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. - Physics

Advertisements
Advertisements
Numerical

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?

Advertisements

Solution

Area of each plate of the parallel plate capacitor, A = 6 × 10−3 m2

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

Supply voltage, V = 100 V

Capacitance C of a parallel plate capacitor is given by,

`"C" = (in_0"A")/"d"`

Where,

`in_0` = Permittivity of free space

= 8.854 × 10−12 N−1 m−2 C−2

∴ `"C" = (8.854 xx 10^-12 xx 6 xx 10^-3)/(3 xx 10^-3)`

= `17.71 xx 10^-12  "F"`

= 17.71 pF

Potential V is related with the charge q and capacitance C as

`"V" = "q"/"C"`

∴ q = VC

= `100 xx 17.71 xx 10^-12`

= 1.771 × 10−9 C

Therefore, the capacitance of the capacitor is 17.71 pF and the charge on each plate is 1.771 × 10−9 C.

Concept: The Parallel Plate Capacitor
  Is there an error in this question or solution?
Chapter 2: Electrostatic Potential and Capacitance - Exercise [Page 86]

APPEARS IN

NCERT Physics Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.8 | Page 86
NCERT Physics Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 8 | Page 87

RELATED QUESTIONS

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


A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?


What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]


 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 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.


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 ______.


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.

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.


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



      Forgot password?
Use app×