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
$\frac{2}{K} = \frac{3}{{{K_1} + {K_2} + {K_3}}} + \frac{1}{{{K_4}}}\;\;\;\;$
$\;\frac{1}{K} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}} + \frac{1}{{{K_3}}} + \frac{3}{{2{K_4}}}$
$K=K_1+K_2+K_3+3K_4$
$K=$ $\frac{2}{3}\left[ {{K_1} + {K_2} + {K_3}} \right] + 2{K_4}$
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.....$m$
A composite parallel plate capacitor is made up of two different dielectric materials with different thickness $\left(t_{1}\right.$ and $\left.t_{2}\right)$ as shown in figure. The two different dielectric material are separated by a conducting foil $F$. The voltage of the conducting foil is $.....V$
The capacitance of a parallel plate capacitor is $5\, \mu F$ . When a glass slab of thickness equal to the separation between the plates is introduced between the plates, the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is
A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3d$. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from $0$ to $3d$
A parallel plate air capacitor has a capacitance of $100\,\mu F$. The plates are at a distance $d$ apart. If a slab of thickness $t(t \le d)$and dielectric constant $5$ is introduced between the parallel plates, then the capacitance will be.......$\mu F$