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Define the Capacitance of a Capacitor. Obtain the Expression for the Capacitance of a Parallel Plate Capacitor in Vacuum in Terms of Plate Area a and Separation D Between the Plates. - Physics

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प्रश्न

Define the capacitance of a capacitor. Obtain the expression for the capacitance of a parallel plate capacitor in vacuum in terms of plate area A and separation d between the plates.

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उत्तर

The capacitance of a capacitor is the amount of charge which creates unit potential difference between collecting plate and condensing plate after giving charge on the collecting plate.

Parallel Plate Capacitor

  • A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.

  • Let be the area of each plate and the separation between them. The two plates have charges Q and −Q.

  • Surface charge density of plate 1, σ = Q/A, and that of plate 2 is σ.

  • Electric field in different regions:

Outer region I,

`E = σ/(2ε_0) - σ/(2ε_0 ) = 0`

In the inner region between plates 1 and 2, the electric fields due to the two charged plates add up. So,

\[E = \frac{\sigma}{2 \epsilon_0} + \frac{\sigma}{2 \epsilon_0} = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}\]

E=σ2ε0+σ2ε0=σε0=Qε0

  • The direction of electric field is from the positive to the negative plate.

  • For uniform electric field, potential difference is simply the electric field multiplied by the distance between the plates, i.e.

`V=Ed = 1/c (Qd)/A`

Capacitance of the parallel plate capacitor in vacuum is

`C = Q/V =(ε_0A)/d `

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The Parallel Plate Capacitor
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2016-2017 (March) Foreign Set 3

संबंधित प्रश्‍न

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


Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere's circuital law to include the term due to displacement current.


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`.


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?

Define the capacitance of a capacitor and its SI unit.


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?


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


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:


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