A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is $A\;metr{e^2}$ and the separation is $t$ $metre$. The dielectric constants are ${k_1}$ and ${k_2}$ respectively. Its capacitance in farad will be
$\frac{{{\varepsilon _0}A}}{t}({k_1} + {k_2})$
$\frac{{{\varepsilon _0}A}}{t}.\frac{{{k_1} + {k_2}}}{2}$
$\frac{{2{\varepsilon _0}A}}{t}({k_1} + {k_2})$
$\frac{{{\varepsilon _0}A}}{t}.\frac{{{k_1} - {k_2}}}{2}$
After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :-
Which one statement is correct ? A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are ${Q_o},\;{V_o},\;{E_o}$ and ${U_o}$ respectively. In order to fill the complete space between the plates a dielectric slab is inserted, the battery is still connected. Now the corresponding values $Q,\;V,\;E$ and $U$ are in relation with the initially stated as
In a capacitor of capacitance $20\,\mu \,F$, the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates, then the new capacitance is......$\mu \,F$
A parallel plate capacitor is to be designed, using a dielectric of dielectric constant $5$, so as to have a dielectric strength of $10^9\;Vm^{-1}$ . If the voltage rating of the capacitor is $12\;kV$, the minimum area of each plate required to have a capacitance of $80\;pF$ is
The plates of a parallel plate capacitor are charged up to $100\, volt$. A $2\, mm$ thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by $1.6\, mm$. The dielectric constant of the plate is