The plates of parallel plate capacitor are charged upto $100\;V$. A $2\,mm$ thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by $1.6\;mm$. The dielectric constant of the plate is
- A$5$
- B$1.25$
- C$4$
- D$2.5$
Similar Questions
Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100\,V$ $D.C.$ source. Capacitor $C _1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $C _2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $.........V.$ (Assuming Dielectric constant $=10$ )
Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100\,V$ $D.C.$ source. Capacitor $C _1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $C _2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $.........V.$ (Assuming Dielectric constant $=10$ )
- [JEE MAIN 2023]
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