मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

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. - Physics

Advertisements
Advertisements

प्रश्न

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.

बेरीज

उत्तर

Given:-

Area of the plates, A = 20 cm2 = 20 × 10−4 m2

Separation between the plates, d = 1 mm = 1 × 10−3 m,

Resistance of the circuit, R = 10 kΩ

The capacitance of a parallel-plate capacitor,

\[C = \frac{\in_0 A}{d}\]

\[   = \frac{8 . 85 \times {10}^{- 12} \times 20 \times {10}^{- 4}}{1 \times {10}^{- 3}}\]

\[   = \frac{8 . 85 \times {10}^{- 12} \times 2 \times {10}^{- 4}}{{10}^{- 3}}\]

\[   = 17 . 7 \times  {10}^{- 12}   F\]

Time constant = CR

= 17.7 × 10−12 × 10 × 103

= 17.7 × 10−8 s

= 0.177 × 10−6 s

= 0.18 μs

shaalaa.com
The Parallel Plate Capacitor
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 10: Electric Current in Conductors - Exercises [पृष्ठ २०२]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
पाठ 10 Electric Current in Conductors
Exercises | Q 62 | पृष्ठ २०२

संबंधित प्रश्‍न

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


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.


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?


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.


 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 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 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 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.


A parallel-plate capacitor has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy of the capacitor 8.9 μs after the connections are made.


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?


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.

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  ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×