A capacitor is half filled with a dielectric $(K=2)$ as shown in figure A. If the same capacitor is to be filled with same dielectric as shown, what would be the thickness of dielectric so that capacitor still has same capacity?
$2 d / 3$
$3 d / 2$
$3 d / 4$
$4 d / 3$
What is dielectric ?
Four identical plates $1, 2, 3$ and $4$ are placed parallel to each other at equal distance as shown in the figure. Plates $1$ and $4$ are joined together and the space between $2$ and $3$ is filled with a dielectric of dielectric constant $k$ $=$ $2$. The capacitance of the system between $1$ and $3$ $\&$ $2$ and $4$ are $C_1$ and $C_2$ respectively. The ratio $\frac{{{C_1}}}{{{C_2}}}$ is
A parallel plate capacitor with plate area $A$ and plate separation $d$ is filled with a dielectric material of dielectric constant $K =4$. The thickness of the dielectric material is $x$, where $x < d$.
Let $C_1$ and $C_2$ be the capacitance of the system for $x =\frac{1}{3} d$ and $x =\frac{2 d }{3}$, respectively. If $C _1=2 \mu F$ the value of $C _2$ is $........... \mu F$
A parallel plate capacitor, partially filled with a dielectric slab of dielectric constant $K$ , is connected with a cell of emf $V\ volt$ , as shown in the figure. Separation between the plates is $D$ . Then
A parallel plate capacitor is to be designed with a voltage rating $1\; k\,V ,$ using a material of dielectric constant $3$ and dielectric strength about $10^{7}\; V\,m ^{-1} .$ (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say $10 \%$ of the dielectric strength. What minimum area (in $cm^2$) of the plates is required to have a capacitance of $50\; pF ?$