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Ncert Solutions Physics Class 12th

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4 weeks ago

Ncert Solutions Physics Class 12th Magnetisation and Magnetic Intensity

What is magnetic inclination or dip?

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Swayam Gupta

Contributor-Level 6

The Earth's magnetic field makes a magnetic inclination (dip) or angle between Earth's overall magnetic field and the horizontal plane at a particular location. It shows how much the magnetic field is tilted from the horizontal.

tan I = B_V / B_H,

where, B_V is the vertical component and B_H is the horizontal component of the Earth's magnetic field.

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4 weeks ago

Ncert Solutions Physics Class 12th Magnetism and Matter

What is Gauss's Law for magnetism?

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Swayam Gupta

Contributor-Level 6

According to Gauss's law for magnetism, the net magnetic flux through any closed surface is zero.

$? B \cdot d A = 0$

It states that all magnetic field lines must enter and exit from a surface. In this way, it differs from electric charges, which can exist independently as positive and negative.

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4 weeks ago

Ncert Solutions Physics Class 12th Magnetism and Matter

What do you understand by term 'magnetism'?

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Swayam Gupta

Contributor-Level 6

Magnetism is the force exerted by a magnet. It comes when the magnet attracts or repulses an object. For example, a magnet sticks to certain objects because the magnetic force or magnetism pulls it.

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a month ago

Ncert Solutions Physics Class 12th Electricity

Where can I get the Current Electricity Class 12 important questions?

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Vikash Kumar Vishwakarma

Contributor-Level 7

If you want the Current Electricity Class 12 important questions, then check the Class 12 Physics previous year question paper pdf. Also, focusing on the Class 12 Physics past 5 years of exam questions to identify the current electricity class 12 important questions. Well, we have framed all the important questions in the NCERT Physics Class 12 Current Electricity solution pdf, which you can easily download through this article.

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a month ago

Ncert Solutions Physics Class 12th CBSE

What approach should I take to solve the Current Electricity Class 12 NCERT Solutions?

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Vikash Kumar Vishwakarma

Contributor-Level 7

Before attempting the NCERT solutions PDF, solve the textbook exercise questions. Hereafter, verify your answer using the NCERT Class 12 Physics Ch 3 solutions pdf. Following a systematic approach to solve a problem will help to score good marks in the CBSE Class 12 board exam.

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Ncert Solutions Physics Class 12th NCERT

Why should I practice NCERT Solutions for Class 12 Physics Chapter 3?

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Vikash Kumar Vishwakarma

Contributor-Level 7

Class 12 Physics Chapter 3 Current Electricity is important for the board exam. The weightage of the chapter is 8-10%. When you practice the Current Electricity class 12 NCERT solutions, you get an overview of your preparation. If you are lacking in any concept, then it helps to build the conceptual knowledge.

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a month ago

Ncert Solutions Physics Class 12th Ohm's Law

What are the important topics covered in Class 12 Physics Chapter 3 solutions?

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Vikash Kumar Vishwakarma

Contributor-Level 7

The topics focused on while preparing the Current Electricity NCERT solutions are:

Ohm's law
Drift velocity
Resistivity vs temperature
Combination of cells
Kirchhoff's rules and
Wheatstone bridge

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2 months ago

Ncert Solutions Physics Class 12th NCERT

What are the important topics included in NCERT Class 12 Physics Chapter 13 Nuclei?

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Khushi Jaiswal

Contributor-Level 10

Important topics in NCERT class 12 Physics chapter 13 nuclear include atomic structure, nuclear binding energy, radioactivity, Nuclear Fission, and nuclear fusion. These concepts are fundamental to understanding the behaviour and properties of atomic and nuclear matter and are crucial for both the board exam and competitive entrance test.

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3 months ago

Ncert Solutions Physics Class 12th Electrostatic Potential and Capacitance

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

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

Where $\hat{n}$ is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of $\hat{n}$ is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is $σ \hat{n}$ / $ε_{0}$ .

(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 lo

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

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

(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.]

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alok kumar singh

Contributor-Level 10

2.16 Electric field on one side of a charged body is $E_{1}$ and electric field on the other side of the same body is $E_{2} .$ If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

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$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

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New answer posted

3 months ago

Ncert Solutions Physics Class 12th Electrostatic Potential and Capacitance

2.31 (a) Two large conducting spheres carrying charges $Q_{1}$ and $Q_{2}$ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by $Q_{1} Q_{2}$ /4 $π ε_{0} r^{2}$ , where r is the distance between their centres?

(b) If Coulomb's law involved 1/ $r^{3}$ dependence (instead of 1/ $r^{2}$ ), 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 t

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(b) If Coulomb's law involved 1/ $r^{3}$ dependence (instead of 1/ $r^{2}$ ), 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).

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alok kumar singh

Contributor-Level 10

2.31 The force between two conducting spheres is not exactly given by the expression $Q_{1} Q_{2}$ /4 $? ?_{0} r^{2}$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^{3}}$ dependence, instead of $\frac{1}{r^{2}}$ , on r

Yes. If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not the velocity.

Whenever the electron completes an orbit, either circular or elli

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2.31 The force between two conducting spheres is not exactly given by the expression $Q_{1} Q_{2}$ /4 $? ?_{0} r^{2}$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^{3}}$ dependence, instead of $\frac{1}{r^{2}}$ , on r

Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

No. Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

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