Welcome to HSLC Guru. This page presents the complete Class 12 Physics Chapter 1 Question Answer on Electric Charges and Fields for the ASSEB (Assam State School Education Board) English medium learners. The notes follow the prescribed NCERT framework adopted by ASSEB and cover every textual concept, derivation, NCERT exercise solution, and additional practice question (MCQ, fill in the blanks, true/false, very short, short and long answer types) needed for HS Final examinations and for competitive entrance preparation.

Chapter Summary

Electrostatics is the branch of physics that studies charges at rest and the forces, fields, and potentials they generate. Historically Greek philosopher Thales of Miletus observed that amber rubbed with fur attracts light objects; the word “electricity” itself originates from the Greek word “elektron” meaning amber. Modern electrostatics is built on three pillars: the experimental quantitative law given by Charles Augustin de Coulomb, the field concept introduced by Michael Faraday, and the integral relation between flux and enclosed charge stated by Carl Friedrich Gauss.

An electric charge is an intrinsic property of elementary particles which gives rise to the electromagnetic interaction. There are two types of charges, positive and negative, and like charges repel while unlike charges attract. Charges follow three fundamental properties: additivity (the total charge of an isolated system is the algebraic sum of all individual charges), conservation (the total charge of an isolated system never changes with time, although charge can be transferred), and quantisation (any observed charge $q$ must be an integral multiple of the elementary charge $e=1.602\times10^{-19}$ C, that is $q=ne$ where $n=0,\pm1,\pm2,\ldots$).

Coulomb’s law describes the electrostatic force between two stationary point charges. The force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. It acts along the line joining the charges. Mathematically, the vector form is given below.

$$\vec{F}_{12} = \frac{1}{4\pi\epsilon_0}\,\frac{q_1 q_2}{r^{2}}\,\hat{r}_{12}$$

The constant $1/(4\pi\epsilon_0)\approx 9\times 10^{9}$ N·m²/C² in vacuum, and $\epsilon_0=8.854\times10^{-12}$ C²/(N·m²) is the permittivity of free space. In a medium of relative permittivity $\epsilon_r$ (also called dielectric constant $K$), the force is reduced by a factor $K$, so $F_{med}=F_{vac}/K$.

The principle of superposition states that the total force on any one charge due to a system of charges is the vector sum of the individual forces exerted by each of the other charges on that charge. The presence of any third charge does not alter the pairwise force between the other two; the law is linear.

$$\vec{F}_{1} = \sum_{i=2}^{N}\frac{1}{4\pi\epsilon_0}\frac{q_1 q_i}{r_{1i}^{2}}\,\hat{r}_{1i}$$

The electric field $\vec{E}$ at a point in space is the force experienced by a unit positive test charge $q_0$ placed at that point, in the limit $q_0\to 0$ so as not to disturb the source distribution. Its SI unit is N/C or V/m.

$$\vec{E} = \lim_{q_0 \to 0}\frac{\vec{F}}{q_0}\qquad ; \qquad \vec{E}_{point} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^{2}}\hat{r}$$

Electric field lines are continuous imaginary directed curves drawn such that the tangent at each point gives the direction of $\vec{E}$ at that point. They start from a positive charge and end on a negative charge (or extend to infinity). Lines are continuous, never cross each other (because the field has a unique direction at every point), and the density of lines crossing a unit area perpendicular to them gives the magnitude of the field. They contract longitudinally (giving rise to attraction) and exert sideways pressure on each other (giving rise to repulsion).

An electric dipole consists of two equal and opposite point charges $+q$ and $-q$ separated by a small distance $2a$. Its dipole moment is the vector $\vec{p}=q\,\vec{d}$, directed from $-q$ to $+q$, where $\vec{d}=2\vec{a}$. The SI unit is C·m. A small unit, the Debye, is also used: $1\,\text{D}=3.336\times 10^{-30}$ C·m. Important fields produced by a dipole are summarised below.

$$\vec{E}_{axial}=\frac{1}{4\pi\epsilon_0}\,\frac{2\vec{p}}{r^{3}}\qquad (r\gg a)$$

$$\vec{E}_{equatorial}=-\frac{1}{4\pi\epsilon_0}\,\frac{\vec{p}}{r^{3}}\qquad (r\gg a)$$

The axial field is twice the equatorial field at the same distance and is parallel to $\vec{p}$, whereas the equatorial field is anti-parallel to $\vec{p}$. The dipole field falls off as $1/r^{3}$, faster than the $1/r^{2}$ point-charge field, because the contributions from $+q$ and $-q$ partially cancel at large distances.

When placed in a uniform external field $\vec{E}$, an electric dipole experiences no net translational force but a torque tending to align $\vec{p}$ with $\vec{E}$.

$$\vec{\tau}=\vec{p}\times\vec{E}\qquad ; \qquad \tau=pE\sin\theta$$

For continuous distributions, charge is not localised on individual particles but smeared over a region. We therefore define linear charge density $\lambda=dq/dl$ for charge along a line, surface density $\sigma=dq/dA$ for charge spread over a surface, and volume density $\rho=dq/dV$ for charge in a volume. The total field at a point is obtained by integrating the contributions $d\vec{E}=(1/4\pi\epsilon_0)(dq/r^{2})\hat{r}$ over the entire distribution.

The electric flux through an area $\vec{A}$ is the dot product of the field and the area vector. Flux is a scalar quantity, has unit N·m²/C, and represents the number of field lines threading the surface.

$$\Phi_E=\vec{E}\cdot\vec{A}=EA\cos\theta\qquad ;\qquad \Phi_E=\int\vec{E}\cdot d\vec{A}$$

Gauss’s law states that the total electric flux through any closed surface (called a Gaussian surface) is equal to $1/\epsilon_0$ times the net electric charge enclosed by that surface. The law is universal, true for any charge distribution and any closed surface, but is most useful when the charge distribution has spherical, cylindrical or planar symmetry.

$$\oint_{S}\vec{E}\cdot d\vec{A}=\frac{q_{enc}}{\epsilon_0}$$

Three standard applications of Gauss’s law are presented in the textbook. (i) For an infinitely long straight wire of uniform linear charge density $\lambda$, a coaxial cylindrical Gaussian surface gives $E=\lambda/(2\pi\epsilon_0 r)$, directed radially. (ii) For an infinite plane sheet of uniform surface charge density $\sigma$, a Gaussian “pillbox” gives $E=\sigma/(2\epsilon_0)$, directed normal to the sheet on both sides; the field is uniform and independent of distance. (iii) For a uniformly charged thin spherical shell of radius $R$ carrying total charge $q$, the field outside ($r>R$) is $E=q/(4\pi\epsilon_0 r^{2})$ as if all charge were at the centre, while inside ($r

Key Formulas at a Glance

Quantity	Formula	SI Unit
Quantisation of charge	$q=ne,\;e=1.6\times10^{-19}$ C	coulomb (C)
Coulomb force (vacuum)	$F=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r^2}$	newton (N)
Coulomb’s constant	$k=\dfrac{1}{4\pi\epsilon_0}=9\times10^{9}$	N·m²/C²
Permittivity of free space	$\epsilon_0=8.854\times10^{-12}$	C²/N·m²
Force in dielectric	$F_{med}=F_{vac}/K$	N
Electric field (point)	$E=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r^2}$	N/C or V/m
Field on dipole axis	$E_{ax}=\dfrac{1}{4\pi\epsilon_0}\dfrac{2p}{r^3}$	N/C
Field on dipole equator	$E_{eq}=\dfrac{1}{4\pi\epsilon_0}\dfrac{p}{r^3}$	N/C
Dipole moment	$\vec{p}=q\vec{d}$, $|\vec{p}|=q(2a)$	C·m
Torque on dipole	$\vec{\tau}=\vec{p}\times\vec{E}$	N·m
Linear density	$\lambda=dq/dl$	C/m
Surface density	$\sigma=dq/dA$	C/m²
Volume density	$\rho=dq/dV$	C/m³
Electric flux	$\Phi_E=\vec{E}\cdot\vec{A}=EA\cos\theta$	N·m²/C
Gauss’s law	$\oint\vec{E}\cdot d\vec{A}=q_{enc}/\epsilon_0$	N·m²/C
Field of infinite line	$E=\lambda/(2\pi\epsilon_0 r)$	N/C
Field of infinite sheet	$E=\sigma/(2\epsilon_0)$	N/C
Field outside spherical shell	$E=q/(4\pi\epsilon_0 r^2)$	N/C
Field inside spherical shell	$E=0$	N/C

Diagram 1: Two Point Charges (Coulomb’s Law)

Diagram 2: Electric Dipole and Field Points

Diagram 3: Gaussian Cylinder around Line Charge

Diagram 4: Field Lines of an Electric Dipole

NCERT Exercise Solutions

Q1.1. What is the force between two small charged spheres having charges of $2\times10^{-7}$ C and $3\times10^{-7}$ C placed 30 cm apart in air?

Answer: Using Coulomb’s law $F=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r^2}$ with $q_1=2\times10^{-7}$ C, $q_2=3\times10^{-7}$ C and $r=0.30$ m: $F=\dfrac{9\times10^{9}\times 6\times10^{-14}}{0.09}=6\times10^{-3}$ N. Approximately $F\approx 6.0\times10^{-3}$ N (≈ $5.99\times10^{-3}$ N), repulsive since both are positive.

Q1.2. The electrostatic force on a small sphere of charge $0.4\,\mu$C due to another small sphere of charge $-0.8\,\mu$C 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: (a) $r^{2}=\dfrac{kq_1 q_2}{F}=\dfrac{9\times10^{9}\times 0.4\times10^{-6}\times 0.8\times10^{-6}}{0.2}=1.44\times10^{-2}$, so $r=0.12$ m $= 12$ cm. (b) By Newton’s third law the force on the second sphere has the same magnitude, $0.2$ N, but acts in the opposite direction (attractive).

Q1.3. Check that the ratio $ke^{2}/(Gm_e m_p)$ is dimensionless. Look up a Table of physical constants and determine the value of this ratio. What does the ratio signify?

Answer: The ratio is dimensionless because $[ke^{2}]$ = $[\text{N}\cdot\text{m}^2]$ and $[Gm_e m_p]$ = $[\text{N}\cdot\text{m}^2]$. Substituting values: $\dfrac{(9\times10^{9})(1.6\times10^{-19})^{2}}{(6.67\times10^{-11})(9.11\times10^{-31})(1.67\times10^{-27})}\approx 2.27\times10^{39}$. The ratio shows that the electrostatic force between an electron and a proton is about $10^{39}$ times stronger than the gravitational force between them.

Q1.4. (a) Explain the meaning of the statement “electric charge of a body is quantised”. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic, that is, large scale charges?

Answer: (a) The statement means that any free charge $q$ on a body can only take values that are integral multiples of the elementary charge $e=1.6\times10^{-19}$ C; fractional charges (other than the bound quark charges) are not observed. (b) At macroscopic scales the charges involved are enormously larger than $e$ (a microcoulomb already contains about $6\times10^{12}$ electronic charges), so the discreteness becomes invisible and the charge appears to vary continuously.

Q1.5. 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 law of conservation of charge.

Answer: Before rubbing both bodies are electrically neutral. During contact, electrons are transferred from one to the other (silk gains electrons and becomes negatively charged, glass loses electrons and becomes positively charged), but no charge is created or destroyed. The algebraic sum of the charges on the two bodies remains zero, exactly as conservation of charge requires.

Q1.6. Four point charges $q_A=2\,\mu$C, $q_B=-5\,\mu$C, $q_C=2\,\mu$C and $q_D=-5\,\mu$C are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of $1\,\mu$C placed at the centre of the square?

Answer: The diagonally opposite charges are equal in magnitude and sign ($q_A=q_C$ and $q_B=q_D$) and are equidistant from the centre. The forces they produce on the central charge are equal in magnitude and opposite in direction, so they cancel pairwise. Hence the net force on the $1\,\mu$C charge at the centre is exactly zero.

Q1.7. (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) A field line is the path that a small positive test charge would follow in the field. Because the electric field is well-defined and continuous in a charge-free region, the path cannot break suddenly; a break would mean the charge experiences no force at that point and then resumes motion arbitrarily, which is impossible. (b) If two field lines crossed, then at the point of crossing the electric field would have two different directions simultaneously, which is impossible because the field at a point can have only one direction.

Q1.8. Two point charges $q_A=3\,\mu$C and $q_B=-3\,\mu$C are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude $1.5\times10^{-9}$ C is placed at this point, what is the force experienced by the test charge?

Answer: (a) Distance of O from each charge is $0.10$ m. Each charge produces a field of magnitude $E=kq/r^{2}=9\times10^{9}\times 3\times10^{-6}/(0.1)^{2}=2.7\times10^{6}$ N/C. Both fields point from A to B (away from $+3\,\mu$C and toward $-3\,\mu$C), so they add: $E_O=5.4\times10^{6}$ N/C, directed from A to B. (b) $F=qE=1.5\times10^{-9}\times 5.4\times10^{6}=8.1\times10^{-3}$ N. Since the test charge is negative, the force is opposite to $\vec{E}$, that is, from B to A.

Q1.9. A system has two charges $q_A=2.5\times10^{-7}$ C and $q_B=-2.5\times10^{-7}$ C located at points A: $(0,0,-15\,\text{cm})$ and B: $(0,0,+15\,\text{cm})$. What is the total charge and electric dipole moment of the system?

Answer: Total charge $=q_A+q_B=0$. Dipole moment $|\vec{p}|=q\times 2a=2.5\times10^{-7}\times 0.30=7.5\times10^{-8}$ C·m. The dipole moment is directed from $-q$ at B to $+q$ at A, i.e. along the negative $z$-axis.

Q1.10. An electric dipole with dipole moment $4\times10^{-9}$ C·m is aligned at $30^\circ$ with the direction of a uniform electric field of magnitude $5\times10^{4}$ N/C. Calculate the magnitude of the torque acting on the dipole.

Answer: $\tau=pE\sin\theta=4\times10^{-9}\times 5\times10^{4}\times\sin 30^{\circ}=4\times10^{-9}\times 5\times10^{4}\times 0.5=1\times10^{-4}$ N·m.

Q1.11. A polythene piece rubbed with wool is found to have a negative charge of $3\times10^{-7}$ C. (a) Estimate the number of electrons transferred. From which to which? (b) Is there a transfer of mass from wool to polythene?

Answer: (a) $n=q/e=3\times10^{-7}/1.6\times10^{-19}=1.875\times10^{12}$ electrons transferred from wool to polythene. (b) Yes, but only a tiny mass: $\Delta m=n\cdot m_e=1.875\times10^{12}\times 9.11\times10^{-31}\approx 1.71\times10^{-18}$ kg, which is utterly negligible.

Q1.12. (a) Two insulated charged copper spheres A and B have their centres separated by 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5\times10^{-7}$ C? The radii of A and B are negligible compared to the distance. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer: (a) $F=k q^{2}/r^{2}=9\times10^{9}\times(6.5\times10^{-7})^{2}/(0.5)^{2}=1.521\times10^{-2}$ N. (b) Charges become $2q$, distance becomes $r/2$: $F’=k(2q)^{2}/(r/2)^{2}=16 F=16\times 1.521\times10^{-2}=0.2434$ N.

Q1.13. Suppose the spheres A and B in the previous question have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Answer: When the uncharged sphere C touches A, the charge equalises so each becomes $q/2$. When C (carrying $q/2$) then touches B (carrying $q$), the total $3q/2$ shares equally so B is left with $3q/4$. Final charges: $q_A=q/2$, $q_B=3q/4$, distance $r$ unchanged. New force $F’=k(q/2)(3q/4)/r^{2}=(3/8)F=(3/8)\times1.521\times10^{-2}=5.7\times10^{-3}$ N.

Q1.14. Figure 1.33 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?

Answer: Particles 1 and 2 deflect upward (towards the negative plate), so they carry positive charge. Particle 3 deflects downward, so it carries negative charge. Particle 3 shows the largest deflection for the same field length, indicating the highest specific charge $|q|/m$.

Q1.15. Consider a uniform electric field $\vec{E}=3\times10^{3}\hat{i}$ N/C. (a) What is the flux of this field through a square of side 10 cm whose plane is parallel to the $yz$-plane? (b) What is the flux through the same square if the normal to its plane makes a $60^{\circ}$ angle with the $x$-axis?

Answer: Area $A=(0.10)^{2}=10^{-2}$ m². (a) $\Phi=EA\cos 0^{\circ}=3\times10^{3}\times10^{-2}=30$ N·m²/C. (b) $\Phi=EA\cos 60^{\circ}=30\times 0.5=15$ N·m²/C.

Q1.16. 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: The cube encloses no charge and the field is uniform, so flux entering one face equals flux leaving the opposite face. Net flux through the closed surface is zero.

Q1.17. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0\times10^{3}$ N·m²/C. (a) What is the net charge inside the box? (b) If the net outward flux were zero, could you conclude that there were no charges inside the box? Why or why not?

Answer: (a) By Gauss’s law $q=\epsilon_0\Phi=8.854\times10^{-12}\times 8.0\times10^{3}=7.08\times10^{-8}$ C $\approx 0.07\,\mu$C. (b) No. Zero net flux only implies that the algebraic sum of all enclosed charges is zero, not that there are no charges inside; equal positive and negative charges could be present.

Q1.18. A point charge of $+10\,\mu$C is at a distance of 5 cm directly above the centre of a square of side 10 cm. What is the magnitude of the electric flux through the square?

Answer: Imagine the square as one face of a cube of side 10 cm with the charge at its centre. By symmetry the total flux $q/\epsilon_0$ is shared equally by the six faces. $\Phi=\dfrac{q}{6\epsilon_0}=\dfrac{10\times10^{-6}}{6\times 8.854\times10^{-12}}\approx 1.88\times10^{5}$ N·m²/C.

Q1.19. A point charge of $2.0\,\mu$C is at the centre of a cubic Gaussian surface of edge 9.0 cm. What is the net electric flux through the surface?

Answer: $\Phi=\dfrac{q}{\epsilon_0}=\dfrac{2.0\times10^{-6}}{8.854\times10^{-12}}=2.26\times10^{5}$ N·m²/C. The flux is independent of the size or shape of the surface; only the enclosed charge matters.

Q1.20. A point charge causes an electric flux of $-1.0\times10^{3}$ N·m²/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Answer: (a) Flux is independent of the radius of the Gaussian surface; it remains $-1.0\times10^{3}$ N·m²/C. (b) $q=\epsilon_0\Phi=8.854\times10^{-12}\times(-1.0\times10^{3})=-8.85\times10^{-9}$ C $\approx -8.85$ nC.

Q1.21. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre is $1.5\times10^{3}$ N/C and points radially inward, what is the net charge on the sphere?

Answer: Outside a conducting sphere the field behaves as if the charge were concentrated at the centre: $q=4\pi\epsilon_0 r^{2}E=(0.2)^{2}\times 1.5\times10^{3}/9\times10^{9}=6.67\times10^{-9}$ C. The field points inward, so the charge is negative: $q=-6.67$ nC.

Q1.22. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of $80.0\,\mu$C/m². (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Answer: Radius $R=1.2$ m. (a) $q=\sigma\times 4\pi R^{2}=80\times10^{-6}\times 4\pi (1.2)^{2}=1.45\times10^{-3}$ C. (b) $\Phi=q/\epsilon_0=1.45\times10^{-3}/8.854\times10^{-12}=1.64\times10^{8}$ N·m²/C.

Q1.23. An infinite line charge produces a field of $9\times10^{4}$ N/C at a distance of 2 cm. Calculate the linear charge density.

Answer: $E=\lambda/(2\pi\epsilon_0 r)$ gives $\lambda=2\pi\epsilon_0 r E=2\pi\times 8.854\times10^{-12}\times 0.02\times 9\times10^{4}=10^{-7}$ C/m $=0.1\,\mu$C/m.

Q1.24. Two large, thin metal plates are parallel and close to each other. On their inner faces the plates have surface charge densities of opposite signs and of magnitude $17.0\times10^{-22}$ C/m². What is the electric field (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Answer: (a) and (b): In the outer regions the fields due to the two oppositely charged sheets cancel, so $E=0$. (c) Between the plates the fields add: $E=\sigma/\epsilon_0=17.0\times10^{-22}/8.854\times10^{-12}=1.92\times10^{-10}$ N/C, directed from the positive sheet to the negative one.

Additional Multiple Choice Questions (MCQ)

1. The SI unit of electric charge is — (a) ampere  (b) coulomb  (c) volt  (d) ohm

Answer: (b) coulomb.

2. The value of $1/(4\pi\epsilon_0)$ in SI units is — (a) $9\times10^{-9}$ (b) $9\times10^{9}$ (c) $8.854\times10^{-12}$ (d) $1.6\times10^{-19}$

Answer: (b) $9\times10^{9}$ N·m²/C².

3. Charge on $1$ electron is — (a) $+1.6\times10^{-19}$ C (b) $-1.6\times10^{-19}$ C (c) $-9.1\times10^{-31}$ C (d) $0$

Answer: (b) $-1.6\times10^{-19}$ C.

4. Which of the following is NOT a property of charge? (a) Conservation (b) Quantisation (c) Additivity (d) Velocity dependence

Answer: (d) Velocity dependence (charge is invariant).

5. An electric dipole placed at $30^{\circ}$ to a non-uniform electric field experiences — (a) only a force (b) only a torque (c) neither (d) a torque and a translational force

Answer: (d) a torque and a translational force.

6. The closed surface used in Gauss’s law is called a — (a) Gaussian curve (b) Faraday surface (c) Gaussian surface (d) flux surface

Answer: (c) Gaussian surface.

7. Force per unit positive charge at a point is called — (a) electric flux (b) electric field (c) electric potential (d) charge density

Answer: (b) electric field.

8. Eight dipoles are kept inside a closed cube. The total electric flux through the cube is — (a) $8q/\epsilon_0$ (b) $16q/\epsilon_0$ (c) $q/\epsilon_0$ (d) zero

Answer: (d) zero, because each dipole has zero net charge.

9. The axial field of a dipole at distance $r$ is related to the equatorial field at the same distance by — (a) $E_{ax}=2E_{eq}$ (b) $E_{ax}=E_{eq}/2$ (c) equal (d) opposite and equal

Answer: (a) $E_{ax}=2E_{eq}$, with the equatorial field directed opposite to $\vec{p}$.

10. Three equal positive charges are placed at the corners of an equilateral triangle. The net force on a charge of any of the three charges is — (a) zero (b) along the median (c) along one side (d) cannot be determined

Answer: (b) along the median pointing outward (the net force on the central point would be zero, but on a corner charge it is along the bisector).

11. Electric field due to an infinite line of charge at distance $r$ varies as — (a) $1/r$ (b) $1/r^{2}$ (c) $1/r^{3}$ (d) $r$

Answer: (a) $1/r$.

12. Field due to an infinite charged plane sheet of density $\sigma$ is — (a) $\sigma/\epsilon_0$ (b) $\sigma/(2\epsilon_0)$ (c) $2\sigma/\epsilon_0$ (d) $\sigma\epsilon_0$

Answer: (b) $\sigma/(2\epsilon_0)$.

13. Inside a uniformly charged thin spherical shell, the electric field is — (a) maximum at the centre (b) zero everywhere (c) $kQ/R^{2}$ (d) varies linearly

Answer: (b) zero everywhere inside.

14. The unit of electric flux is — (a) N/C (b) N·m²/C (c) C/m² (d) V/m

Answer: (b) N·m²/C.

15. The electric dipole moment is measured in — (a) coulomb (b) coulomb·metre (c) newton·metre (d) farad

Answer: (b) coulomb·metre (also expressed in Debye).

16. Two point charges $q$ and $-q$ are kept $2a$ apart. The dipole moment is — (a) $q\cdot a$ (b) $q\cdot 2a$ (c) $2q\cdot a$ (d) zero

Answer: (b) $p=q(2a)$.

17. A test charge $q_0$ should be — (a) very large (b) negative only (c) infinitesimally small (d) any value

Answer: (c) infinitesimally small, so it does not disturb the source distribution.

18. Two charges repel each other with $100$ N. If one is increased by $10\%$ and the other decreased by $10\%$, the new force is — (a) $100$ N (b) $110$ N (c) $99$ N (d) $81$ N

Answer: (c) $99$ N (since $1.1\times 0.9=0.99$).

19. Charges $+2\,\mu$C and $+6\,\mu$C repel with $12$ N. If each loses $4\,\mu$C, the new force is — (a) $+12$ N (b) $+4$ N (c) $0$ N (d) $-4$ N (attractive of magnitude 4 N)

Answer: (d) Attractive force of $4$ N (charges become $-2\,\mu$C and $+2\,\mu$C, product becomes $-4$ in units relative to original $+12$ which had product 12; ratio $-4/12$ gives $-4$ N).

20. Field lines emerge from a positive charge and terminate on — (a) another positive charge (b) a negative charge or infinity (c) the same charge (d) the centre of the dipole

Answer: (b) a negative charge or extend to infinity.

Fill in the Blanks

1. The SI unit of electric charge is __________.

Answer: coulomb (C).

2. The smallest amount of free charge so far observed is the magnitude of __________.

Answer: the electronic charge $e=1.6\times10^{-19}$ C.

3. Coulomb’s law in vector form between charges $q_1$ and $q_2$ is $\vec{F}=$ __________.

Answer: $\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r^{2}}\hat{r}$.

4. The electric field due to a point charge falls off as __________ with distance.

Answer: $1/r^{2}$ (inverse square).

5. The electric field due to a dipole at large distances falls off as __________.

Answer: $1/r^{3}$.

6. The torque on an electric dipole in a uniform field $\vec{E}$ is given by __________.

Answer: $\vec{\tau}=\vec{p}\times\vec{E}$.

7. A dipole placed in a uniform field experiences zero net __________ but a non-zero __________.

Answer: force; torque.

8. Gauss’s law states that $\oint\vec{E}\cdot d\vec{A}=$ __________.

Answer: $q_{enc}/\epsilon_0$.

9. The electric field inside a conductor in electrostatic equilibrium is __________.

Answer: zero.

10. The electric field at distance $r$ from an infinite line of linear charge density $\lambda$ is __________.

Answer: $\lambda/(2\pi\epsilon_0 r)$.

11. The electric field on either side of an infinite charged plane sheet of density $\sigma$ is __________.

Answer: $\sigma/(2\epsilon_0)$.

12. The principle which allows us to add forces from many charges is called the __________.

Answer: superposition principle.

True / False

1. Charge is conserved in every physical process.

Answer: True.

2. A neutral body has no electrons in it.

Answer: False. It contains equal numbers of protons and electrons.

3. Electric field lines from positive charges form closed loops.

Answer: False. Electrostatic field lines never form closed loops; they begin on positive charges and terminate on negative charges or at infinity.

4. The flux through a closed surface depends only on the enclosed charge.

Answer: True (Gauss’s law).

5. A dipole in a non-uniform field experiences only torque and no force.

Answer: False. It experiences both a torque and a net translational force.

6. Two electric field lines can intersect each other.

Answer: False. They cannot, because the field has a unique direction at every point.

7. Inside a charged spherical shell the electric field is zero.

Answer: True.

8. The Coulomb force between two charges depends on the medium between them.

Answer: True; the medium reduces the force by the dielectric constant $K$.

9. The unit of $\epsilon_0$ is $C^{2}/(N\cdot m^{2})$.

Answer: True.

10. Electric flux is a vector quantity.

Answer: False. Electric flux is a scalar.

Very Short and Short Answer Questions

Q1. Define electric charge and write its SI unit.

Answer: Electric charge is the intrinsic property of a particle by virtue of which it produces and experiences electric and magnetic effects. SI unit is the coulomb (C).

Q2. State the law of conservation of charge.

Answer: The total electric charge of an isolated system remains constant; charge can be transferred from one body to another but cannot be created or destroyed.

Q3. What is meant by quantisation of charge?

Answer: Any free charge $q$ must be an integral multiple of the elementary charge $e=1.6\times10^{-19}$ C, that is $q=ne$ where $n$ is an integer.

Q4. State Coulomb’s law.

Answer: The electrostatic force between two stationary point charges in vacuum is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them, acting along the line joining the charges.

Q5. Define electric field at a point.

Answer: Electric field at a point is the force experienced per unit positive test charge placed at that point in the limit $q_0\to 0$. Its SI unit is N/C.

Q6. Write four properties of electric field lines.

Answer: (i) They begin at positive charges and end at negative charges or at infinity. (ii) They never intersect. (iii) Tangent to a line gives the direction of $\vec{E}$. (iv) Density of lines per unit perpendicular area indicates the field’s strength.

Q7. Define electric dipole and dipole moment.

Answer: An electric dipole is a pair of equal and opposite charges separated by a small distance. Its dipole moment is $\vec{p}=q\vec{d}$, directed from the negative charge to the positive charge; SI unit C·m.

Q8. Write the expression for the field on the axis of a short dipole.

Answer: $\vec{E}_{ax}=\dfrac{1}{4\pi\epsilon_0}\dfrac{2\vec{p}}{r^{3}}$, valid for $r\gg a$.

Q9. Why is the field of a dipole zero at large distances when both charges are present, while it is non-zero for a single charge?

Answer: The fields of $+q$ and $-q$ point in opposite directions and partially cancel; the net field falls as $1/r^{3}$ rather than $1/r^{2}$, so it decreases faster but does not vanish unless the dipole moment itself is zero.

Q10. What is meant by linear, surface and volume charge densities?

Answer: $\lambda=dq/dl$ (charge per unit length), $\sigma=dq/dA$ (charge per unit area) and $\rho=dq/dV$ (charge per unit volume).

Q11. Define electric flux and state its SI unit.

Answer: Electric flux through an area is the dot product of the electric field and the area vector, $\Phi=\vec{E}\cdot\vec{A}$. SI unit: N·m²/C (or V·m).

Q12. State Gauss’s law.

Answer: The total electric flux through any closed surface in vacuum equals $1/\epsilon_0$ times the net charge enclosed by that surface: $\oint\vec{E}\cdot d\vec{A}=q_{enc}/\epsilon_0$.

Q13. Why is the electric field zero inside a hollow conducting sphere?

Answer: Charges on a conductor reside on its outer surface and the interior encloses no charge. Applying Gauss’s law to a Gaussian surface inside the sphere gives zero flux and hence zero field everywhere inside.

Q14. Show that for an infinite plane charged sheet, $E=\sigma/(2\epsilon_0)$.

Answer: Take a cylindrical Gaussian “pillbox” with its axis perpendicular to the sheet. By symmetry, $\vec{E}$ is perpendicular to the sheet and equal on the two flat faces. Flux through the curved surface is zero. Total flux $=2EA$ and enclosed charge $=\sigma A$. Hence $2EA=\sigma A/\epsilon_0$, giving $E=\sigma/(2\epsilon_0)$.

Q15. What is a Gaussian surface? Give two examples used in the textbook.

Answer: Any closed mathematical surface chosen for applying Gauss’s law. Examples: a coaxial cylinder around an infinite line of charge; a concentric sphere around a point charge or charged shell.

Long Answer Questions

Q1. Derive an expression for the electric field on the axis of a short electric dipole.

Answer: Consider a dipole $-q$ at A and $+q$ at B with separation $2a$, having dipole moment $p=q\cdot 2a$. Let P be a point on the axis at distance $r$ from the centre O, on the side of $+q$. Field at P due to $+q$ is $E_{+}=kq/(r-a)^{2}$, directed from B to P. Field due to $-q$ is $E_{-}=kq/(r+a)^{2}$, directed from P to A. Net axial field $E_{ax}=E_{+}-E_{-}=kq\bigl[\frac{1}{(r-a)^{2}}-\frac{1}{(r+a)^{2}}\bigr]=kq\cdot\frac{4ar}{(r^{2}-a^{2})^{2}}$. For $r\gg a$, $(r^{2}-a^{2})^{2}\approx r^{4}$ giving $E_{ax}\approx \dfrac{1}{4\pi\epsilon_0}\dfrac{2p}{r^{3}}$, in the direction of $\vec{p}$.

Q2. Derive the expression for the electric field at a point on the equatorial plane of a short dipole.

Answer: Let P be on the perpendicular bisector at distance $r$ from O. Distance from P to either charge is $\sqrt{r^{2}+a^{2}}$. Each charge produces field $kq/(r^{2}+a^{2})$. The components along the dipole axis combine while the perpendicular components cancel. The axial component from each is $\cos\theta$ times the field magnitude, where $\cos\theta=a/\sqrt{r^{2}+a^{2}}$. Therefore $E_{eq}=2\cdot\frac{kq}{r^{2}+a^{2}}\cdot\frac{a}{\sqrt{r^{2}+a^{2}}}=\frac{kq\cdot 2a}{(r^{2}+a^{2})^{3/2}}=\frac{kp}{(r^{2}+a^{2})^{3/2}}$. For $r\gg a$, $E_{eq}\approx kp/r^{3}=\dfrac{1}{4\pi\epsilon_0}\dfrac{p}{r^{3}}$, anti-parallel to $\vec{p}$.

Q3. Using Gauss’s law, derive the electric field due to an infinitely long straight uniformly charged wire.

Answer: Let the wire have linear charge density $\lambda$. Choose a coaxial cylindrical Gaussian surface of radius $r$ and length $L$. By symmetry the field is radial and uniform on the curved surface; the flux through the flat ends is zero. Total flux $=E\cdot 2\pi r L$. Enclosed charge $=\lambda L$. By Gauss’s law, $E\cdot 2\pi r L=\lambda L/\epsilon_0$. Therefore $E=\lambda/(2\pi\epsilon_0 r)$, directed radially outward (for $\lambda>0$).

Q4. Use Gauss’s law to find the electric field outside and inside a uniformly charged thin spherical shell.

Answer: Let the shell have radius $R$ and total charge $q$. (i) Outside ($r>R$): take a concentric sphere of radius $r$. By symmetry, $E$ is radial and uniform on the surface; $\oint\vec{E}\cdot d\vec{A}=E\cdot 4\pi r^{2}=q/\epsilon_0$, giving $E=q/(4\pi\epsilon_0 r^{2})$ — exactly as if all charge were at the centre. (ii) Inside ($r

Q5. Derive the torque acting on an electric dipole placed in a uniform electric field and discuss the cases of stable and unstable equilibrium.

Answer: A dipole with moment $\vec{p}$ in field $\vec{E}$ experiences forces $+q\vec{E}$ on $+q$ and $-q\vec{E}$ on $-q$. The forces are equal and opposite so the net force is zero, but they form a couple. Magnitude of torque $\tau=qE\cdot 2a\sin\theta=pE\sin\theta$. In vector form $\vec{\tau}=\vec{p}\times\vec{E}$. Stable equilibrium: $\theta=0$ ($\vec{p}\parallel\vec{E}$, $\tau=0$ and any disturbance produces a restoring torque). Unstable equilibrium: $\theta=180^{\circ}$ ($\vec{p}$ anti-parallel to $\vec{E}$, $\tau=0$ but any disturbance grows).

Glossary

Term	Meaning
Electric charge	Intrinsic property of matter giving rise to electromagnetic interaction.
Coulomb (C)	SI unit of charge; charge transferred by 1 ampere in 1 second.
Elementary charge ($e$)	Magnitude of the charge of a proton or electron, $1.6\times10^{-19}$ C.
Quantisation	Property that observed charge is an integer multiple of $e$.
Conservation of charge	Total charge of an isolated system is constant.
Coulomb’s law	Force between two stationary point charges $F=kq_1 q_2/r^{2}$.
Permittivity of free space ($\epsilon_0$)	$8.854\times10^{-12}$ C²/(N·m²).
Dielectric constant ($K$)	Ratio $\epsilon/\epsilon_0$; reduces the Coulomb force in a medium.
Superposition principle	Total force/field is the vector sum of contributions from each source.
Electric field ($\vec{E}$)	Force per unit positive test charge.
Field line	Curve whose tangent gives the direction of $\vec{E}$.
Electric dipole	Pair of equal and opposite charges separated by a small distance.
Dipole moment ($\vec{p}$)	$q\vec{d}$, directed from $-q$ to $+q$.
Debye	Unit of dipole moment, $1\text{ D}=3.336\times10^{-30}$ C·m.
Charge density ($\lambda,\sigma,\rho$)	Linear, surface, and volume charge per unit length, area, volume.
Electric flux ($\Phi_E$)	$\vec{E}\cdot\vec{A}$; number of field lines through an area.
Gaussian surface	Imaginary closed surface used to apply Gauss’s law.
Gauss’s law	$\oint\vec{E}\cdot d\vec{A}=q_{enc}/\epsilon_0$.
Torque on dipole	$\vec{\tau}=\vec{p}\times\vec{E}$, aligning $\vec{p}$ with $\vec{E}$.
Stable equilibrium of dipole	$\vec{p}\parallel\vec{E}$.
Unstable equilibrium of dipole	$\vec{p}$ anti-parallel to $\vec{E}$.

Numerical Practice Problems

N1. Two point charges of $+5\,\mu$C and $+10\,\mu$C are placed $20$ cm apart in air. Find the magnitude and nature of the force between them.

Answer: $F=k\dfrac{q_1 q_2}{r^{2}}=9\times10^{9}\times\dfrac{5\times10^{-6}\times 10\times10^{-6}}{(0.2)^{2}}=11.25$ N. As both charges are positive the force is repulsive.

N2. Two charges $q_1=+3\,\mu$C and $q_2=-3\,\mu$C are separated by $2a=4$ cm. Find the dipole moment and the field at $r=20$ cm on the axis (assume short dipole).

Answer: $p=q\cdot 2a=3\times10^{-6}\times 0.04=1.2\times10^{-7}$ C·m. $E_{ax}=\dfrac{1}{4\pi\epsilon_0}\dfrac{2p}{r^{3}}=9\times10^{9}\times\dfrac{2\times 1.2\times10^{-7}}{(0.20)^{3}}=2.7\times10^{5}$ N/C, parallel to $\vec{p}$.

N3. A charge of $1\,\mu$C is placed at the centre of a cube of edge $10$ cm. Find the flux through one face.

Answer: Total flux $=q/\epsilon_0=1\times10^{-6}/8.854\times10^{-12}=1.13\times10^{5}$ N·m²/C. By symmetry, flux through one face $=\Phi_{tot}/6=1.88\times10^{4}$ N·m²/C.

N4. An infinite plane sheet has uniform surface charge density $\sigma=2\,\mu$C/m². Find the electric field at any point near the sheet.

Answer: $E=\dfrac{\sigma}{2\epsilon_0}=\dfrac{2\times10^{-6}}{2\times 8.854\times10^{-12}}=1.13\times10^{5}$ N/C, directed normal to the sheet on either side.

N5. An electron and a proton are separated by $1$ Å $(=10^{-10}$ m$)$. Find the magnitude of the electrostatic force between them.

Answer: $F=\dfrac{ke^{2}}{r^{2}}=\dfrac{9\times10^{9}\times (1.6\times10^{-19})^{2}}{(10^{-10})^{2}}=2.30\times10^{-8}$ N (attractive).

N6. Two pith balls each carrying charge $q$ are suspended from a common point by silk threads of length $L$. At equilibrium each thread makes a small angle $\theta$ with the vertical. Show that $q^{2}=16\pi\epsilon_0 mgL^{2}\sin^{2}\theta\tan\theta$.

Answer: Each ball is in equilibrium under three forces: weight $mg$ downward, tension $T$ along the thread, and Coulomb repulsion $F=kq^{2}/r^{2}$ horizontally where $r=2L\sin\theta$. Resolving: $T\cos\theta=mg$ and $T\sin\theta=F$. Dividing: $\tan\theta=F/mg=kq^{2}/(mg r^{2})$. Therefore $q^{2}=mg r^{2}\tan\theta/k=mg(2L\sin\theta)^{2}\tan\theta\cdot 4\pi\epsilon_0=16\pi\epsilon_0 mgL^{2}\sin^{2}\theta\tan\theta$.

N7. A spherical conductor of radius $0.10$ m has surface charge density $1\times10^{-6}$ C/m². Find the electric field just outside the conductor.

Answer: Just outside a conductor $E=\sigma/\epsilon_0=10^{-6}/8.854\times10^{-12}=1.13\times10^{5}$ N/C, normal to the surface.

N8. A point charge of $+4\,\mu$C is placed at one corner of a cube of side $5$ cm. Find the flux through the cube.

Answer: A charge at a corner is shared by eight identical cubes meeting at that corner, so only $1/8$ of the total flux $q/\epsilon_0$ passes through this single cube. $\Phi=q/(8\epsilon_0)=4\times10^{-6}/(8\times 8.854\times10^{-12})=5.65\times10^{4}$ N·m²/C.

N9. A dipole of moment $1\times10^{-29}$ C·m is held perpendicular to a uniform field of $10^{5}$ N/C. Calculate the work done in rotating the dipole through $180^{\circ}$ from this position.

Answer: $W=pE(\cos\theta_1-\cos\theta_2)$ where $\theta_1=90^{\circ}$ and $\theta_2=270^{\circ}$. $W=10^{-29}\times 10^{5}\times(0-0)=0$. Alternatively expressed via potential energy $U=-pE\cos\theta$: rotating from $90^{\circ}$ to $270^{\circ}$ gives $\Delta U=-pE(\cos 270^{\circ}-\cos 90^{\circ})=0$, so net work done is zero.

N10. An infinite line of charge has $\lambda=5\times10^{-6}$ C/m. Find the electric field at a perpendicular distance of $10$ cm.

Answer: $E=\lambda/(2\pi\epsilon_0 r)=5\times10^{-6}/(2\pi\times 8.854\times10^{-12}\times 0.10)=8.99\times10^{5}$ N/C, directed radially outward.

Important Points to Remember

Two types of charges exist: positive and negative; like charges repel and unlike charges attract.

Charge is conserved, additive and quantised; the elementary charge is $e=1.6\times10^{-19}$ C.

Coulomb’s law gives $F=kq_1q_2/r^{2}$ with $k=9\times10^{9}$ N·m²/C² in vacuum.

The Coulomb force in a medium is reduced by the dielectric constant $K$.

The principle of superposition lets us add forces and fields vectorially.

Electric field $\vec{E}=\vec{F}/q_0$ in the limit $q_0\to 0$.

Field of a point charge: $E=kq/r^{2}$.

Dipole moment $\vec{p}=q\vec{d}$ points from $-q$ to $+q$.

For a short dipole, $E_{ax}=2kp/r^{3}$ and $E_{eq}=-kp/r^{3}$.

Torque on a dipole in a uniform field: $\vec{\tau}=\vec{p}\times\vec{E}$.

Net force on a dipole in a uniform field is zero; in a non-uniform field there is a net force.

Electric flux $\Phi=\vec{E}\cdot\vec{A}$; SI unit N·m²/C.

Gauss’s law: $\oint\vec{E}\cdot d\vec{A}=q_{enc}/\epsilon_0$.

Field of an infinite straight line of charge: $E=\lambda/(2\pi\epsilon_0 r)$.

Field of an infinite plane sheet of charge: $E=\sigma/(2\epsilon_0)$.

Field outside a uniformly charged spherical shell: $E=q/(4\pi\epsilon_0 r^{2})$ (acts as a point charge).

Field inside a uniformly charged spherical shell: $E=0$.

Just outside a conductor: $E=\sigma/\epsilon_0$ normal to the surface.

Conceptual Questions and Discussion

C1. Why does a charged comb attract small bits of paper that are themselves uncharged?

Answer: The charged comb produces an electric field that polarises the molecules of paper. The induced charges on the paper bits are arranged so that the unlike charge is closer to the comb than the like charge. The attractive force on the closer (unlike) charge is greater than the repulsive force on the farther (like) charge, leading to a net attraction.

C2. Why is the electric field always perpendicular to the surface of a conductor in electrostatic equilibrium?

Answer: If the field had a tangential component, free electrons in the conductor would experience a force along the surface and would continue to move, contradicting the assumption of electrostatic equilibrium. Hence the field at the surface of a conductor must be entirely normal to the surface.

C3. Why is the electric field due to a uniformly charged spherical shell zero inside it but not zero outside?

Answer: Inside the shell, every Gaussian sphere encloses no charge, so by Gauss’s law the field is zero. Outside, the Gaussian surface encloses the entire charge $q$, and by symmetry the field is the same as that of a point charge at the centre.

C4. Two identical metallic spheres carrying charges $+q$ and $-3q$ are kept some distance apart and a force $F$ acts between them. They are touched and replaced at the same distance. Find the new force.

Answer: When touched, the total charge $-2q$ distributes equally so each sphere carries $-q$. The new force is $F’=k(-q)(-q)/r^{2}=kq^{2}/r^{2}$. The original force was $F=k(q)(3q)/r^{2}=3kq^{2}/r^{2}$ (attractive). Thus $F’=F/3$ and now repulsive.

C5. A charge $Q$ is placed at the centre of a cube. What is the flux through any one face?

Answer: By symmetry the total flux $Q/\epsilon_0$ is shared equally by the six faces, hence flux through one face $=Q/(6\epsilon_0)$.

C6. Distinguish between an electric dipole and a magnetic dipole.

Answer: An electric dipole consists of two equal and opposite electric charges separated by a small distance, while a magnetic dipole consists of a current loop or two equal and opposite “magnetic poles” (which never exist in isolation). The electric dipole field falls as $1/r^{3}$, similar to a magnetic dipole field, but the underlying charges of an electric dipole are observable monopoles whereas magnetic monopoles have not been observed.

C7. Why is Gauss’s law in electrostatics a more general statement than Coulomb’s law?

Answer: Gauss’s law follows from Coulomb’s law combined with the principle of superposition, but it is also valid for time-dependent fields, where the static form of Coulomb’s law fails. It applies to any closed surface and any distribution of enclosed charge, making it a fundamental Maxwell equation.

C8. What is the difference between $\sigma/(2\epsilon_0)$ and $\sigma/\epsilon_0$?

Answer: $\sigma/(2\epsilon_0)$ is the field due to an infinite charged sheet (each side contributes $\sigma/(2\epsilon_0)$). $\sigma/\epsilon_0$ is the field just outside a charged conductor, where the field exists only on the outside (the field inside the conductor is zero), so the full $\sigma/\epsilon_0$ appears on the outer side.

C9. Two infinite plane parallel sheets carry surface densities $+\sigma$ and $-\sigma$. Find the field in the three regions: left of both, between them, and right of both.

Answer: Each sheet alone produces a field $\sigma/(2\epsilon_0)$ on either side, directed away from a positive sheet and toward a negative one. (i) Left of both: the two fields cancel; $E=0$. (ii) Between: the two fields add; $E=\sigma/\epsilon_0$, directed from $+\sigma$ to $-\sigma$. (iii) Right of both: the two fields cancel; $E=0$.

C10. The number of electric field lines through a given closed surface is doubled. What can you conclude about the enclosed charge?

Answer: By Gauss’s law, the flux is proportional to the enclosed charge. Doubling the field lines (flux) means the enclosed charge has doubled.

Frictional Electricity, Conductors and Insulators

When two bodies are rubbed together, electrons are transferred from one to the other. The body that loses electrons becomes positively charged and the body that gains electrons becomes negatively charged. This phenomenon is called frictional or tribo-electricity. The triboelectric series lists materials in order of their tendency to lose electrons. Common pairs and their resulting charges include glass and silk (glass becomes positive, silk negative), ebonite and fur (ebonite becomes negative, fur positive), and amber and wool (amber becomes negative, wool positive).

Conductors are materials that contain a large number of free charge carriers and therefore allow electric current to flow easily; metals, electrolytes, and ionised gases are conductors. Insulators (or dielectrics) lack free carriers; charges are localised and do not move easily. Wood, glass, plastic and dry air are common insulators. Semiconductors like silicon and germanium have an intermediate behaviour; their conductivity rises with temperature or doping.

Charging by induction allows a body to be charged without contact. A charged rod is brought near (but not touching) an isolated metallic sphere. The free electrons of the sphere redistribute, accumulating opposite charge on the near face and like charge on the far face. While the rod is still nearby, the sphere is connected to ground through a conducting wire; the like charge flows away to ground. The wire is removed first, then the rod, leaving the sphere with a net charge opposite to that of the rod. Induction does not deplete the inducing rod; the inducing rod retains its charge throughout.

Diagram 5: Charging by Induction

Continuous Charge Distributions: Worked Examples

W1. Find the field on the axis of a uniformly charged ring of radius $R$ carrying total charge $Q$, at a distance $x$ from its centre.

Answer: Take an element $dq$ on the ring. Its contribution at the axial point P at distance $r=\sqrt{R^{2}+x^{2}}$ from the element is $dE=k\,dq/r^{2}$. By symmetry, components perpendicular to the axis cancel. The axial component is $dE_{x}=dE\cos\theta=dE\cdot x/r$. Integrating: $E=\int dE_{x}=\dfrac{k x}{r^{3}}\int dq=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qx}{(R^{2}+x^{2})^{3/2}}$. The field is zero at the centre ($x=0$) and falls as $1/x^{2}$ for $x\gg R$.

W2. A long, thin wire of length $L$ carries a uniform linear charge density $\lambda$. Find the field at a perpendicular distance $r$ from its midpoint, in the limit $L\gg r$.

Answer: Treat the wire as effectively infinite when $L\gg r$. By Gauss’s law applied to a coaxial cylinder, $E=\lambda/(2\pi\epsilon_0 r)$, directed radially outward. The same result follows from direct integration with the limits extended to $\pm\infty$.

W3. A solid sphere of radius $R$ carries a uniform volume charge density $\rho$. Find the field at distance $r$ from the centre, both inside ($rR$).

Answer: Outside ($r>R$): the entire charge $Q=(4/3)\pi R^{3}\rho$ acts as if at the centre, giving $E=Q/(4\pi\epsilon_0 r^{2})=\rho R^{3}/(3\epsilon_0 r^{2})$. Inside ($r

This concludes the ASSEB Class 12 Physics Chapter 1 study material on Electric Charges and Fields. Continue your preparation with subsequent chapters from the same series at HSLC Guru.