A parallel plate capacitor of capacitance 10 F is charged to 1 µC. The charging battery is removed and then the separation between the plates is doubled. Work done during the process is

• 5 mJ

• 0.05 mJ

• 1 mJ

• 10 mJ

62.

The potential at a point P which is forming a corner of a square of side 93 mm with charges, Q₁ = 33 nC, Q₂ = − 51 nC, Q₃ = 47 nC located at the other three corners is already

• 16 kV

• 4 kV

• 400 V

• 160 V

The capacitors A and B have identical geometry. A material with a dielectric constant 3 is present between the plates of B. The potential difference across A and B are respectively

• 2.5 V, 7.5 V

• 2 V, 8 V

• 8 V, 2 V

• 7.5 V, 2.5 V

Four 10 F capacitors are connected to a 500 V supply as shown in the figure. The equivalent capacitance of the network is

• 40 μF

• 20 μF

• 13.3 μF

• 10 μF

Small rain drops of the same size are charged to potential V volts each. If n such drops coalesce to form a single drop, then the potential of the bigger drop is

• n^1/3 V

• n^2/3 V

• nV

• n^3/2 V

A parallel plate capacitor has plate of area A and separation d. It is charged to a potential difference V₀. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

• $\frac{{\mathrm{\epsilon }}_0{{\mathrm{AV}}_0}^2}{3\mathrm{d}}$

• $\frac{{\mathrm{\epsilon }}_0{{\mathrm{AV}}_0}^2}{2\mathrm{d}}$

• $\frac{{\mathrm{\epsilon }}_0{{\mathrm{AV}}_0}^2}{4\mathrm{d}}$

• $\frac{{\mathrm{\epsilon }}_0{{\mathrm{AV}}_0}^2}{\mathrm{d}}$

A small sphere is charged to a potential of 50 V and a big hollow sphere is charged to potential of 100 V. Charge will flow from the smaller sphere to the bigger one when

• the smaller one is placed inside the bigger one and connected by a wire

• the bigger one is placed inside the smaller one and connected by means of a wire

• bigger one placed by the side of the smaller one and connected by a wire

• smaller one placed by the side of the bigger one

In the circuit, the potential difference across PQ will be nearest to

• 9.6 V

• 6.6 V

• 4.8 V

• 3.2 V

In the circuit as shown in the figure, the effective capacitance between A and B is

• 2μ

• 2μ

• 8μ

• 4μ

Capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value, if a dielectric slab of thickness $\mathrm{t} = \frac{\mathrm{d}}{2}$ is inserted between the plates [d is the separation between the plates]. The dielectric constant of the slab is

• 4

• 8

• 2

• 6