parallel plate capacitor with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the figure. The dielectric constants are $K_1$ and $K_2$ respectively. The capacitance will be:

$A$ dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

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..........$C^{2} N^{-1} m^{-2}$.
$(\varepsilon_{0} = 8.85 \times 10^{-12} C^{2} N^{-1} m^{-2})$

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