$A$ frictionless dielectric plate $S$ is placed on a frictionless table $T$. $A$ charged parallel plate capacitor $C$ (whose plates are also frictionless) is kept near it. The plate $S$ is partially between the plates of the capacitor. When the plate $S$ is released, what happens?

  • A
    It will remain stationary on the table.
  • B
    It is pulled by the capacitor and will pass through to the other end.
  • C
    It is pulled into the space between the plates and will remain there.
  • D
    All the above statements are false.

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$A$ copper slab of thickness $d/2$ is inserted between the plates of a parallel plate capacitor,where $d$ is the distance between its two plates. If the capacitance of the capacitor without the copper slab is $C$ and with the copper slab is $C'$,find the ratio $C'/C$.

Two parallel plates have equal and opposite charges. When the space between them is evacuated,the electric field between the plates is $2 \times 10^5 \ V/m$. When the space is filled with a dielectric,the electric field becomes $1 \times 10^5 \ V/m$. Find the dielectric constant of the dielectric material.

In the adjoining figure,capacitors $(1)$ and $(2)$ have a capacitance $C$ each. When a dielectric of dielectric constant $K$ is inserted between the plates of one of the capacitors,the total charge flowing through the battery is:

Two identical capacitors $A$ and $B$ are connected in series to a battery of $E$.$M$.$F$. $E$. Capacitor $B$ contains a slab of dielectric constant $K$. $Q_A$ and $Q_B$ are the charges stored in $A$ and $B$. When the dielectric slab is removed,the corresponding charges are $Q_A^{\prime}$ and $Q_B^{\prime}$. Then:

$A$ capacitor is connected to a $10\,V$ battery. The charge on the plates is $10\,\mu C$ when the medium between the plates is air. The charge on the plates becomes $100\,\mu C$ when the space between the plates is filled with oil. The dielectric constant of the oil is:

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