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]
- A
$40$
- B
$50$
- C
$55$
- D
$65$
Similar Questions
Two dielectric slabs of constant ${K_1}$ and ${K_2}$ have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor
Two dielectric slabs of constant ${K_1}$ and ${K_2}$ have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor
A parallel plate condenser has a capacitance $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $'k'$ of the oil is
A parallel plate condenser has a capacitance $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $'k'$ of the oil is
A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^2$ and they are $0.6 \mathrm{~cm}$ apart. When a slab of dielectric having area $12 \mathrm{~cm}^2$ and $0.6 \mathrm{~cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \mathrm{~cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\left.\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$
A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^2$ and they are $0.6 \mathrm{~cm}$ apart. When a slab of dielectric having area $12 \mathrm{~cm}^2$ and $0.6 \mathrm{~cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \mathrm{~cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $\left.\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$
- [JEE MAIN 2024]
A parallel plate capacitor Air filled with a dielectric whose dielectric constant varies with applied voltage as $K = V$. An identical capacitor $B$ of capacitance $C_0$ with air as dielectric is connected to voltage source $V_0 = 30\,V$ and then connected to the first capacitor after disconnecting the voltage source. The charge and voltage on capacitor.
A parallel plate capacitor Air filled with a dielectric whose dielectric constant varies with applied voltage as $K = V$. An identical capacitor $B$ of capacitance $C_0$ with air as dielectric is connected to voltage source $V_0 = 30\,V$ and then connected to the first capacitor after disconnecting the voltage source. The charge and voltage on capacitor.
A parallel plate capacitor with air between the plates has capacitance of $9\ pF$. The separation between its plates is '$d$'. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_1 = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $k_2 = 6$ and thickness $\frac{2d}{3}$ . Capacitance of the capacitor is now.......$pF$
A parallel plate capacitor with air between the plates has capacitance of $9\ pF$. The separation between its plates is '$d$'. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_1 = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $k_2 = 6$ and thickness $\frac{2d}{3}$ . Capacitance of the capacitor is now.......$pF$
- [AIEEE 2008]