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Explain Briefly the Process of Charging a Parallel Plate Capacitor When It is Connected Across a D.C. Battery

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प्रश्न

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

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उत्तर

Consider a parallel plate capacitor connected across a d.c. battery as shown in the figure. The electric current will flow through the circuit. As the charges reach the plate, the insulating gap does not allow the charges to move further; hence, positive charges get deposited on one side of the plate and negative charges get deposited on the other side of the plate. As the voltage begins to develop, the electric charge begins to resist the deposition of further charge. Thus, the current flowing through the circuit gradually becomes less and then zero till the voltage of the capacitor is exactly equal but opposite to the voltage of the battery. This is how the capacitor gets charged when it is connected across a d.c. battery.

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2014-2015 (March) Panchkula Set 3

संबंधित प्रश्न

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.


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.


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.   


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


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.

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