$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is introduced between the plates which results in

  • A
    increase in the potential difference across the plates and reduction in stored energy but no change in the charge on the plates
  • B
    decrease in the potential difference across the plates and reduction in the stored energy but no change in the charge on the plates
  • C
    reduction of charge on the plates and increase of potential difference across the plates
  • D
    increase in stored energy but no change in potential difference across the plates

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Similar Questions

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

Write the formula for the capacitance of a capacitor having a dielectric constant $K = 2$.

Consider the arrangement shown in the figure. The total energy stored is $U_1$ when the key $K$ is closed. Now,the key $K$ is opened,and two dielectric slabs of relative permittivity $\epsilon_r$ are introduced between the plates of the two capacitors. The slabs fit tightly between the plates. The total energy stored is now $U_2$. Then,the ratio $U_1/U_2$ is:

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$A$ parallel plate air capacitor has capacity $C$ and distance of separation between plates is $d$. If a conducting sheet of thickness $\frac{2d}{3}$ is inserted between the plates,the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is (in $:1$)

As shown in the figure,a dielectric slab of thickness $d/2$ and dielectric constant $K$ is placed between the plates of a parallel plate capacitor of plate separation $d$. The capacitor is charged to a potential $V$ using a battery. If the dielectric slab is pulled out after disconnecting the battery from the capacitor,the final potential difference across the plates of the capacitor is

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