A parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates then :
some charge from the capacitor will flow back into the source.
some extra charge from the source will flow back into the capacitor.
the electric field intensity between the two plate does not change.
$B$ and $C$ both
The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate $ = A$)
Three identical capacitors $\mathrm{C}_1, \mathrm{C}_2$ and $\mathrm{C}_3$ have a capacitance of $1.0 \mu \mathrm{F}$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $\mathrm{C}_1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{\mathrm{r}}$. The cell electromotive force (emf) $V_0=8 \mathrm{~V}$. First the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged, $S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $\mathrm{C}_3$ is found to be $5 \mu \mathrm{C}$. The value of $\varepsilon_{\mathrm{r}}=$. . . .
(image)
On which the extant of polarisation depend ?
Figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant $3$. What will be the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric
The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of $\in _r = 6$ is inserted in it then energy and capacity becomes (Assume charge on plates remains constant)