A parallel plate capacitor is made of two square plates of side $a$, separated by a distance $d\,(d < < a)$. The lower triangular portion is filled with a dielectric of dielectric constant $K$, as shown in the figure. Capacitance of this capacitor is
$\frac{{K{\varepsilon _0}{a^2}}}{{d\left( {K - 1} \right)}}\,\ln \,K$
$\frac{{K{\varepsilon _0}{a^2}}}{{2d\left( {K + 1} \right)}}$
$\frac{{K{\varepsilon _0}{a^2}}}{d}\,\ln \,K$
$\frac{1}{2}\frac{{K{\varepsilon _0}{a^2}}}{d}$
The capacitance of an air filled parallel plate capacitor is $10\,p F$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $40 \times {10^{ - 12}}farads$. The dielectric constant of wax is
Assertion : A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant $K$ is introduced between the plates. The energy which is stored becomes $K$ times.
Reason : The surface density of charge onthe plate remains constant or unchanged.
A parallel plate capacitor with plate area $A$ and plate separation $d =2 \,m$ has a capacitance of $4 \,\mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K =3$ (as shown in figure) will be .........$ \mu \,F$
Consider the arrangement shown in figure. The total energy stored is $U_1$ when key is closed. Now the key $K$ is made off (opened) and two dielectric slabs of relative permittivity ${ \in _r}$ are introduced between the plates of the two capacitors. The slab tightly fit in between the plates. The total energy stored is now $U_2$. Then the ratio of $U_1/U_2$ is
Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is