The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

Two square-shaped metal plates of side $1 \,m$, kept $0.01 \,m$ apart in air, form a parallel plate capacitor. It is connected to a battery of $500 \,V$. The plates of the capacitor are then immersed in an insulating oil by lowering the plates vertically with a speed of $0.001 \,m/s$. If the dielectric constant of the oil is $11$, then the current drawn from the battery during this process is:

$A$ parallel plate capacitor of capacitance $C_1$ with a dielectric slab in between its plates is connected to a battery. It has a potential difference $V_1$ across its plates. When the dielectric slab is removed,keeping the capacitor connected to the battery,the new capacitance and potential difference are $C_2$ and $V_2$ respectively. Then:

$A$ parallel plate capacitor with air between the plates has a capacitance of $1.0 \text{ pF}$. If the distance between the plates is doubled and the space between them is filled with a dielectric substance, the capacitance becomes $2.0 \text{ pF}$. Then the value of the dielectric constant of the dielectric substance is . . . . . . .

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.

$A$ parallel plate capacitor with air between the plates has a capacitance of $12 \mu F$. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant $4$,what will be the new capacitance of the capacitor (in $\mu F$)?