A parallel-plate capacitor of area $A,$ plate separation $d$ and capacitance $C$ is filled with four dielectric materials having dielectric constants $K_1,K_2,K_3$ and $K_4$ as shown in the figure. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $K$ is given by
$\frac{2}{K} = \frac{3}{{{K_1} + {K_2} + {K_3}}} + \frac{1}{{{K_4}}}\;\;\;\;$
$\;\frac{1}{K} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}} + \frac{1}{{{K_3}}} + \frac{3}{{2{K_4}}}$
$K=K_1+K_2+K_3+3K_4$
$K=$ $\frac{2}{3}\left[ {{K_1} + {K_2} + {K_3}} \right] + 2{K_4}$
Two dielectric slabs of constant ${K_1}$ and ${K_2}$ have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor
Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100\,V$ $D.C.$ source. Capacitor $C _1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $C _2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $.........V.$ (Assuming Dielectric constant $=10$ )
Capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$, dielectric constant of paper is $2.5$ and width of paper is $400\,mm$, then length of foil will be.....$m$
If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become
Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in figure. Capacitor $2$ contains a dielectric slab of dielectric constant k as shown. $Q_1$ and $Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q’_1$ and $Q’_2$. Then