The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2\,\mu \,F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu \,F$

A capacitor has capacitance $5 \mu F$ when it's parallel plates are separated by air medium of thickness $d$. A slab of material of dielectric constant $1.5$ having area equal to that of plates but thickness $\frac{ d }{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be $..........\mu F$

A parallel plate capacitor is charged to a potential difference of $100\,V$ and disconnected from the source of $emf$ . A slab of dielectric is then inserted between the plates. Which of the following three quantities change?

$(i)$  The potential difference

$(ii)$ The capacitance

$(iii)$ The charge on the plates

A parallel plate capacitor is formed by two plates each of area $30 \pi\, cm ^{2}$ separated by $1\, mm$. A material of dielectric strength $3.6 \times 10^{7} \,Vm ^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6}\, C$, the value of dielectric constant of the material is

$\left\{ Use : \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}\right\}$

The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively ${C_o}$ and ${W_o}$. If the air is replaced by glass (dielectric constant $= 5$ ) between the plates, the capacity of the plates and the energy stored in it will respectively be

Two parallel plates have equal and opposite charge. 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 dielectric, the electric field becomes $1 \times {10^5}\,V/m$. The dielectric constant of the dielectric material