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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.

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


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?


A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?


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?

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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 of plate area 40 cm2 and separation between the plates 0.10 mm, is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.


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 ______.


Two charges – q each are separated by distance 2d. A third charge + q is kept at mid point O. Find potential energy of + q as a function of small distance x from O due to – q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium.


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