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qUESTIONS

The capacitance of a capacitor becomes 7/6 times its original value, if a dielectric slab of thickness t = 2d/3 is introduced between the plates, where ‘d’ is distance of separation between the plates. The dielectric constant of the slab is

Two condensers of capacities ‘C’ and ‘2C’ are connected in parallel and then in series with third condenser of capacity ‘3C’. The combination is charged to ‘V’ volt. The charge on the condenser of capacity ‘C’ is

Two finite parallel charged plates having air between them have charge densities +σ and -σ respectively and they are separated by a small distance. If the permittivity of the air is &, then the electric field between the two plates will be

The capacitance of a parallel plate capacitor with air as medium is 3 µF. With the introduction of a dielectric medium between the plates, the capacitance becomes 15 µF. The permittivity of the medium in SI unit is (ε0= 8.85 × 10^-12 SI unit)

A capacitor of unknown capacity is connected across a battery of V volt. The charge stored in it is ‘Q’ C. When potential across a capacitor is reduced by V’ volt, the charge stored in it becomes Q’ C. The potential V is

The capacity of a pure capacitor is 1F. In d.c. circuits, its effective resistance will be

A parallel plate capacitor is charged by a battery and then disconnected. The dielectric slab of constant ‘K’ is then inserted between the plates. Then potential difference between the plates (K>1)

If the equivalent capacitance between A and B of the combination of capacitors shown in figure is 3C, the capacitor C’ is equal to

Two capacitors C and C/2 are connected to a battery of V volt as shown below. The work done in charging both the capacitor fully is

An electric dipole is as shown in figure. The electric potential at point P due to the dipole is: ( ε0= Permittivity of free space)

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