Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf, $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$, then the amount of charge which will flow through the battery is (neglect internal resistance of the battery)
$\frac{k+1}{2(k-1)} \cdot C E$
$\frac{k-1}{2(k+1)} \cdot C E$
$\frac{k-2}{k+2} \cdot C E$
$\frac{k+2}{k-2} \cdot C E$
The capacity of a parallel plate condenser is $5\,\mu F$. When a glass plate is placed between the plates of the conductor, its potential becomes $1/8^{th}$ of the original value. The value of dielectric constant will be
Eight small drops, each of radius $r$ and having same charge $q$ are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is
A combination of parallel plate capacitors is maintained at a certain potential difference When a $3\, mm$ thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by $2.4\, mm$. Find the dielectric constant of the slab.
Three different dielectrics are filled in a parallel plate capacitor as shown. What should be the dielectric constant of a material, which when fully filled between the plates produces same capacitance?
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