If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become
If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become
- A
Half
- B
Two times
- C
Four times
- D
Remains the same
Similar Questions
. Three identical capacitors $C _1, C _2$ and $C _3$ have a capacitance of $1.0 \mu F$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C _1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{ r }$. The cell electromotive force (emf) $V_0=8 V$. 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 $C _3$ is found to be $5 \mu C$. The value of $\varepsilon_{ r }=$. . . . .
. Three identical capacitors $C _1, C _2$ and $C _3$ have a capacitance of $1.0 \mu F$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C _1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{ r }$. The cell electromotive force (emf) $V_0=8 V$. 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 $C _3$ is found to be $5 \mu C$. The value of $\varepsilon_{ r }=$. . . . .
- [IIT 2018]
After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :-
After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :-
Two capacitors, each having capacitance $40\,\mu F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24\,\mu F$. The value of $K$ will be.
Two capacitors, each having capacitance $40\,\mu F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24\,\mu F$. The value of $K$ will be.
- [JEE MAIN 2022]
A parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$ what will be capacitance if distance between plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?.....$\mu F$
A parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$ what will be capacitance if distance between plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?.....$\mu F$
A parallel plate capacitor has plate area $40\,cm ^2$ and plates separation $2\,mm$. The space between the plates is filled with a dielectric medium of a thickness $1\,mm$ and dielectric constant $5$ . The capacitance of the system is :
A parallel plate capacitor has plate area $40\,cm ^2$ and plates separation $2\,mm$. The space between the plates is filled with a dielectric medium of a thickness $1\,mm$ and dielectric constant $5$ . The capacitance of the system is :
- [JEE MAIN 2023]