Capacitor

A 10 $$\mu$$F capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :

Match List I with List II. List - I List - II (a) Capacitance, C (i) $${M^1}{L^1}{T^{ - 3}}{A^{ - 1}}$$ (b) Permittivity of free space, $${\varepsilon _0}$$ (ii) $${M^{ - 1}}{L^{ - 3}}{T^4}{A^2}$$ (c) Permeability of free space, $${\mu _0}$$ (iii) $${M^{ - 1}}{L^{ - 2}}{T^4}{A^2}$$ (d) Electric field, E (iv) $${M^1}{L^1}{T^{ - 2}}{A^{ - 2}}$$ Choose the correct answer from the options given below

Capacitance of an isolated conducting sphere of radius R₁ becomes n times when it is enclosed by a concentric conducting sphere of radius R₂ connected to earth. The ratio of their radii $$\left( {{{{R_2}} \over {{R_1}}}} \right)$$ is :

Given below are two statements: One is labeled as Assertion A and the other is labeled as Reason R. Assertion A : Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one. Reason R : Capacitance of metallic spheres depend on the radii of spheres In light of the above statements, choose the correct answer from the options given below.

A capacitance of 2 $$\mu$$F is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 $$\mu$$F capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:

Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :

A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the 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 has dielectric constant $${k_1}$$ $$=3$$ and thickness $${d \over 3}$$ while the other one has dielectric constant $${k_2} = 6$$ and thickness $${{2d} \over 3}$$. Capacitance of the capacitor is now

A parallel plate capacitor is made of two circular plates separated by a distance $$5$$ $$mm$$ and with a dielectric of dielectric constant $$2.2$$ between them. When the electric field in the dielectric is $$3 \times {10^4}\,V/m$$ the charge density of the positive plate will be close to:

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be :

The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is :-

Voltage rating of a parallel plate capacitor is 500V. Its dielectric can withstand a maximum electric field of 10⁶ V/m. The plate area is 10^–4 m². What is the dielectric constant is the capacitance is 15 pF? (given $$\varepsilon$$₀ = 8.86 × 10^–12 C²/Nm²)

A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab 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 the energy stored in the capacitor be two times the initial energy stored?

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 $${3 \over 4}$$d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C₀) is given by the following relation :

A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as : $$\varepsilon (x) = {\varepsilon _0} + kx$$, for $$\left( {0 < x \le {d \over 2}} \right)$$ $$\varepsilon (x) = {\varepsilon _0} + k(d - x)$$, for $$\left( {{d \over 2} \le x \le d} \right)$$

If q_f is the free charge on the capacitor plates and q_b is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge q_b an be expressed as :

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :

Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness $${d \over 2}$$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?