Capacitor

The distance between two plates of a capacitor is $$\mathrm{d}$$ and its capacitance is $$\mathrm{C}_{1}$$, when air is the medium between the plates. If a metal sheet of thickness $$\frac{2 d}{3}$$ and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes $$\mathrm{C}_{2}$$. The ratio $$\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$$ 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 $$\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}$$)

A fully charged capacitor has a capacitance $$'C'$$. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity $$'s'$$ and mass $$'m'.$$ If the temperature of the block is raised by $$'\Delta T',$$ the potential difference $$'v'$$ across the capacitance is

A parallel plate capacitor is made by stacking $$n$$ equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is $$'C'$$ then the resultant capacitance is

Two capacitors $${C_1}$$ and $${C_2}$$ are charged to $$120$$ $$V$$ and $$200$$ $$V$$ respectively. It is found that connecting them together the potential on each one can be made zero. Then

Three capacitors each of 4 $$\mu$$F are to be connected in such a way that the effective capacitance is 6 $$\mu$$F. This can be done by connecting them :

Effective capacitance of parallel combination of two capacitors C₁ and C₂ is 10 μF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C₂ is 4 times that of C₁. If these capacitors are connected in series, their effective capacitance will be :

Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :

Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be :

Consider the combination of 2 capacitors C₁ and C₂ with C₂ > C₁, when connected in parallel, the equivalent capacitance is $${{15} \over 4}$$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $${{{C_2}} \over {{C_1}}}$$.