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NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance is the most important & scoring chapter in class 12 physics. Students can get more subject knowledge about chapter Electrostatic Potential and Capacitance from these NCERT Solutions of class 12 physics. We presented step by step solutions for each question given in NCERT Class 12 Physics Textbook as per CBSE Board guidelines.

Thus, referring to the NCERT Solutions for 12th Physics Ch 2 is the smart way to ace up board & competitive exam preparation. NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance Exercises and Added Exercises free download PDF format links are provided here. So, download 12th Physics NCERT Exercises Questions for all concepts of Electrostatic Potential and Capacitance chapter in Hindi & English for a better practice.

Class 12 Physics NCERT Solutions Chapter 2 Electrostatic Potential and Capacitance

One of the scoring chapters in 12th physics is chapter 2 Electrostatic Potential and Capacitance. It offers the best weightage for numeric problems and derivations. You can find the effective numerical problems on the topic capacitance of a complex combination of capacitors in the final exams from this chapter. Also, you will deal with some of the important topics like electrostatic potential and its various factors, the difference between the two electric fields, etc in ch 2 Physics.

All the topics & subtopics important answered questions are covered in the NCERT Solutions Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance PDF. So, refer to the 12th physics NCERT Exercises Questions provided here on Ch 2 Electrostatic Potential and Capacitance & score well.

Class	12
Subject	Physics
Book	Physics
Chapter Number	2
Chapter Name	Electrostatic Potential and Capacitance

NCERT Exercise Questions of Class 12 Physics Chapter Electrostatic Potential and Capacitance

Class 12 NCERT Physics Solutions for Chapter 2 Electrostatic Potential and Capacitance Questions are given here for all students of CBSE board & other state boards. They can refer to these NCERT solutions for better understanding and clarification of the chapter Electrostatic Potential and Capacitance. Here, we have provided online as well as the PDF version of NCERT solutions for Electrostatic Potential and Capacitance.

Question 1.
Two charges 5 x 10^-8 C and -3 x 10^-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero. Ans. Let the potential be zero at 0, then
Answer:
Let the potential be zero at 0, then
v_A + V_B= 0,

where V_A is electric potential due to charge q_A and V_B is the electric potential due to charge q_B.

Question 2.
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the center of the hexagon.
Answer:
Total potential at O is given by,

Question 3.
Two charges 2μC and -2μC are placed at points A and B, 6 cm apart.
(a) Identify an equipotential surface of the system. % What is the direction of the electric field at every through the mid-point. On this plane, potential is zero everywhere.
(b) The direction of electric field is from positive to negative charge i.e. A to B, which is infact perpendicular to the equipotential plane.
Answer:
(a) The eguipotential surface is the plane pependicuIarto the line AB joining the two charges and passing through the mid-point. On this plane, potential is zero everywhere.
(b) The direction of electric field is from positive to negative charge i.e. A to B, which is infact perpendicular to the equipotential plane.

Question 4.
A spherical conductor of radius 12 cm has a charge of 1.6 x 10^-7 C distributed uniformly on its surface. What is the electric field
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the center of the sphere?
Answer:
(a) inside a conductor, the electric field is zero because the charge resides on the surface of a conductor.
(b) Electric field just outside the sphere is given by

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF
(1 pF = 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?
Answer:

Question 6.
Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?
Answer:

Question 7.
Three capacitors of capacitances 2 pF, 3 pF and 4pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Answer:
(a) Total capacitance
= c₁ + c₂ + c₃
= 2 + 3+ 4 = 9pF.
(b) Using C = qv we get q = CV
∴ q_x = C₁V = 2 x 10-¹² x 100
= 2 x 10-¹⁰ C = 200 pC
q₂ = c₂V
= 3 x 10^-12 x 100
= 3 x 10-¹⁰ C = 300 pC
q₃ = c₃v
= 4 x 10-¹² x 100
= 4 x 10-¹⁰ C = 400 pC

Question 8.
In a parallel plate capacitor with air between the plates, each plate has an area of
6 x 10^-3 in² 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 ?
Answer:

Question 9.
Explain what would happen if in the capacitor given in Q. 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.
Answer:

Question 10.
A 12 pF capacitor is connected to a 50 V battery.How much electrostatic energy is stored in the capacitor ?
Answer:
E = 12 CV² = 12 x 12 x 10-¹² x 50 x 50 = 1.5 2 2 x 10-⁸J.

Question 11.
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from die supply and is connected  to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process ?
Ans.
Here, C₁ = 600pF = 6 x 10^-10 F, C₂ =
6 x 10-¹⁰ F, V₁ = 200V, V₂ = 0

Question 12.
A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2 x 10^-9 C from a point P(0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point
R(0, 6 cm, 9 cm.)
Answer:
The work done by electrostatic force on a charge is independent of the path followed by the charge. It depends only on the initial and final positions of the charge.

Question 13.
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the center of the cube.
Answer:
(1) Distance of the center of the cube from vertex is half of the diagonal of the cube

(2) From symmetry, it is clear that electric field at center of the cube is zero.

Question 14.
Two tiny spheres carrying charges 1.5 μc and 2.5 μc are located 30 cm apart. Find the potential and electric field :
(a) at the mid point of the line joining the two charges,
(b) at a point 10 cm from this mid-point in a plane normal to the line and passing through the mid­point.
Answer:

Question 15.
A spherical conducting shell of inner radius r_l and outer radius r₂ has a charge Q.
(a) A charge q is placed at the center of the shell. What is the surface charge density on the inner and outer surfaces of the shell ?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape ? Explain.
Answer:
(a) Charge Q appears on the outer surface.

When charge q is placed at the center, it induces – q charge on the inner surface and +q on the outer surface.
.’. charge density of the inner surface,

and charge density of the outer surface,

Consider a cavity of irregular shape with net charge to be zero inside it. Let a closed loop be partially inside and the rest outside the cavity. The field inside the conductor is zero, so some work is done by the field to carry a test charge in the closed loop, but this is against the provisions of an electrostatic field because as per Gauss’ law, the net charge inside a Gaussian surface must be zero. Thus, there cannot be field lines inside the cavity irrespective of its shape.

Question 16.
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

where n^ is a unit vector normal to the surface at a point and a is the surface charge density at that point. (The direction of h is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σn^ /ε₀
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.[Hint. For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
Answer:
Consider a sheet of charge having charge density a. E on either side of the sheet, perpendicular to the plane of sheet, has same magnitude at all points equidistant from the sheet.
Electric field intensity on the left side of the sheet,

The electric field tangential to the plate is continuous throughout.

Question 17.
long charged cylinder of linear charged density A. is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders ?
Answer:
A cylinder P has linear charge density, λ, length l and radius r₁
The charge on cylinder P, q = XL A hollow co-axial conducting cylinder of length / and radius r₂ surrounds the cylinder P. Charge on cylinder Q = – q.

Consider a Gaussian surface in the form of a cylinder of radius rand length /. The electric flux through the curved surface of the Gaussian surface,

Question 18.
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å :
(a) Estimate the potential Energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a) ?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A separation?
Answer:

Question 19.
If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion H⁺₂ .In the ground state of an H⁺₂, the two protons are separated by roughly 1.5 Å, and the electron is roughly Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Answer:

Question 20.
Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric field at the surfaces of the two spheres ? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Answer:
Two charged conducting spheres of radii a and b connected by a wire will reach to same potential.

Clearly electric charge density for the pointed surface will be more because a flat surface can be equated to a spherical surface of large radius and a pointed portion to a spherical surface of small radius.

Question 21.
Two charges -q and + q are located at points (0, 0, -a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a > > 1.
(c) How much work is done in moving a small test charge from the point
(5, 0, 0) to (-7, 0, 0) along the x-axis ? Does the answer change if the path of the test charge between the same points is not along the x-axis ?
Answer:

The point (x, y, 0) is perpendicular to Z-axis, there­fore, potential at (x, y, 0) is zero.
(b) Consider P to be the point of observation at a distance r from the center (O) of the electric dipole.
Let OP make an angle 0 with the dipole moment p→
and r_1, r₂ be the distances of point P from – q charge and + q charge respectively. Potential at P due to – q charge,



Answer does not change because in electrostatics, the work done does not depend upon actual path, it simply depends upon the initial and final positions.

Question 22.
Figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a > > l, and contrast your results with that due to an electric dipole, and an electric monopole
(i.e., a single charge.)

Answer:

(2) Due to electric dipole, the potential is of 1/r² type.
(3) Due to an electric monopole, the potential is of 1/r type.

Question 23.
An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Answer:
Let N capacitors be used in m rows when each row has n capacitor i.e. N = mn
In series

Question 24.
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 realise 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.)
Answer:

Question 25.
Obtain the equivalent capacitance of the network in figure. For a 300 V supply, determine the charge and voltage across each capacitor. (C.B.S.E. 2008)

Answer:
The equivalent circuit is as shown below :

Potential difference across C₄ is in the ratio 2:1
i.e., 200 V
.’. Charge on C₄ = C₄V₄
= 100 x 200 x 10^-12 = 2 x 10^-8 C
Potential difference across C₁= 100 V
Charge on C₁ = C₁x V₁
=100 x 100 x 10-¹² = 1 x 10^-8 C
Potential difference across C₂ and C₃ is 50 v each
∴ Charge on C₂ or C₃ = C₂V₂
= 200 X 50 x 10^-12 = 10-⁸ C.

Question 26.
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 Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Answer:

Question 27.
A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?
Answer:

Question 28.
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor and E is the magnitude of electric field between the plates. Explain the origin of the factor (1/2).
Answer:
Let F be the force on each plate of the capacitor. If the distance between the plates of the capacitor is increased by dx, then work done = F dx. This work done is stored as the potential energy of the capacitor. The increase in the volume of capacitor = A dx

Question 29.
A spherical capacitor consists of two concentric spherical conductors held in position by suitable insulating supports (Figure.) Show that the capacitance of a spherical capacitor is given by

where r_x and r₂ are the radii of outer and inner spheres, respectively.
Answer:
It consists of two concentric spherical shells A and B of radii a and b with charge +q and charge -q respectively. (Outer sphere is grounded)

Question 30.
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 pC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere ?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Answer:

Question 31.
Answer carefully :
(a)Two large conducting spheres carrying charges Q₁ and Q₂ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q₁Q₂/4Πε₀r², where r is the distance between their centers ?
(b) If Coulomb’s law involved 1/r³ dependence (instead of 1/r²), would Gauss’s law be still true ?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron ? What if the orbit is elliptical ?
(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there ?
(f) What meaning would you give to the capacitance of a single conductor ?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (=6).
Answer:
(a) No, because the distribution of charge on the conducting spheres becomes non-uniform when they are brought closer to each other. Coulomb’s law is valid for point charges only or for spheres on which charge distribution is uniform.
(b) If dependence of Coulomb’s law involves 1/r³ dependence then Gauss’ law in the existing state will not be true.
(c) It is true only if field line is a straight line.
(d) Whatever may be the shape of the closed orbit, it is zero.
(e) Potential is continuous.
(f) It means that a single conductor is a capacitor whose other plate can be considered to be at infinity.
(g) A water molecule has permanent dipole moment
so its dielectric constant is high.
Mica does not have polar molecules.

Question 32.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 pC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects, (i.e., bending of field lines at the end)
Answer:
Capacitance of a cylindrical capacitor is given by


Question 33.
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 ?
Answer:
10% of the given field i.e. 10⁷ V m¹ gives E = 0.1 X 10⁷ Vnr¹

Question 34.
Describe schematically the equi-potential surfaces corresponding to
(a) a constant electric field in the z-direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer:
(a) A plane parallel to xy plane.
(b) Plane parallel to xy plane but the planes having different fixed potential will become closer with the increase in field intensity.
(c) Concentric spheres with origin as center.
(d) A time dependent changing shape nearer to grid which slowly becomes planar and parallel to the grid at far off distances from the grid.

Question 35.
In a van de Graaff type generator a spherical metal shell is to be a 15 X 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 x 10⁷ vm^_1. What is the minimum radius of the spherical shell required ? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.) (C.B.S.E. 2008)
Answer:
Minimum radius of the shell of van de Graaff generator is given by the relation

Question 36.
A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q₂ .Show that if q₁ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge
q₂ on the shell is.
Answer:
Charge resides on the outer surface of a conductor. So the charge on inner sphere will flow towards the shell through the conducting wire. Moreover, from Gauss’ law, no electric field exists inside a Gaussian surface and also the charges enclosed by a closed surface only contribute towards the field. So q₂ does not matter in this case.If is positive, potential difference is also positive.

Question 37.
Answer the following:
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decrases with altitude. Near the surface of the earth, the field is about 100 Vm^_1. Why then do we not get an electric shock as we step out of our house into the open ? (Assume the house to be a stell cage so there is no field inside !)
(b) A man fixes outside his house one evening a two meter high insulating slab carrying on its top a large aluminium sheet of area 1 m². Will he get an electric shock if he touches the metal sheet next morning ?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral ? In other words, what keeps the atmosphere charged ?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning ?
[Hint. The earth has an electric field of about 100 Vm^-1 at its surface in the downward direction, corresponding to a surface charge density = -10^-9 C m^-2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about +1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.]
Answer:
(a) Our body and the earth surface become equipotential (at equal potential). It means there is no potential difference between the earth and our body. Hence no current flows through our body and therefore we do not experience an electric shock.

(b) The aluminium sheet and the earth form a capacitor with the insulating slab as dielectric. The down pour of the atmospheric charge will raise the potential of the sheet of aluminium. When we touch the aluminium sheet, charge will flow to the earth through our body. This flow of charge constitutes on electric current and we will experience a shock.

(c) No doubt the atmosphere continuously gets charged due to lightning, thunder storms but simultaneously it gets discharged through normal weather zones. This keeps the system balanced.

(d) Electrical energy of the atmosphere appears as light, sound and heat energies during thunder storms and lightning.

Chapter 2 Electrostatic Potential and Capacitance Physics Class XII NCERT Exercise Questions & Solutions hold special importance in Engineering entrance exams also. NCERT Solutions of 12th Physics chapter 2 have 16 topics along with few subtopics which deal with the electrostatics concept. Therefore, Students are recommended to refer to all topics & subtopics conceptual explanation from NCERT Physics Books and then start solving the questions with ease.

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NCERT Solutions for Class 11 Maths Chapter 1 here are prepared by subject experts after enormous research. Class 11 Maths NCERT Solutions helps students to solve problems efficiently. NCERT Solutions prevailing are as per the latest CBSE Curriculum and covers all the concepts in NCERT Textbooks. You can find NCERT Solutions Class 11 Maths Chapter 1 Exercises 1.1, 1.2, 1.3, 1.4, 1.5, 1.6 from here. Download the NCERT Solutions of Class 11 Maths Chapter 1 Sets PDF through the quick links available.

We strive our best to provide you the step by step solutions for all the NCERT Problems so that it is easy for students to understand. Get a good grip on the Sets with our Class 11 NCERT Maths Ch 1 Solutions provided covering all the topics in detail. To help you out, we have listed the NCERT Solutions for 11th Class Maths Sets in different mediums such as English, Hindi.

Class 11 Maths NCERT Solutions Chapter 1 Sets

Class 11 Maths Chapter 1 of NCERT has basic definitions and Operations involving Sets. In general, Sets are used to define the concepts of functions and relations. It is necessary to have basic knowledge of Sets as they are needed in the further study of sequences, geometry, probability. Sets is the easiest chapter among all the chapters and can be quite scoring. You will learn about Empty, Finite, Infinite, Equal, Subsets in this Chapter. You will also have Operations on Sets like Difference of Sets, Intersection of Sets and Union of Sets, etc. covered.

Avail the CBSE 11th Class NCERT Solutions for Chapter 1 Sets Ex 1.1, Ex 1.2, Ex 1.3, Ex 1.4, Ex 1.5, Ex 1.6, and brush up your fundamentals.

Class	11
Book	Mathematics
Subject	Maths
Chapter Number	1
Name of the Chapter	Sets

Download NCERT Solutions for Class 11 Maths Sets – Solved Exercises

In order to score more marks in the board exams practice NCERT Solutions of Class 11 Maths Sets on a frequent basis. Students of CBSE, UP, MP, Gujarat and other boards can follow the NCERT Solutions prepared as per the latest syllabus. You can rely on these NCERT Maths Class 11 Solutions for Sets during your practice as all of them are prepared by subject expertise. Apart from NCERT Solutions, you will get Notes, Study Material, Important questions, Formulas, Exemplar Problems, etc. regarding the Sets.

• Sets Class 11 Ex 1.1
• Sets Class 11 Ex 1.2
• Sets Class 11 Ex 1.3
• Sets Class 11 Ex 1.4
• Sets Class 11 Ex 1.5
• Sets Class 11 Ex 1.6
• Sets Class 11 Miscellaneous Exercise

Start Practicing the NCERT Maths Class 11 Solutions of Sets PDF at the earliest and attempt well in the board exams. For the sake of your convenience, we even listed the Class 11 Maths Ch 1 NCERT Solutions in both English & Hindi Mediums. Make the most out of them and stand out from the rest of the crowd.

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