The capacity of a parallel plate condenser is $10\,\mu F$ without dielectric. Dielectric of constant $2$ is used to fill half the distance between the plates, the new capacitance in $\mu F$ is

A parallel plate capacitor of capacitance $200 \,\mu {F}$ is connected to a battery of $200 \, {V} .$ A dielectric slab of dielectric constant $2$ is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ......$ J.$

A parallel plate capacitor has plates with area $A$ and separation $d$ . A battery charges the plates to a potential difference $V_0$ . The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced, is

A parallel plate capacitor Air filled with a dielectric whose dielectric constant varies with applied voltage as $K = V$. An identical capacitor $B$ of capacitance $C_0$ with air as dielectric is connected to voltage source $V_0 = 30\,V$ and then connected to the first capacitor after disconnecting the voltage source. The charge and voltage on capacitor.

The radii of the inner and outer spheres of a condenser are $9\,cm$ and $10\,cm$ respectively. If the dielectric constant of the medium between the two spheres is $6$ and charge on the inner sphere is $18 \times {10^{ - 9}}\;coulomb$, then the potential of inner sphere will be, if the outer sphere is earthed........$volts$

Two thin dielectric slabs of dielectric constants $K_1$ and $K_2$ $(K_1 < K_2)$ are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field $E$ between the plates with distance $d$ as measured from plate $P$ is correctly shown by