A parallel plate capacitor with air between the plate has a capacitance of $15 pF$. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant $3.5.$ Then the capacitance becomes $\frac{ x }{4}\,pF$.The value of $x$ is $............$
$10.5$
$1.05$
$105$
$108$
A capacitor has capacitance $5 \mu F$ when it's parallel plates are separated by air medium of thickness $d$. A slab of material of dielectric constant $1.5$ having area equal to that of plates but thickness $\frac{ d }{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be $..........\mu F$
A parallel plate capacitor has plates with area $A$ and separation $d$ . A battery charges the plates to a potential difference $V_0$ . The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced, is
How does the polarised dielectric modify the original external field inside it ?
A parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. A dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is........$\mu C$
A parallel plate capacitor is filled with $3$ dielectric materials of same thickness, as shown in the sketch. The dielectric constants are such that $k_3 > k_2 > k_1$. Let the magnitudes of the electric field in and potential drops across each dielectric be $E_3$, $E_2$,$ E_1$, $\Delta V_3$, $\Delta V_2$ and $\Delta V_1$, respectively. Which one of the following statement is true ?