Important Questions for CBSE Class 12 Physics Capacitance

# Important Questions for CBSE Class 12 Physics Capacitance

Class 12 students can access the important questions for Chapter 2 of Physics in PDF format here. Scoring well in class 12 board exams is crucial for every student. Therefore, these important questions for Chapter 2 – Electrostatic Potential and Capacitance are provided to help students revise main topics and confidently excel in their exams. The PDF of important questions for Physics Class 12 Chapter 2 can be downloaded from Infinity Learn, allowing students to refer to them at their convenience.

Fill Out the Form for Expert Academic Guidance!

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

 CBSE Class 12 Physics Important Questions S.No Chapter No Chapter Name 1 Chapter 1 Electric Charges and Fields 2 Chapter 2 Electrostatic Potential and Capacitance 3 Chapter 3 Current Electricity 4 Chapter 4 Moving Charges and Magnetism 5 Chapter 5 Magnetism And Matter 6 Chapter 6 Electromagnetic Induction 7 Chapter 7 Alternating Current 8 Chapter 8 Electromagnetic Waves 9 Chapter 9 Ray Optics and Optical Instruments 10 Chapter 10 Wave Optics 11 Chapter 11 Dual Nature of Radiation and Matter 12 Chapter 12 Atoms 13 Chapter 13 Nuclei 14 Chapter 14 Semiconductor Electronic: Material, Devices And Simple Circuits 15 Chapter 15 Communication System

## Electrostatic Potential and Capacitance Important Questions for CBSE Class 12 Physics Capacitance

1.Conductors and Insulators Conductor contains a large number of free charge carriers to conduct electricity while insulator does not contain any free charge carriers to conduct electricity.

Examples of conductors are metals and graphite.

Examples of insulators are plastic rod and nylon.

NOTE Inside a conductor, the electrostatic field is zero.

2.Free Charges and Bound Charges inside a Conductor

(i) In a metal, the outer (valence) electrons are free to move. These electrons are free for moving within the metal but not free to leave the metal. These free electrons are free charges inside a conductor and are cause of conduct of electricity by conductors.

(ii) The bound charges are those positive ions which are made up of the nuclei and the bound electrons remain in their fixed positions.

Some important results regarding electrostatics of conductors are given as below:

(i) Inside a conductor, the electric field is zero.

(ii) The interior of a conductor can have no excess charge in static situation.

(iii) Electric field just outside a charged conductor is perpendicular to the surface of the conductorat every point.

3. Dielectric and Electric Polarisation

(i) In an External Electric Field When a conductor is placed in an external electric field, the free charge carriers adjust itself in such a way that the electric field due to induced charges and external field cancel each other and the net field inside the conductor is zero.
In case of dielectric however, the opposing field so induced does not exactly cancel the external field.

(ii) A net dipole moment is developed by an external field in either case, whether polar or non-polar dielectric. The dipole moment per unit volume is called polarisation and it is denoted by

4. Capacitor and Capacitance A capacitor is a device which is used to store electrostatic potential energy or charge. It comprises of two conductors separated by an insulating medium and capacitance of a conductor is the ability to store the electric charge.

5. Capacitance of a Conductor If charge q is given to an insulated conductor, it leads to increase its electric potential by V such that

where, C is known as capacitance of a conductor. The capacitance depends on the shape, size and geometry of conductor, nature of surrounding medium and presence of other conductor in the neighbourhood of it.Its SI unit is farad (F).

Farad is a very large unit of capacitance. So, pF is usually taken.

6. Principle of Capacitor The capacitance of an insulated conductor is increased significantly when an earthed uncharged conductor is placed near it. Such an arrangement is called capacitor.

9. Parallel Plate Capacitor The most common among all capacitors is parallel plate capacitor. It comprises of two metal plates of area A and separated by distance d filled with air or some other dielectric medium. The capacitance of air filled parallel plate capacitor is given by

When a dielectric of dielectric constant K is filled fully between the plates, then

10. Dielectric When a dielectric slab is introduced between the plates of charged capacitor or in the region of electric field, an electric field Ep induces inside the dielectric due to induced charge on dielectric in a direction opposite to the direction of applied external electric field. Hence, net electric field inside the dielectric gets reduced to (E0 – Ep), where, E0 is external electric field. The ratio of applied external electric field and reduced electric field is known as dielectric constant K of dielectric medium, i.e.

Also Check: CBSE Class 12 Physics Answer Key 2024 | CBSE CBSE Class 12 Physics Paper Analysis 2024

## Previous Year Examination Questions

### 1 Mark Questions

1. The given graph shows the variation of charge q versus potential difference V for two capacitors Cl and C2. Both the capacitors” have same plate separation but plate area of C2 is greater than that Cx .Which line (A or B) corresponds to and why?[All India 2014 C]

Ans. Line B corresponds to C-, because slope (q/v) of B is less than slope of A.

2. A capacitor has been charged by a DC source. What are the magnitude of conduction and displacement current when it is fully charged?[Delhi 2013]

Ans.

3. Distinguish between a dielectric and a conductor. [Delhi 2012]

Ans. Dielectrics are non-conductors and do not have free electrons at all. While conductor has free electrons in its any volume which makes it able to pass the electricity through it.

4. Define the dielectric constant of a medium. What is its unit? [Delhi 2011c]

Ans. Dielectric When a dielectric slab is introduced between the plates of charged capacitor or in the region of electric field, an electric field EP induces inside the dielectric due to induced charge on dielectric in a direction opposite to the direction of applied external electric field. Hence, net electric field inside the dielectric gets reduced to £0 – £P, where £0 is external electric field. The ratio of applied external electric field and reduced electric field is known as dielectric constant K of dielectric medium, i.e.

5. A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?[Foreign 2009]

Ans. If a metal plate is introduced between the plates of a charged parallel plate capacitor. The capacitance of parallel plate capacitor will becomes infinite.

6.In the figure given below X, Y represent parallel plate capacitors having the same area of plates and the same distance of separation between them. What is the relation between the energies stored in the capacitors?

Ans.

7.The following graph shows the variation of charge Q with voltage V for two capacitors K and L In which capacitor is more electrostatic energy stored?

Ans.

8. Define dielectric strength of a dielectric. [Delhi 2008 C]

Ans. The dielectric strength of a dielectric is the maximum value of applied electric field required to just breakdown of the dielectric material.

### 2 Marks Questions

9. A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.[All India 2014]

Ans.

10. Two parallel plate capacitors of capacitances Cx and C2 such that Cx =2C2 are connected across a battery of V volt as shown in the figure Initially, the key (k) is kept closed to fully charge the capacitors.The key is now thrown open and a dielectric slab of dielectric constant K is inserted in the two capacitors to completely fill the gap between the plates. Find the ratio of (i) the net capacitance and (ii) the energies stored in the combination before and after the introduction of the dielectric slab.[Delhi 2014 C]

Ans.

11.Two parallel plate capacitors of capacitances Qand C2 such that q =C2 /2 are connected across a battery of V volts as shown in the figure. Initially, the key (k) is kept closed to fully charge the capacitors.The key is now thrown open and a dielectric slab of dielectric constant K is inserted in the two capacitors to completely fill the gap between the plates. Find the ratio of (i) the net capacitance and (ii) the energies stored in the combination before and after the introduction of the dielectric slab.[Delhi 2014 C]

Ans.

12.Find the charge on the capacitor as Shown in the circuit. [Foreign 2014]

Ans.

13. 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. [Delhi 2013]

Ans.

14.Determine the potential difference across the plates of the capacitor Cx of the network shown in the figure, (assume, E2 >E1 ) [All India 2013]

Ans.

Ans.

Ans.

17. Two identical parallel plate (air) capacitors Cx and C2 have capacitance C each. The space between their plates is now filled with dielectrics as shown in the figure. If the two capacitors still have equal capacitance, they obtain the relation between dielectric constants K, Kx andK2.[Foreign 2011]

Ans.

18. You are given an air filled parallel plate capacitor C1. The space between its plates is now filled with slabs of dielectric constants Kx and K2 as shown in figure. Find the capacitance of the capacitor C2 if area of the plates is A and distance between the plates is d.

Ans.

19. Figure shows a sheet of aluminium foil of negligible thickness placed between the plates of a capacitor. How will its capacitance be affected if

(i) the foil is electrically insulated?

(ii) the foil is connected to the upper plate with a conducting wire?[Foreign 2011]

Ans.

20. (i) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.

(ii) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop abcda. [Delhi 2014]

Ans. (i) The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential.

### 3 Marks Questions

21. (i) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.

(ii) Two charged spherical conductors of radii and 1^ when connected by a conducting plate respectively. Find the ratio of their surface charge densities in terms of their radii. [Delhi 2014]

Ans.

Ans.

Ans.

24. A capacitor of 200 pF is charged by a 300 V battery. The battery is then disconnected and the charged capacitor is connected to another uncharged capacitor of 100 pF. Calculate the difference between the final energy stored in the combined system and the initial energy stored in the single capacitor.[Foreign 2012]

Ans.

Ans.

26. A parallel plate capacitor is charged by a battery. After sometime, the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will

(i)the capacitances of the capacitor,

(ii)potential difference between the plates and

(iii)the energy stored in the capacitors be affected? Justify your answer in each case.[Delhi 2010]

Ans.

27. A parallel plate capacitor, each with plate area A and separation d is charged to a potential difference V. The battery used to charge it remains connected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change if any will take place in

(i)charge on plates?

(ii)electric field intensity between the plates?

Ans.

28.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 will change?

(i)Electric field between the plates

(ii)Capacitance

(iii)Energy stored in the capacitor [Delhi 2010]

Ans.

29. Find the ratio of the potential differences that must be applied across the parallel and the series combination of two identical capacitors so that the energy stored in the two cases becomes the same.
[Foreign 2010]

Ans.

30. (i) How is the electric field due to a charged parallel plate capacitor affected when a dielectric slab is inserted between the plates fully occupying the intervening region?

(ii) A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness 1/2 d,
where d is the separation between the plates. Find the expression for the capacitance when the slab is inserted between the plates. [Foreign 2010]

Ans.

31. (i) Plot a graph comparing the variation of potential V and electric field E due to a point charge 0 as a function of distance R from the point charge.

(ii) Find the ratio of the potential differences that must be applied across the parallel and the series combination of two capacitors, Cl and C2 with their capacitances in the ratio 1 : 2, so that the energy stored in the two cases becomes the same[Foreign 2010]

Ans.

Ans.

33. A parallel plate* capacitor is charged by a battery. After sometime, the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would

(i)the electric field between the plates

(ii)the energy stored in the capacitor be affected? Justify your answer. [All India 2009]

Ans. (i) The total charge on the capacitor remains conserved on introduction of dielectric slab. Also, the capacitance of capacitor increases to K times of original values.

Ans.

Ans.

Ans.

37. A system of capacitors connected as shown in the figure has a total energy of 160 mJ stored in it. Obtain the value of the equivalent capacitance of this system and the value of Z. [All India 2009 c]

Ans.

38. The two plates of a parallel plate capacitor are 4 mm apart. A slab of dielectric constant 3 and thickness 3 mm is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the capacitance of the capacitor becomes (2/3 )rd of its original value. What is the new distance between the plates?[All India 2008 C]

Ans.

### 5 Marks Questions

39.(i) A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates.
How will

(a)its capacitance

(b)electric field between the plates and

(c)energy stored in the capacitor be affected? Justify your answer giving necessary mathematical expressions for each case.

(ii) (a) Draw the electric field lines due to a conducting sphere.

(b) Draw the electric field lines due to a dipole.

Ans.

Ans.

41. (i)Show that in a parallel plate capacitor if the medium between the plates of a capacitor is filled with an insulating substance of dielectric constant K, its capacitance increases.

(ii) Deduce the expression for the energy stored in a capacitor of capacitance C with charge Q.[Delhi 2009 C]

Ans.

42. A small sphere of radius a carrying a positive charge q is placed concentrically inside a larger hollow conducting shell of radius b(b> a). This outer shell has charge Q on it. Show that if these spheres are connected by a conducting wire, charge will always flow from the inner sphere to the outer sphere irrespective of the magnitude of the two charges.

Ans.

Ans. (i) An expression for energy stored in a charged capacitor. (ii) Connected battery ensures same potential difference (V) across capacitor even after reducing the spacing between plates halved and introducing the dielectric slab.

44.Define the terms (i)capacitance of a capacitor (ii)dielectric strength of a dielectric

(iii)When a dielectric is inserted between the plates of a charged parallel plate capacitor fully occupying the intervening region, how does the polarisation of the dielectric medium affect the net electric field? For a linear dielectric,show that the introduction of the dielectric increases its capacitance by a factor K which is a characteristic of the dielectric. [Delhi 2008 C]

Ans.

## Related content

 NCERT Exemplar for Class 6 Maths Solutions CBSE Notes for Class 8 Maths CBSE Notes for Class 7 Science CBSE Notes for Class 8 Science Lines and Angles Class 9 Extra Questions Maths Chapter 6 CBSE Notes for Class 7 Maths Class 7 CBSE Notes AMU Class 11 Entrance Exam Sample Papers CBSE Notes Class 4 Maths Harappan Civilization