हिंदी

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3m2 and the separation between the plates is 3 mm.

Advertisements
Advertisements

प्रश्न

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3m2 and the separation between the plates is 3 mm.

  1. Calculate the capacitance of the capacitor.
  2. If this capacitor is connected to 100 V supply, what would be the charge on each plate?
  3. How would charge on the plates be affected, if a 3 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected?
योग
Advertisements

उत्तर

Here, A = 6 × 10–3m2, d = 3 mm = 3 × 10–3m

  1. Capacitance, C = `(∈_0A)/d`
    = `((8.85 xx 10^-12 xx 6 xx 10^-3))/(3 xx 10^-3)`
    = 17.7 × 10–12F
  2. Charge, Q = CV
    = 17.7 × 10–12 × 100
    = 17.7 × 10–10C
  3. New charge Q' = KQ
    = 6 × 17.7 × 10–10
    = 1.062 × 10–8C
shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2013-2014 (March) Foreign Set 1

संबंधित प्रश्न

Draw a neat labelled diagram of a parallel plate capacitor completely filled with dielectric.


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.


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 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 filled with a dielectric material of resistivity ρ and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant of the discharge is independent of all geometrical parameters like the plate area or separation between the plates. Find this time constant.


A parallel plate air condenser has a capacity of 20µF. What will be a new capacity if:

1) the distance between the two plates is doubled?

2) a marble slab of dielectric constant 8 is introduced between the two plates?


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.


In a parallel plate capacitor, the capacity increases if ______.


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  ______.


A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×