NCERT Solutions

Subject experts at Shiksha has designed these Class 12 CBSE Notes in concise and lucid language. We offer complete coverage of topics and subtopics of all chapters of class 12 physics. We have covered following Class 12 Physics Chapters and their subtopics.

• Class 12 Physics Chapter 1 Electric Charges and Fields

• Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

• Class 12 Physics Chapter 3 Current Electricity

• Class 12 Physics Chapter 4 Moving Charges and Magnetism

• Class 12 Physics Chapter 5 Magnetism and Matter

• Class 12 Physics Chapter 6 Electromagnetic Induction

• Class 12 Physics Chapter 7 Alternating Current

• Class 12 Physics Chapter 8 Electromagnetic Waves

• Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

• Class 12 Physics Chapter 10 Wave Optics

• Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

• Class 12 Physics Chapter 12 Atoms

• Class 12 Physics Chapter 13 Nuclei

• Class 12 Physics Chapter 14 Semiconductor Electronics Materials, Devices and Simple Circuits

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,

$\stackrel{?}{E_1}$ = $\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}$ ………….(i)

Where,

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

$\sigma$ = surface charge density at that point

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

$\stackrel{?}{E_2}$ = $\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}$ ………….(ii)

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

$\stackrel{?}{E_2}-\stackrel{?}{E_1}=\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}+\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}=\frac{\sigma }{{\epsilon }_0}\stackrel{?}{n}$ ……(iii)

$\mathrm{S}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{}\mathrm{a}\mathrm{}\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r},\mathrm{}\stackrel{?}{{\mathrm{E}}_1}=0,$

$\stackrel{?}{E}$ = $\stackrel{?}{E_2}$ = $\frac{\sigma }{{\epsilon }_0}\stackrel{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{\sigma }{{\epsilon }_0}\stackrel{?}{n}$

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.

2.31 The force between two conducting spheres is not exactly given by the expression ${\mathit{Q}}_1{\mathit{Q}}_2$ /4 $\mathit{?}{\mathit{?}}_0{\mathit{r}}^2$ , 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 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.