मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

What is the Area of the Plates of a 2 F Parallel Plate Capacitor, Given that the Separation Between the Plates is 0.5 Cm? - Physics

Advertisements
Advertisements

प्रश्न

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

संख्यात्मक
Advertisements

उत्तर

Capacitance of a parallel capacitor, V = 2 F

Distance between the two plates, d = 0.5 cm = 0.5 × 10−2 m

Capacitance of a parallel plate capacitor is given by the relation,

C = `(in_0"A")/"d"`

A = `("Cd")/in_0`

Where,

`in_0` = Permittivity of free space = 8.85 × 10−12 C2 N−1 m−2

∴ A = `(2 xx 0.5 xx 10^-2)/(8.85 xx 10^-12)` = 1130 km2

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.

shaalaa.com
The Parallel Plate Capacitor
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

संबंधित प्रश्‍न

Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery


In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?


The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

(a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.


Show that the force on each plate of a parallel plate capacitor has a magnitude equal to `(1/2)` QE, where Q is the charge on the capacitor, and E is the magnitude of the electric field between the plates. Explain the origin of the factor `1/2`.


 A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness \[\frac{3d}{4}\]. Find the ratio of the capacitance with dielectric inside it to its capacitance without the dielectric. 


A ray of light falls on a transparent sphere with centre C as shown in the figure. The ray emerges from the sphere parallel to the line AB. Find the angle of refraction at A if the refractive index of the material of the sphere is \[\sqrt{3}\].


A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor. 


A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness 2d/3, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.


A parallel-plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change:

(i) electric field between the plates

(ii) capacitance, and

(iii) energy stored in the capacitor


Define the capacitance of a capacitor and its SI unit.


A parallel-plate capacitor with plate area 20 cm2 and plate separation 1.0 mm is connected to a battery. The resistance of the circuit is 10 kΩ. Find the time constant of the circuit.


Answer the following question.
Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.


Solve the following question.
A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.   


For a one dimensional electric field, the correct relation of E and potential V is _________.


Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?


A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations:

  1. Key K is kept closed and plates of capacitors are moved apart using insulating handle.
  2. Key K is opened and plates of capacitors are moved apart using insulating handle.

Choose the correct option(s).

  1. In A: Q remains same but C changes.
  2. In B: V remains same but C changes.
  3. In A: V remains same and hence Q changes.
  4. In B: Q remains same and hence V changes.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×