"Electric Charge and Electric Field"

About

Electricity begins with charge — the fundamental property of matter that causes it to experience a force in an electromagnetic field. This chapter introduces electric charge, Coulomb's law for the force between charges, the concept of electric field, and Gauss's theorem — a powerful tool for calculating electric fields of symmetric charge distributions.

Key Concepts

15.1 Electric Charge

Charge is a fundamental property of matter. Two types: positive and negative.

Properties of charge:

• Quantisation: $q=ne$, where $e=1.6\times 10^{-19}$ C (elementary charge)
• Conservation: Total charge of an isolated system remains constant
• Additivity: Total charge = algebraic sum of individual charges
• Charge is a scalar quantity. SI unit: coulomb (C)

Charging by friction: When glass is rubbed with silk, electrons transfer from glass to silk → glass becomes positive, silk becomes negative. Charges are equal in magnitude, opposite in sign.

Charging by conduction: When a charged conductor touches an uncharged identical conductor, charge distributes equally between them.

15.2 Coulomb's Law

The force between two point charges $q_1$ and $q_2$ separated by distance $r$:

$F=k\frac{q_1{q}_2}{r^2}$

Where $k=\frac{1}{4\pi {\epsilon }_0}=9\times 10^9$ N⋅m²/C²

${\epsilon }_0=8.85\times 10^{-12}$ C²/N⋅m² (permittivity of free space)

In vector form: ${\vec{F}}_{12}=k\frac{q_1{q}_2}{r^2}{\hat{r}}_{12}$

• Like charges repel; unlike charges attract
• Obeys Newton's third law: ${\vec{F}}_{12}=-{\vec{F}}_{21}$
• Force acts along the line joining the charges (central force)

15.3 Electric Field

The electric field $\vec{E}$ at a point is the force experienced per unit positive test charge:

$\vec{E}=\frac{\vec{F}}{q_0}$

SI unit: N/C or V/m

Field due to a point charge: $E=k\frac{q}{r^2}$

Electric field lines:

• Start from positive charges, end on negative charges
• Never intersect
• Closer lines → stronger field
• Tangent to field line gives direction of $\vec{E}$

15.4 Electric Dipole

An electric dipole consists of two equal and opposite charges $\pm q$ separated by distance $2a$.

Dipole moment: $\vec{p}=q\times 2\vec{a}$

Direction: from negative to positive charge. SI unit: C⋅m

Field on axial line: $E=\frac{2kp}{r^3}$ (for $r\gg a$)

Field on equatorial line: $E=\frac{kp}{r^3}$ (for $r\gg a$)

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

15.5 Electric Flux

${\varphi }_E=\vec{E}\cdot \vec{A}=EA\cos \theta$

SI unit: N⋅m²/C

15.6 Gauss's Theorem

The total electric flux through any closed surface equals $\frac{1}{{\epsilon }_0}$ times the net charge enclosed.

$\oint \vec{E}\cdot d\vec{A}=\frac{q_{\text{enclosed}}}{{\epsilon }_0}$

Applications:

1. Field due to infinitely long charged wire: $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$
2. Field due to infinite charged sheet: $E=\frac{\sigma }{2{\epsilon }_0}$
3. Field due to charged spherical shell: $E=k\frac{q}{r^2}$ (outside), $E=0$ (inside)

INTEXT QUESTIONS 15.1

Q1. A glass rod when rubbed with silk cloth acquires a charge q = +3.2 × 10⁻¹⁷ C. (i) Is silk cloth also charged? (ii) What is the nature and magnitude of the charge on silk cloth?

Ans: (i) Yes, silk cloth is also charged. When glass is rubbed with silk, electrons transfer from glass to silk. By conservation of charge, both acquire equal and opposite charges.

(ii) Nature: Negative. Magnitude: 3.2 × 10⁻¹⁷ C (equal to the charge on glass).

Q2. There are two identical metallic spheres A and B. A is given a charge +Q. Both spheres are then brought in contact and then separated. (i) Will there be any charge on B? (ii) What will the magnitude of charge on B?

Ans: (i) Yes, there will be charge on B. When identical conductors touch, charge distributes equally due to electrostatic repulsion.

(ii) Magnitude of charge on B = +Q/2. Total charge +Q distributes equally between the two identical spheres.

Q3. A charged object has q = 4.8 × 10⁻¹⁶ C. How many units of fundamental charge are there on the object? (Take e = 1.6 × 10⁻¹⁹ C)

Ans: $N=\frac{q}{e}=\frac{4.8\times 10^{-16}}{1.6\times 10^{-19}}=3.0\times 10^3=3000$

3000 units of fundamental charge.

INTEXT QUESTIONS 15.2

Q1. Two charges q₁ = 16 μC and q₂ = 9 μC are separated by a distance 12 m. Determine the magnitude of the force experienced by q₁ due to q₂ and also the direction of this force. What is the direction of the force experienced by q₂ due to q₁?

Ans: $F=k\frac{q_1{q}_2}{r^2}=\frac{9\times 10^9\times 16\times 10^{-6}\times 9\times 10^{-6}}{12^2}$ $=\frac{9\times 10^9\times 144\times 10^{-12}}{144}=9\times 10^{-3}\text{ N}=0.009\text{ N}$

Both charges are positive → force is repulsive. Force on q₁ is directed away from q₂. By Newton's third law, force on q₂ is also repulsive, directed away from q₁.

Q2. Three point charges of equal magnitude q are placed at three corners of a right angle triangle. AB = AC. What is the magnitude and direction of the force exerted on –q?

Ans: Charges: A (+q), B (+q), C (−q). AB = AC = a, angle A = 90°.

Force on C due to A ($F_1$): attractive, along CA, $F_1=k{q}^2/a^2$

Force on C due to B ($F_2$): attractive, along CB, $F_2=k{q}^2/a^2$

$F_1$ and $F_2$ are perpendicular (angle between CA and CB = 90°).

Resultant: $F=\sqrt{F_1^2+F_2^2}=\sqrt{2}\times \frac{k{q}^2}{a^2}$

Direction: Toward point A, making 45° with both CA and CB.

Terminal Exercise

1. State the properties of electric charge. Explain quantisation and conservation of charge.

2. State Coulomb's law. Write its vector form and explain.

3. Compare Coulomb's law with Newton's law of gravitation. State two similarities and two differences.

4. Define electric field and electric field intensity. Derive the expression for the electric field due to a point charge.

5. Sketch electric field lines for: (a) an isolated positive charge, (b) an isolated negative charge, (c) an electric dipole, (d) two equal positive charges.

6. Define electric dipole and dipole moment. Derive the expression for the electric field at a point on the axial line of a dipole.

7. Define electric flux. State and prove Gauss's theorem.

8. Using Gauss's theorem, derive the expression for electric field due to: (a) an infinitely long charged wire, (b) an infinite plane sheet of charge.

9. A point charge of 2 μC is placed at the centre of a cube of side 10 cm. Find the electric flux through: (a) the entire cube, (b) one face of the cube.

10. Two point charges of +3 μC and −3 μC are placed 20 cm apart. Find the electric field at a point 15 cm from the centre of the dipole on: (a) the axial line, (b) the equatorial line.

11. A small ball of mass 0.1 g carries a charge of 5 × 10⁻⁸ C. What electric field must be applied to balance the weight of the ball? (g = 10 m/s²)

12. A proton and an electron are placed in a uniform electric field. Compare: (a) the forces experienced, (b) the accelerations produced.

Worked Examples

Example 1: Coulomb's Law

Problem: Two charges of 5 μC and −2 μC are 3 m apart. Find the force between them.

Solution: $F=9\times 10^9\times \frac{5\times 10^{-6}\times 2\times 10^{-6}}{3^2}$ $=9\times 10^9\times \frac{10\times 10^{-12}}{9}=0.01\text{ N (attractive)}$

Example 2: Electric Field

Problem: Find the electric field 2 m away from a point charge of 4 μC.

Solution: $E=k\frac{q}{r^2}=9\times 10^9\times \frac{4\times 10^{-6}}{4}=9\times 10^3\text{ N/C}$

Example 3: Gauss's Theorem

Problem: An infinite line charge has linear charge density $\lambda =2\times 10^{-6}$ C/m. Find E at 0.5 m from the line.

Solution: $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{2\times 10^{-6}}{2\pi \times 8.85\times 10^{-12}\times 0.5}$ $=\frac{2\times 10^{-6}}{2.78\times 10^{-11}}\approx 7.2\times 10^4\text{ N/C}$

Common Mistakes

1. Forgetting that charge is quantised: $q$ is always an integer multiple of $e=1.6\times 10^{-19}$ C.
2. Using Coulomb's law without the sign of charges for direction: Use signs only for magnitude; determine direction from whether charges attract or repel.
3. Confusing electric field with electric force: $E=F/q$ — field exists independent of the test charge.
4. Applying Gauss's law without proper symmetry: Choose Gaussian surface matching the symmetry of the charge distribution.
5. Forgetting that field inside a conductor is zero in electrostatic equilibrium.

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