A parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$ its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is
A parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$ its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is
- A
${K_1} + {K_2}$
- B
$\frac{{{K_1}{K_2}}}{{{K_1} - {K_2}}}$
- C
$\frac{{{K_1} + {K_2}}}{{{K_1}{K_2}}}$
- D
$\frac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}$
Similar Questions
The radii of the inner and outer spheres of a condenser are $9\,cm$ and $10\,cm$ respectively. If the dielectric constant of the medium between the two spheres is $6$ and charge on the inner sphere is $18 \times {10^{ - 9}}\;coulomb$, then the potential of inner sphere will be, if the outer sphere is earthed........$volts$
The radii of the inner and outer spheres of a condenser are $9\,cm$ and $10\,cm$ respectively. If the dielectric constant of the medium between the two spheres is $6$ and charge on the inner sphere is $18 \times {10^{ - 9}}\;coulomb$, then the potential of inner sphere will be, if the outer sphere is earthed........$volts$
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation
- [JEE MAIN 2021]
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A container has a base of $50 \mathrm{~cm} \times 5 \mathrm{~cm}$ and height $50 \mathrm{~cm}$, as shown in the figure. It has two parallel electrically conducting walls each of area $50 \mathrm{~cm} \times 50 \mathrm{~cm}$. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant $3$ at a uniform rate of $250 \mathrm{~cm}^3 \mathrm{~s}^{-1}$. What is the value of the capacitance of the container after $10$ seconds? [Given: Permittivity of free space $\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]
A container has a base of $50 \mathrm{~cm} \times 5 \mathrm{~cm}$ and height $50 \mathrm{~cm}$, as shown in the figure. It has two parallel electrically conducting walls each of area $50 \mathrm{~cm} \times 50 \mathrm{~cm}$. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant $3$ at a uniform rate of $250 \mathrm{~cm}^3 \mathrm{~s}^{-1}$. What is the value of the capacitance of the container after $10$ seconds? [Given: Permittivity of free space $\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]
- [IIT 2023]