Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes
Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes
- A
$\frac{{{\varepsilon _0}A}}{{d + t\left( {1 - \frac{1}{k}} \right)}}$
- B
$\frac{{{\varepsilon _0}A}}{{d + t\left( {1 + \frac{1}{k}} \right)}}$
- C
$\frac{{{\varepsilon _0}A}}{{d - t\left( {1 - \frac{1}{k}} \right)}}$
- D
$\frac{{{\varepsilon _0}A}}{{d - t\left( {1 + \frac{1}{k}} \right)}}$
Similar Questions
Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in figure. Capacitor $2$ contains a dielectric slab of dielectric constant k as shown. $Q_1$ and $Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q’_1$ and $Q’_2$. Then
Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in figure. Capacitor $2$ contains a dielectric slab of dielectric constant k as shown. $Q_1$ and $Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q’_1$ and $Q’_2$. Then
A parallel plate capacitor of plate area $A$ and plate seperation $d$ is charged to potential difference $V$ and then the battery is disconnected. Aslab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q, E$ and $W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then
A parallel plate capacitor of plate area $A$ and plate seperation $d$ is charged to potential difference $V$ and then the battery is disconnected. Aslab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q, E$ and $W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then
A parallel plate capacitor is made of two plates of length $l$, width $w$ and separated by distance $d$. A dielectric slab ( dielectric constant $K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F = -\frac{{\partial U}}{{\partial x}}$ where $U$ is the energy of the capacitor when dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is $Q$ then the force on the dielectric when it is near the edge is
A parallel plate capacitor is made of two plates of length $l$, width $w$ and separated by distance $d$. A dielectric slab ( dielectric constant $K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F = -\frac{{\partial U}}{{\partial x}}$ where $U$ is the energy of the capacitor when dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is $Q$ then the force on the dielectric when it is near the edge is
- [JEE MAIN 2014]
In the figure a capacitor is filled with dielectrics. The resultant capacitance is
In the figure a capacitor is filled with dielectrics. The resultant capacitance is
A parallel plate capacitor has plate of length $'l',$ width $'w'$ and separation of plates is $'d'.$ It is connected to a battery of emf $V$. A dielectric slab of the same thickness '$d$' and of dielectric constant $k =4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored$?$
A parallel plate capacitor has plate of length $'l',$ width $'w'$ and separation of plates is $'d'.$ It is connected to a battery of emf $V$. A dielectric slab of the same thickness '$d$' and of dielectric constant $k =4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored$?$
- [JEE MAIN 2020]