Electricity is one of the most fascinating branches of physics. The study of electric charges, their properties, and the fields they create is fundamental to understanding modern technology and natural phenomena. In Class 12 Physics, the chapter Electric Charges and Fields introduces students to the concept of charges, Coulomb’s law, electric fields, flux, Gauss’s law, and more.
In this article, we will cover every topic of this chapter in detail with clear explanations, definitions, formulas, and examples.
Class 12 Physics: Electric Charges and Fields Notes | Complete Guide with FAQs
Electric Charges
Definition of Electric Charge
Electric charge is a fundamental property of matter due to which it experiences a force when placed in an electric or magnetic field. There are two types of charges:
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Positive Charge: Carried by protons.
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Negative Charge: Carried by electrons.
Charge is represented by q and its SI unit is Coulomb (C).
Properties of Electric Charges
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Additivity of Charge: Total charge on a system is the algebraic sum of all charges.
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Charge is Conserved: The net charge in an isolated system remains constant.
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Charge is Quantized: Charge exists in integral multiples of the elementary charge e = 1.6 × 10⁻¹⁹ C.
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Charge is Invariant: It does not change with the motion of the body.
Conductors and Insulators
Conductors
Materials that allow the flow of electric charge easily.
Examples: Metals like copper, aluminum.
Insulators
Materials that do not allow the flow of electric charges.
Examples: Rubber, glass, plastic.
Coulomb’s Law
Statement
Coulomb’s law gives the force between two point charges. It states:
The electrostatic force between two stationary point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Mathematically,
F=kq1q2r2F = k \frac{q_1 q_2}{r^2}
Where:
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F = Electrostatic force
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q₁, q₂ = Charges
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r = Distance between charges
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k = Coulomb’s constant 14πϵ0\frac{1}{4 \pi \epsilon_0}
Principle of Superposition
The total force on a charge due to a number of charges is the vector sum of the individual forces exerted by each charge.
Electric Field
Definition
The region around a charged object where its effect can be felt is called the electric field.
Mathematically,
E⃗=F⃗q\vec{E} = \frac{\vec{F}}{q}
Properties of Electric Field
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Electric field is a vector quantity.
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Direction of E is along the force experienced by a positive test charge.
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Superposition principle applies.
Electric Field Lines
Characteristics of Electric Field Lines
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They start from positive charges and end on negative charges.
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They never intersect each other.
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The density of lines represents the strength of the field.
Electric Dipole
Definition
A pair of equal and opposite charges separated by a small distance is called an electric dipole.
Dipole Moment
p⃗=q×2l\vec{p} = q \times 2l
Where 2l is the distance between charges.
Torque on an Electric Dipole
When a dipole is placed in a uniform electric field, it experiences a torque given by
τ=pEsinθ\tau = pE \sin \theta
Electric Flux
Definition
Electric flux measures the total number of electric field lines passing through a surface.
ϕ=E⃗⋅A⃗\phi = \vec{E} \cdot \vec{A}
Gauss’s Law
Statement
The total electric flux through a closed surface is equal to 1ϵ0\frac{1}{\epsilon_0} times the charge enclosed by the surface.
ϕ=qϵ0\phi = \frac{q}{\epsilon_0}
Applications of Gauss’s Law
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Electric field due to a point charge
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Electric field due to a line charge
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Electric field due to a plane sheet
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Electric field due to a spherical shell
Electric Field due to Various Charge Distributions
Point Charge
E=14πϵ0qr2E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}
Infinite Line of Charge
E=λ2πϵ0rE = \frac{\lambda}{2 \pi \epsilon_0 r}
Where λ = linear charge density.
Uniformly Charged Sphere
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Outside Sphere: E=14πϵ0qr2E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}
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Inside Sphere: E=0E = 0
Electrostatic Potential and Potential Energy
Electrostatic Potential
The work done per unit charge in bringing a test charge from infinity to a point in an electric field.
V=WqV = \frac{W}{q}
Potential Energy of a System of Charges
For two charges:
U=14πϵ0q1q2rU = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r}
Equipotential Surfaces
Surfaces where the potential remains the same at every point.
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Electric field is always perpendicular to equipotential surfaces.
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Work done along an equipotential surface is zero.
Relation Between Electric Field and Potential
E=−dVdrE = – \frac{dV}{dr}
Capacitance and Capacitors (Brief Introduction)
Capacitance is the ability of a system to store charge per unit potential difference.
C=QVC = \frac{Q}{V}
Summary Table for Quick Revision
| Concept | Formula | SI Unit |
|---|---|---|
| Coulomb’s Law | F=kq1q2r2F = k \frac{q_1 q_2}{r^2} | Newton (N) |
| Electric Field | E=FqE = \frac{F}{q} | N/C |
| Electric Potential | V=WqV = \frac{W}{q} | Volt (V) |
| Electric Flux | ϕ=E⃗⋅A⃗\phi = \vec{E} \cdot \vec{A} | N·m²/C |
| Torque on Dipole | τ=pEsinθ\tau = pE \sin \theta | N·m |
Objective Questions-Electric Charges and Fields
Q1. Which of the following is the SI unit of electric charge?
(a) Volt
(b) Ampere
(c) Coulomb
(d) Ohm
Answer: The SI unit of electric charge is Coulomb (C).
Q2. Electric charges are of how many types?
(a) One
(b) Two
(c) Three
(d) Four
Answer: Electric charges are of two types – positive and negative.
Q3. Which law governs the force between two point charges?
(a) Ohm’s Law
(b) Coulomb’s Law
(c) Faraday’s Law
(d) Gauss’s Law
Answer: The force between two point charges is governed by Coulomb’s Law.
Q4. What is the value of Coulomb’s constant kk?
(a) 9×109 Nm2/C29 × 10^9 \, Nm^2/C^2
(b) 3×108 m/s3 × 10^8 \, m/s
(c) 6.67×10−11 Nm2/kg26.67 × 10^{-11} \, Nm^2/kg^2
(d) 1.6×10−19 C1.6 × 10^{-19} \, C
Answer: The value of Coulomb’s constant is 9×109 Nm2/C29 × 10^9 \, Nm^2/C^2.
Q5. What is the SI unit of electric field?
(a) Newton
(b) N/C
(c) Coulomb
(d) Ampere
Answer: The SI unit of electric field is N/C (Newton per Coulomb).
Q6. Electric field lines originate from which type of charge?
(a) Negative
(b) Neutral
(c) Positive
(d) Both Positive & Negative
Answer: Electric field lines originate from positive charges.
Q7. What is the formula for electric flux?
(a) ϕ=E⋅A\phi = E \cdot A
(b) ϕ=qϵ0\phi = \frac{q}{\epsilon_0}
(c) ϕ=q⋅E\phi = q \cdot E
(d) ϕ=14πϵ0\phi = \frac{1}{4 \pi \epsilon_0}
Answer: Electric flux formula is ϕ=E⋅A\phi = E \cdot A.
Q8. In an electric dipole, the charges are:
(a) Equal & Opposite
(b) Equal & Same Sign
(c) Unequal & Opposite
(d) Unequal & Same Sign
Answer: In an electric dipole, the charges are equal and opposite.
Q9. What is the SI unit of potential difference?
(a) Ampere
(b) Volt
(c) Newton
(d) Joule
Answer: The SI unit of potential difference is Volt (V).
Q10. Gauss’s law relates electric flux to:
(a) Current
(b) Charge
(c) Resistance
(d) Magnetic Field
Answer: Gauss’s law relates electric flux to charge.
Q11. The electric field inside a conductor is:
(a) Infinite
(b) Zero
(c) Maximum
(d) Minimum
Answer: The electric field inside a conductor is zero.
Q12. Electric potential energy between two charges is directly proportional to:
(a) Distance
(b) Inverse of Distance
(c) Product of Charges
(d) Square of Charges
Answer: Electric potential energy is directly proportional to the product of charges.
Q13. The work done in moving a charge on an equipotential surface is:
(a) Maximum
(b) Zero
(c) Infinite
(d) Variable
Answer: The work done is zero.
Q14. The torque on an electric dipole in a uniform electric field is given by:
(a) pEsinθpE \sin \theta
(b) pEcosθpE \cos \theta
(c) qEsinθqE \sin \theta
(d) qEcosθqE \cos \theta
Answer: Torque on an electric dipole is pEsinθpE \sin \theta.
Q15. Electric field is the gradient of:
(a) Potential
(b) Charge
(c) Force
(d) Flux
Answer: Electric field is the gradient of potential.
Q16. A closed surface enclosing no charge will have electric flux:
(a) Zero
(b) Infinite
(c) Maximum
(d) Constant
Answer: Electric flux will be zero.
Q17. What is the direction of electric field lines?
(a) Negative to Positive
(b) Positive to Negative
(c) Circular
(d) Random
Answer: Electric field lines go from positive to negative.
Q18. SI unit of electric flux is:
(a) N/C
(b) N·m²/C
(c) V/m
(d) C/m²
Answer: SI unit of electric flux is N·m²/C.
Q19. Principle of superposition applies to:
(a) Only one charge
(b) Two charges
(c) Multiple charges
(d) No charges
Answer: Principle of superposition applies to multiple charges.
Q20. Which of the following is not a property of electric charges?
(a) Additivity
(b) Conservation
(c) Quantization
(d) Amplification
Answer: Amplification is not a property of electric charges.
Short Answer Questions-Electric Charges and Fields
Q1. Define electric charge and state its properties.
Answer: Electric charge is a fundamental property of matter responsible for electric forces. Its properties include additivity, conservation, quantization, and invariance.
Q2. State and explain Coulomb’s law.
Answer: Coulomb’s law states that the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Q3. Write two properties of electric field lines.
Answer: Electric field lines start from positive charges and end on negative charges, and they never intersect each other.
Q4. What is electric flux? Write its formula.
Answer: Electric flux measures the number of electric field lines passing through a surface and is given by ϕ=E⃗⋅A⃗\phi = \vec{E} \cdot \vec{A}.
Q5. State Gauss’s law.
Answer: Gauss’s law states that the total electric flux through a closed surface equals 1ϵ0\frac{1}{\epsilon_0} times the charge enclosed within it.
Long Answer Questions-Electric Charges and Fields
Q1. Derive the expression for electric field due to a point charge.
Answer: Consider a point charge qq at a distance rr. The electric field at any point is given by Coulomb’s law as E=14πϵ0qr2E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}. The direction is along the line joining the charge and the point.
Q2. Explain the concept of an electric dipole and derive the torque on it in a uniform electric field.
Answer: An electric dipole consists of two equal and opposite charges separated by a distance. When placed in a uniform electric field, it experiences a torque τ=pEsinθ\tau = pE \sin \theta which tends to align the dipole with the field.
Q3. Derive the relation between electric field and electric potential.
Answer: The electric field is related to the potential difference as E=−dVdrE = -\frac{dV}{dr}. It shows that the electric field is the negative gradient of potential.
Q4. Using Gauss’s law, derive the expression for the electric field due to a uniformly charged spherical shell.
Answer: For a point outside the shell, the entire charge is considered at the center, giving E=14πϵ0qr2E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}. For a point inside, electric field is zero as total flux inside is zero.
Q5. Explain equipotential surfaces and their properties.
Answer: Equipotential surfaces have the same electric potential at all points. Electric field is always perpendicular to these surfaces, and no work is done in moving a charge along the surface.
Conclusion
The chapter Electric Charges and Fields builds the foundation for understanding electricity and electrostatics in Physics. Concepts like Coulomb’s Law, Electric Field, Gauss’s Law, and Potential not only help in board exams but also play a vital role in competitive exams like JEE and NEET.
We are Completed Electric Charges and Fields Notes.
Electrostatic Potential and Capacitance Class 12 Physics Notes | Detailed Explanation