Advertisements
Advertisements
प्रश्न
The plates of a parallel-plate capacitor are made of circular discs of radii 5⋅0 cm each. If the separation between the plates is 1⋅0 mm, what is the capacitance?
Advertisements
उत्तर
The capacitance of a parallel-plate capacitor is given by `C = (∈_0A)/d`
Here ,
A = Area of the plate
d = Distance between the parallel plates
Given :
`A = pir^2 = pi xx (5 xx 10^-2)^2`
`d = 1.0 xx 10^-3 "m"`
`∈_0 = 8.85 xx 10^-12 F/m`
`therefore C = (∈_0A)/d`
= `(8.85 xx 10^-12 xx 3.14 xx 25 xx 10^-4)/10^-3`
= `6.95 xx 10^-5 uF`
APPEARS IN
संबंधित प्रश्न
A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change?
(i) charge stored by the capacitor.
(ii) Field strength between the plates.
(iii) Energy stored by the capacitor.
Justify your answer in each case.
Find the equivalent capacitance of the network shown in the figure, when each capacitor is of 1 μF. When the ends X and Y are connected to a 6 V battery, find out (i) the charge and (ii) the energy stored in the network.

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports. Show that the capacitance of a spherical capacitor is given by
C = `(4piin_0"r"_1"r"_2)/("r"_1 - "r"_2)`

where r1 and r2 are the radii of outer and inner spheres, respectively.
As `C = (1/V) Q` , can you say that the capacitance C is proportional to the charge Q?
A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is
A thin metal plate P is inserted between the plates of a parallel-plate capacitor of capacitance C in such a way that its edges touch the two plates . The capacitance now becomes _________ .

Find the equivalent capacitance of the system shown in figure between the points a and b.

Find the equivalent capacitances of the combinations shown in figure between the indicated points.




Find the capacitance of the combination shown in figure between A and B.

The two square faces of a rectangular dielectric slab (dielectric constant 4⋅0) of dimensions 20 cm × 20 cm × 1⋅0 mm are metal-coated. Find the capacitance between the coated surfaces.
A parallel-plate capacitor of capacitance 5 µF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?
A parallel-plate capacitor has plate area 100 cm2 and plate separation 1⋅0 cm. A glass plate (dielectric constant 6⋅0) of thickness 6⋅0 mm and an ebonite plate (dielectric constant 4⋅0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.
Consider an assembly of three conducting concentric spherical shell of radii a, b and c as shown in figure Find the capacitance of the assembly between the points Aand B.

Suppose the space between the two inner shells is filled with a dielectric of dielectric constant K. Find the capacitance of the system between A and B.

The figure show a network of five capacitors connected to a 10V battery. Calculate the charge acquired by the 5μF capacitor.

- Charge on each capacitor remains same and equals to the main charge supplied by the battery.
- Potential difference and energy distribute in the reverse ratio of capacitance.
- Effective capacitance is even les than the least of teh individual capacitances.
A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as ε = αU where α = 2V–1. A similar capacitor with no dielectric is charged to U0 = 78V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.
The displacement current of 4.425 µA is developed in the space between the plates of the parallel plate capacitor when voltage is changing at a rate of 106 Vs-1. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor is x × 10-3 m. The value of x is ______.
(Permittivity of free space, ε0 = 8.85 × 10-12C2N-1m-2).
A capacitor with capacitance 5µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done?
