Download 12th CBSE Physics NCERT Answers of Chapter 1 Electric Charges and Fields (Q 9-16)

Chapter 1, Electric Charges and Fields

NCERT Answers of 12th CBSE Physics Textbook Exercise Questions

Question 1.9: A system has two charges qA = 2.5 x 10-7 C and qB = -2.5 x 10-7 C located at points A (0, 0, -15 cm) and B (0, 0, +15 cm) respectively. What is the total charge and electric dipole moment of the system ?
Answer: The charges qA and qB are located at points A (0, 0, -15 cm) and B (0, 0, +15 cm) on z –axis as shown in the figure here. They form an electric dipole, where,
q = total charge =?
p = electric dipole moment of the system =?
p = q.2a
Now,
q = qA + qB = 2.5 x 10-7 + (-2.5 x 10-7) = 0
Also, 2a = AB = dipole length = OA + OB
= 15 + 15 = 30 cm = 0.30 m.
Hence, p = either charge x dipole length
               = qA x AB = 2.5 x 10-7 x 0.30
               = 7.5 x 10-8 cm.
The electric dipole moment is directed from B to A i.e., along negative z-axis.
https://www.cbsencertsolution.com/2018/08/download-12-cbse-physics-ncert-answers-of-chapter-1-electric-charges-and-fields.html
Question 1.10: An electric dipole with dipole moment 4 x 10-9 C m is aligned at 300 with the direction of a uniform electric field of magnitude 5 x 104 NC-1. Calculate the magnitude of the torque acting on the dipole.
Answer: Here, p= 4 x 10-9 Cm, E = 5 x 104 NC-1, θ = 300, t =?
Using the formula, t = pE sinθ, we get
t  = 4 x 10-9 x 5 x 104 x sin 300
   = 20 x 10-5 x ½
   = 10-4 Nm.
Question 1.11: A polythene piece rubbed with wool is found to have a negative charge of 3 x 10-7C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene ?
Answer: (a) Here, q = total charge transferred = -3 x 10-7 C. Charge on an electron, e = -1.6 x 10-19 C.
n = no. of electrons transferred =?
As the polythene piece rubbed with wool is found to attain –ve charge, so the electrons are transferred from wool to polythene piece.
From quantisation of charge, we know that q = ne
CBSE Guide NCERT Solutions - Download Class12 CBSE Physics NCERT Solutions of Electric Charges and Fields




(b) Yes, there is a transfer of mass from wool to polythene as electrons are material particles and are transferred from wool to polythene piece.
m = mass of each electron = 9.1 x 10-31 kg, n = 1.875x 1012
M = total mass transferred to polythene =?
    = m x n = 9.1 x 10-31 x 1.875 x 1012 = 1.71 x 10-18 kg or, 2 x 10-18 kg.
Question 1.12: (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 x 10-7 C? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion of each sphere is charged double the above amount, and the distance between them is halved?
Answer: (a) Charge on 1st sphere, A = qA = 6.5 x 10-7 C
Charge on 2nd sphere, B = qB = 6.5 x 10-7 C
Distance between sphere A and B = 50 cm = 0.5 m = d
We know that,
NCERT Solutions of Class 12 CBSE Physics - Electric Charges and Fields








(b) If each sphere is charged double the amount, then,
qA = qB = 2 x 6.5 x 10-7 C = 13 x 10-7 C
and, r = 1/2 x 50 cm = 0.25 m.
We know that,
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Question 1.13: Suppose the spheres A and B of previous exercise have identical sizes. A third sphere of the same size but unchanged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the force of repulsion between A and B ?
Answer: Initial charges on spheres A and B are qA = qB = 6.5 x 10-7 C
r = Distance between the spheres A and B = 0.5 m.
All the spheres will have equal charges on being brought in contact because they are of the same size.
When 3rd sphere C having charge qc = zero is brought in contact with 1st sphere A, then;
Charge on A = Charge on C = q’1 say

























Question 1.14: Figure given below shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio ?
Download NCERT Answers of Electric Charges and Fields, Class 12 CBSE Physics

Answer












Question 1.15: Consider a uniform electric field E = 3 x 103 i N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ? (b) What is the flux through the same square if the normal to its plane makes at 60o angle with the x-axis ?
Answer




























Question 1.16: What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Answer: Net flux over the cube is zero because the number of lines entering the cube of side 20 cm is same as the number of lines leaving the cube.

 12th CBSE Physics - Electric Charges and Fields - earlier and next posts
# NCERT Solutions (Exercise Question 1.17 - 1.24)
# CBSE Guide - Study materials, Additional MCQ type, Objective & Long Questions, Numerical Problems etc. 


Download NCERT Solutions of Class 12 CBSE Physics Chapter 1, Electric Charges and Fields (Q 1-8)

Electric Charges and Fields

NCERT Solutions of Class XII, CBSE Physics Chapter 1 Textbook Exercise Questions

Question 1.1: What is the force between two small charged spheres having charges of 2 x 10-7 C and 3x10-7 C placed 30 cm apart in air ?
Answer: Here, q1 = 2x10-7 C, q2 = 3x10-7 C, r = 0.30m, F =?
Using the relation,
https://www.cbsencertsolution.com/2018/08/download-ncert-solutions-of-class-12-cbse-physics-chapter-1-electric-charges-and-fields.html







Question 1.2: The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge -0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres ? (b) What is the force on the second sphere due to the first ?
Answer: Given that,
q1 = 0.4 µC = 0.4x10-6 C, q2 = -0.8 µC = -0.8x10-6 C,
F= electrostatic force between q1 and q2 = 0.2 N.
(a) r=?
Using the relation,
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(b) Force on q2 due to q1 =?
We know that electrostatic forces always, appear in priors and follow Newton’s 3rd law of motion.
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Question 1.3: Check that the ratio ke2 / Gmemp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does this ratio signify ?
Answer:
Dimensions of e2 are = [C2]
Dimensions of k are = [Nm2 c2] = [ML3 T-2 C-2]
Dimensions of G are = [M-1 L3 T-2]
Dimensions of me = [M]
https://www.cbsencertsolution.com/2018/08/download-ncert-solutions-of-class-12-cbse-physics-chapter-1-electric-charges-and-fields.html


























Question 1.4: (a) Explain the meaning of the statement ‘electric charge of a body is quantized’.
(b) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large scale charges?
Answer:
(a) The meaning of the statement ‘electric charge of a body is quantized’ is that the charge on it is always some integral multiple of elementary charge of an electron or a proton (= e in magnitude) i.e., discrete packets called quanta or packets of charge. Mathematically, the charge on a body can be expressed as -
q = ± ne where,
n = an integer, e = magnitude of the charge of an electron or proton = 1.6x10-19 C.
A fraction of the fundamental charge e has never been observed in free state.

(b) In practice, the charge on a charged body at macroscopic level is very large while, the charge on an electron is very small. When electrons are added to or removed from a body, the change taking place in the total charge on the body is so small that the charges seems to be varying in a continuous manner. Thus, quantization of electric charge can be ignored at macroscopic level i.e., when dealing with a large scale charged body.

Question 1.5: When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the other law of conservation of charge.
Answer: Initially both the glass rod and silk cloth are electrically neutral before rubbing. In other words, net charge on the glass rod and silk cloth is zero. When the glass rod is rubbed with silk cloth, thus, glass rod becomes positively charged and silk cloth negatively charged. The positive charge on the glass rod is exactly equal to the negative charge on the silk cloth, so net charge on the system is again zero. Thus, the appearance of charge on the glass rod and silk cloth is in accordance with the law of conservation of charge as the total charge of the isolated system is constant. Similarly, when ebonite rod is rubbed with fur, they acquire –ve and +ve charges respectively and net charge is zero again.
Thus, we conclude that charge is neither created nor destroyed but it is merely transferred from one body to another which is consistent with the law of conservation of charge.

Question 1.6: Four points charges qA = 2 µC, qB = -5 µC, qC= 2 µC and qD = -5 µC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square ?
Answer: Consider the square ABCD of each side 10 cm and centre O. The charge of 1 µC is placed at O. Now clearly,
OA = OB = OC = OD
AB = BC = 10 cm = 0.1 m
https://www.cbsencertsolution.com/2018/08/download-ncert-solutions-of-class-12-cbse-physics-chapter-1-electric-charges-and-fields.html





























Question 1.7: (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not ?
(b) Explain why two field lines never cross each other at any point ?
Answer:
(a) The electrostatic line of force is the path tangent at every point of which gives the direction of electric field at that point. The direction of electric field generally changes from point to point. So, the lines of force are generally curved lines. Further, they are continuous curves and cannot have sudden breaks if it is so, then it will indicate the absence of electric field at the break points.

b) The electric lines of force never cross each other because if they do so, then, at the point of their intersection, we can draw two tangents which give two directions of electric field at that point which is not possible.
https://www.cbsencertsolution.com/2018/08/download-ncert-solutions-of-class-12-cbse-physics-chapter-1-electric-charges-and-fields.html
Question 1.8: Two point charges qA = 3 µC and qB = -3 µC are located 20 cm apart in vacuum.
(a) What is the electric field at the mid-point O of the line AB joining the two charges ?
(b) If a negative test charge of magnitude 1.5 x 10-29 C is placed at this point, what is the force experienced by the test charge ?
Answer:
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Here, charge at point A, qA = +3 µC = 3x10-6 C. 
Charge at point B, qB = -3 µC = -3x10-6 C.
r = AB = 20 cm = 0.2 m.
Let O be the mid-point of the line AB, then,
OA = OB = r/2 = 0.2/2 = 0.1 m.
https://www.cbsencertsolution.com


























(b) Force on a negative charge of magnitude 1.5 x 10-9 C is given by the formula.
https://www.cbsencertsolution.com/2018/08/download-ncert-solutions-of-class-12-cbse-physics-chapter-1-electric-charges-and-fields.html













 12th CBSE Physics - Electric Charges and Fields - Next Posts (coming soon)

# NCERT Solutions (Exercise Question 1.17 - 1.24)

# CBSE Guide - Study materials, Additional MCQ type, Objective & Long Questions, Numerical Problems etc. 



CBSE Class 7 Sanskrit Vyakran | NCERT Sanskrit Grammar | Vyanjan Sandhi | व्यंजन संधि

Cbse Ncert Sanskrit Grammar संस्कृत व्याकरण
Sanskrit Sandhi Prakarnm (व्यंजन संधि) 

(१) व्यंजन संधि ==> स्वर अथवा व्यंजन वर्णों के मेल से जो विकार उत्पन्न होता है, उसे व्यंजन संधि कहते हैं। जैसे - 
उत् + लेखः = उल्लेखः 

(२) वर्ग के प्रथम वर्ण के बाद वर्ग का तृतीय, चतुर्थी वर्ण या, य्, र्, ल्, व्, ह् अथवा स्वर वर्ण आये तो प्रथम वर्ण अपने वर्ग के तृतीय वर्ण में बदल जायेगा| यथा - 
क ==> ग्, च् ==> ज्, ट् ==>ड्, त् ==> द्, प् ==> ब् 
दिक् + अम्बरः = दिगम्बरः, 
दिक् + गजः = दिग्गजः, 
अप् + जम्  = अब्ज्म्, 
अच् + अन्तः = अजन्तः, 
जगत् + बन्धुः = जगद्बन्धुः  

(३) "स्" और तवर्ग के बाद श् और चवर्ग आए तो इसप्रकार बदलाव होते हैं - 
स् ==> श, तवर्ग ==> चवर्ग 
रामस् + शेते = रामश्शेते, 
सत् + चारित्रः = सच्चरित्रः,
"स्" और तवर्ग के बाद ष, तवर्ग रहे तो तवर्ग का टवर्ग और स का ष हो जाता है| यथा -
आकृष् + तः = आकृष्टः

(४) अगर झ्, भ्, ध्, ढ़, घ्, ज्, ब्, ग्, ड्, ढ, क्, और प के बाद 'श्' हो तो 'श' के स्थान में 'छ्' हो जाता है और 'त्' या 'द' वर्ण का 'च्' हो जाता है|    
तद् + शिरः = तच्छिरः, 
तद् + शिवः = तच्छिवः 

(५) अगर 'म' के बाद कोई व्यंजन वर्ण रहे तो 'म' का अनुस्वार हो जाता है। जैसे -
सम् + गमः = संगम्,
सम् + सारः = संसारः

(६) अगर 'म' के बाद कोई स्पर्श वर्ण (कवर्ग से पवर्ग तक) हो तो 'म' का अनुस्वार या फिर परवर्ती वर्ण का पंचम हो जाता है।
सम् + कल्पः = संकल्पः,
मृत्युम् + जयः = मृत्युञ्जयः

(७) सम् उपसर्ग के बाद अगर 'कृ' धातु से बना कोइ शब्द आये तो 'म' का अनुस्वार हो जाता है और 'स्' का आगम हो जाता है| जैसे -
सम् + कारः = संस्कारः,
सम् + कृतिः = संस्कृतिः,
सम् + कृतः = संस्कृतः

 CBSE NCERT Sanskrit Grammar (Vyakran) 

To be continued: Next will be some Home Work / Exercise with their answers 
    

CBSE Sanskrit Grammar | NCERT Sanskrit Vyakran Sandhi Prakarnam | संधि-प्रकरणम्

Cbse Ncert Sanskrit Grammar संस्कृत व्याकरण
Sanskrit Sandhi Prakarnm (संधि-प्रकरणम्)

Question : स्वर संधि किसे कहते हैं?
Answer: दो स्वरों के मिलने से जो विकार उत्पन्न होता है, उसे स्वर संधि कहते हैं।
जैसे - अ + अ = आ, कार्य + अधीन = कार्याधीन।

Question : स्वर संधि के कितने भेद हैं?
Answer: स्वर संधि के पाँच भेद हैं -
क) दीर्घ संधि।
ख) वृद्धि संधि।
ग) गुण संधि।
घ) यण संधि।
ङ) अयादि संधि।

Question : संस्कृत मे संधि के कितने प्रकार हैं (संधि प्रकरणम्) ?
Answer:
क) दीर्घ संधिः।
ख) वृद्धि संधिः।
ग) गुण संधिः।
घ) यण संधिः।
ङ) अयादि संधिः।
च) पूर्वरूपसंधिः।
छ) पररूपसंधिः। 
ज) प्रकृतिभावसंधिः (संधि - निषेध) 

Question: दीर्घ संधि किसे कहते हैं?
Answer: अ, इ, उ, ऋ के बाद समान स्वर आए तो दोनों मिलाकर दीर्घ हो जाता है।
उदाहरण - अ + अ = आ, राम + अनुजः = रामानुजः।
आ + आ = आ, विद्या + आलयः = विद्यालयः।
इ + इ = ई, रवि + इन्द्रः = रवीन्द्रः।
उ + उ = ऊ, भानु + उदयः = भानूदयः।
ऋ + ऋ = ऋ, पितृ + ऋणम् = पितृणम् 

Question: गुण संधि किसे कहते हैं?
Answer: अगर 'अ' या 'आ' के बाद 'इ' या 'ई' आए तो दोनों मिलकर 'ए' हो जाता है और 'अ' या 'आ' के बाद 'उ' या 'ऊ' आए तो दोनों मिलकर 'ओ' हो जाता है। साथ ही 'अ' या 'आ' के बाद 'ऋ' आए तो दोनों मिलकर 'अर्' हो जाता है।
जैसे - अ + इ = ए, देव + इन्द्रः = देवेन्द्रः। 
आ + ई = ए, रमा + ईशः = रमेशः।
अ + उ = ओ, सूर्य + उदयः = सूर्योदयः। 
अ + ऋ = अर्, देव + ऋषिः = देवर्षिः। 

Question: वृद्धि संधि किसे कःते हैं? 
Answer: 'अ' या 'आ' के बाद 'ए' या 'ऎ' आये तो दोनों मिलकर 'ऐ' हो जाता है और  'अ' या 'आ' के बाद 'ओ' या 'औ' आये तो दोनों मिलकर 'औ' हो जाता है।
उदहारण - अ + ओ = औ, ग्राम + ओकः = ग्रामौकः।
आ + ओ = औ, गंगा + ओधः = गंगौधः।
अ + ए = ऐ, तत्र + एव = तत्रैव।
आ + ए = ऐ, यथा + एव = यथैव।

Question: यण संधि किसे कह्ते हैं?
Answer: अगर 'इ', 'उ', 'ऋ', 'लृ' के बाद कोइ भिन्न स्वर आये तो 'इ' का 'य', 'उ' का 'व्', 'ऋ' का 'र्' और 'लृ' का 'ल्' हो जाता है| जैसे -
प्रति + एकम् = प्रत्येकम्,
अनु + अयः = अन्वयः,
पितृ + आदेश = पित्रादेशः,
लृ + आकृति = लाकृतिः,
सु + आगतम् = स्वागतम्

Question: अयादि संधि किसे कह्ते हैं?
Answer: अगर 'ए', 'ऐ', 'ओ' और 'औ' के बाद कोइ भिन्न स्वर आये तो 'ए' का 'अय्', 'ऐ' का 'आय्', 'ओ' का 'अव', 'औ' का 'आव्' हो जाता है| जैसे -
ने + अनम् = नयनम्,
नै + अकः = नायकः,
पौ + अकः = पावकः                         

Question: पूर्वरूपसंधिः का उदाहरण दें। 
Answer: हरे + अव = हरेsव

Question: पररूपसंधिः का उदाहरण दें। 
Answer: प्र + एषणम् = प्रेषणम्,
उप + ओषति = उपोषति,
अव + ओषति = अवोषति

Question: प्रकृतिभावसंधिः (संधि - निषेध) का उदाहरण दें। 
Answer: कवी + इमौ = कवी इमौ,
विष्णू + एतौ = विष्णू एतौ,
लते + एते = लते एते।
पूर्वपद में द्विवचन के 'ई', 'ऊ', 'ए' होने पर ऐसा होता है।

 CBSE NCERT Sanskrit Grammar (Sandhi) 

Next Vyanjan Sandhi (व्यंजन संधि)
.. 








CBSE Class 12 Physics Part-I | 12th NCERT Physics Solutions | Class XII Physics CBSE Guide | CBSE Physics Notes and NCERT Answers

Welcome to CBSE Guide NCERT Solutions
Here you get -
  • Chapter wise NCERT Solutions, NCERT Answers for CBSE Class XII, Physics (Part - I). 
  • A complete Class XII CBSE Physics Guide carrying CBSE Notes and additional CBSE Questions with their solutions on various topics from Class 12 NCERT Physics Part - I textbook.
  • Click links given under each chapter of NCERT Physics Textbook below.    

CBSE Class 12 NCERT Physics Part - I
(NCERT Solutions – CBSE Guide - CBSE Guess - NCERT Answers - CBSE Notes)

CHAPTER 1: ELECTRIC CHARGES AND FIELDS
Electric Charges, Conductors and Insulators, Charging by Induction, Basic Properties of Electric Charges, Coulomb's Law, Forces between Multiple Charges, Electric Field, Electric Field Lines, Electric Flux, Electric Dipole, Dipole in a Uniform External Field, Continuous Charge Distribution, Gauss's Law, Application of Gauss's Law
CHAPTER 2: ELECTROSTATIC POTENTIAL AND CAPACITANCE
Electrostatic Potential, Potential due to a Point Charge, Potential due to an Electric Dipole, Potential due to a System of Charges, Equipotential Surfaces, Potential Energy in a System of Charges, Potential Energy in an External Field, Electrostatics of Conductors, Dielectrics and Polarisation, Capacitors and Capacitance, The Parallel Plate Capacitor, Effect of Dielectric on Capacitance, Combination of Capacitors, Energy Stored in a Capacitor, Van de Graff Generator
*
CHAPTER 3: CURRENT ELECTRICITY
Electric Current, Electric Currents in Conductors, Ohm's Law, Drift of Electrons and the Origin of Resistivity, Limitations of Ohm's Law, Resistivity of Various Materials, Temperature Dependence of Resistivity, Electrical Energy, Power, Combination of Resistors - Series and Parallel, Cells, emf, Internal Resistance, Cells in Series and in Parallel, Kirchhoff's Laws, Wheatstone Bridge, Meter Bridge, Potentiometer
*
CHAPTER 4: MOVING CHARGES  AND MAGNETISM
Magnetic Force, Motion in a Magnetic Field, Motion in Combined Electric and Magnetic Fields, Magnetic Field due to a Current Element, Biot-Savart Law, Magnetic Field on the Axis of a Circular Current Loop, Ampere's Circuital Law, The Solenoid and The Toroid, Force between two Parallel Currents, The Ampere, Torque on Current Loop, Magnetic Dipole, The Moving Coil Galvanometer
*
CHAPTER 5: MAGNETISM AND MATTER
The Bar Magnet, Magnetism and Gauss's Law, The Earth's Magnetism, Magnetization and Magnetic Intensity, Magnetic Properties of Materials, Permanent Magnets and Electromagnets
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CHAPTER 6: ELECTROMAGNETIC INDUCTION
The Experiments of Faraday and Henry, Magnetic Flux, Faraday's Law of Induction, Lenz's Law and Conservation of Energy, Motional Electromotive Force, Energy Conservation: A Quantitative Study, Eddy Currents, Inductance, AC Generator
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CHAPTER 7: ALTERNATING CURRENT
AC Voltage Applied to a Resistor, Representation of AC Current and Voltage by Rotating Vectors - Phasors, AC Voltage Applied to an Inductor, AC Voltage Applied to a Capacitor, AC Voltage Applied to a Series LCR Circuit, Power in AC Circuit: The Power Factor, LC Oscillations, Transformers
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CHAPTER 8: ELECTROMAGNETIC WAVES
Displacement Current, Electromagnetic Waves, Electromagnetic Spectrum
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