In the adjoining figure, capacitor $(1)$ and $(2)$ have a capacitance $‘C’$ each. When the dielectric of dielectric consatnt $K$ is inserted between the plates of one of the capacitor, the total charge flowing through battery is
$\frac{{KCE}}{{K + 1}}$ from $B $ to $C$
$\frac{{KCE}}{{K + 1}}$ from $C$ to $B$
$\frac{{(K - 1)CE}}{{2(K + 1)}}$ from $B$ to $C$
$\frac{{(K - 1)CE}}{{2(K + 1)}}$ from $C$ to $B$
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........$volts$
In a parallel plate capacitor with air between the plates, each plate has an area of $6 \times 10^{-3}\; m ^{2}$ and the distance between the plates is $3 \;mm$ the capacitance of the capacitor is $17.71 \;pF$. If this capacitor is connected to a $100\; V$ supply, $3\; mm$ thick mica sheet (of dielectric constant $=6$ ) were inserted between the plates,
$(a)$ while the voltage supply remained connected.
$(b)$ after the supply was disconnected.
In the figure a capacitor is filled with dielectrics. The resultant capacitance is
A capacitor of capacity $'C'$ is connected to a cell of $'V'\, volt$. Now a dielectric slab of dielectric constant ${ \in _r}$ is inserted in it keeping cell connected then
A parallel plate capacitor has plate area $40\,cm ^2$ and plates separation $2\,mm$. The space between the plates is filled with a dielectric medium of a thickness $1\,mm$ and dielectric constant $5$ . The capacitance of the system is :