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Electric Charges and Fields

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Chapter Notes

Overview
This chapter introduces the fundamental concept of electric charge, the most basic property of matter that governs all electrical phenomena. It explains how charges are produced, their basic properties (additivity, conservation, and quantisation), and the quantitative law (Coulomb's law) that describes the force between them. The central idea is that a charge creates an electric field in the space around it, and this field exerts a force on any other charge placed within it, forming the bedrock for understanding all of electrostatics.
Key Concepts (17)
Electric Charge
An intrinsic property of matter that gives rise to electric forces and interactions. It exists in two types, conventionally called positive and negative.
e.g. Rubbing a plastic comb with dry hair transfers charge to the comb, allowing it to attract small pieces of paper.
Electrification
The process by which a neutral body acquires a net electric charge, typically through rubbing (friction).
e.g. A glass rod rubbed with silk becomes electrified and can attract light objects like pith balls.
Like and Unlike Charges
Like charges (both positive or both negative) repel each other, while unlike charges (one positive and one negative) attract each other.
e.g. Two glass rods rubbed with silk repel each other, but a glass rod and the silk it was rubbed with attract each other.
Gold-leaf Electroscope
A simple apparatus used to detect the presence and approximate magnitude of electric charge on a body, consisting of a metal rod with two thin gold leaves inside a box.
e.g. When a charged rod touches the metal knob, charge flows to the gold leaves, causing them to diverge due to repulsion.
Conductors
Substances that allow electricity to pass through them easily because they have free electrons that can move within the material. Examples: metals, human body, earth.
e.g. A metal spoon does not get electrified when rubbed because charges leak through our body to the ground.
Insulators
Substances that offer high resistance to the passage of electricity because their electrons are tightly bound. Examples: glass, plastic, nylon, wood.
e.g. A nylon comb gets electrified on combing dry hair because the charge stays at the place it was generated.
Additivity of Charges
The total charge of a system is the algebraic sum of all the individual charges present, taking their signs into account.
e.g. A system with charges +1, +2, -3, +4, and -5 units has a total charge of -1 unit.
Conservation of Charge
The total charge of an isolated system is always conserved; charge can neither be created nor destroyed, only transferred from one body to another.
e.g. When a glass rod is rubbed with silk, the rod loses electrons (becomes positive) and the silk gains the same number of electrons (becomes negative); total charge remains zero.
Quantisation of Charge
All observable charges are integral multiples of the elementary charge 'e', i.e., q = ne, where n is an integer (positive or negative).
e.g. A charge of 3.2 × 10⁻¹⁹ C is possible (n=2), but a charge of 2.5 × 10⁻¹⁹ C is not possible as it is not an integral multiple of e (1.6 × 10⁻¹⁹ C).
Elementary Charge (e)
The smallest unit of free charge, equal to the magnitude of charge on an electron or a proton. Its value is 1.602192 × 10⁻¹⁹ C.
e.g. The charge on an electron is -e, and the charge on a proton is +e.
Coulomb (C)
The SI unit of electric charge. One coulomb is the charge flowing through a wire in 1 second if the current is 1 ampere.
e.g. A charge of 1 C is a very large unit, equivalent to the charge on approximately 6 × 10¹⁸ electrons.
Point Charge
A charged body whose size is very small compared to the distance from it to other charges, allowing its entire charge to be considered as concentrated at a single point.
e.g. Two charged spheres with radii of 1 cm separated by a distance of 1 m can be treated as point charges.
Coulomb's Law
The electrostatic force between two point charges 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 them.
e.g. If the distance between two charges is doubled, the force between them becomes one-fourth of its original value.
Permittivity of Free Space (ε₀)
A physical constant that determines the strength of the electric force in a vacuum. Its value is 8.854 × 10⁻¹² C² N⁻¹ m⁻².
e.g. It appears in the denominator of Coulomb's law: F = (1/4πε₀) * (q₁q₂/r²).
Principle of Superposition
The total force on a given charge due to a system of charges is the vector sum of the individual forces exerted on it by each of the other charges, taken one at a time.
e.g. The force on charge q₁ due to q₂ and q₃ is F₁ = F₁₂ + F₁₃, where F₁₂ and F₁₃ are calculated using Coulomb's law independently.
Electric Field
The electric field at a point in space is defined as the force experienced by a unit positive test charge placed at that point, without disturbing the source charges. It is a vector quantity.
e.g. A positive charge Q creates an electric field that points radially outward. A small positive test charge placed near Q will experience a force away from Q.
Source Charge and Test Charge
The source charge is the charge that produces the electric field. The test charge is a negligibly small positive charge used to detect and measure the electric field without disturbing the source charge.
e.g. To find the electric field due to a large charged sphere (source), we use a very small positive charge (test charge) to probe the field.
Formulas & Equations (7)
F = k * (|q₁q₂|) / r²
Reads as: Force equals k times the product of the magnitudes of the two charges divided by the square of the distance between them.
Variables: F = electrostatic force (N), k = Coulomb's constant (9 × 10⁹ N m² C⁻²), q₁, q₂ = magnitudes of charges (C), r = distance between charges (m).
Use when: Calculating the magnitude of the electrostatic force between two point charges.
⚠ Forgetting to take the absolute values of the charges when calculating the magnitude of the force, or incorrectly applying the sign in the vector form.
F = (1/4πε₀) * (q₁q₂/r²)
Reads as: Force equals one over four pi epsilon-naught times the product of the two charges divided by the square of the distance between them.
Variables: F = electrostatic force (N), ε₀ = permittivity of free space (8.854 × 10⁻¹² C² N⁻¹ m⁻²), q₁, q₂ = charges (C), r = distance between charges (m).
Use when: This is the standard form of Coulomb's law used in calculations, especially when ε₀ is given or when deriving other formulas.
⚠ Using the value of 'k' (9 × 10⁹) when the formula with ε₀ is written, leading to a factor of 4π error.
F₁₂ = (1/4πε₀) * (q₁q₂/r₁₂²) * r̂₁₂
Reads as: Force on charge 1 due to charge 2 equals one over four pi epsilon-naught times the product of the charges divided by the square of the distance between them, multiplied by the unit vector from charge 2 to charge 1.
Variables: F₁₂ = force vector on q₁ due to q₂ (N), r₁₂ = distance between q₁ and q₂ (m), r̂₁₂ = unit vector pointing from q₂ to q₁.
Use when: Writing Coulomb's law in vector form to correctly determine the direction of the force (repulsion or attraction).
⚠ Confusing the direction of the unit vector r̂₁₂ and r̂₂₁, leading to an incorrect direction for the force vector.
q = ne
Reads as: Charge q equals n times e.
Variables: q = total charge on a body (C), n = an integer (positive or negative), e = elementary charge (1.6 × 10⁻¹⁹ C).
Use when: Solving numerical problems based on the quantisation of charge, e.g., finding the number of electrons transferred.
⚠ Forgetting that 'n' must be an integer; a fractional value of 'n' is not physically possible.
E = F / q₀
Reads as: Electric field equals force per unit test charge.
Variables: E = electric field (N/C or V/m), F = electrostatic force experienced by the test charge (N), q₀ = magnitude of the test charge (C).
Use when: Defining the electric field at a point, or calculating the force on a charge placed in a known electric field (F = qE).
⚠ Using the source charge instead of the test charge in the denominator, or forgetting that E is a vector and its direction is the same as the force on a positive test charge.
E = (1/4πε₀) * (Q/r²) * r̂
Reads as: Electric field due to a point charge Q equals one over four pi epsilon-naught times the charge divided by the square of the distance, in the direction of the unit vector r̂.
Variables: E = electric field (N/C), Q = source charge (C), r = distance from the source charge (m), r̂ = unit vector pointing from the source charge to the point.
Use when: Calculating the electric field at a point in space due to a single point charge.
⚠ Forgetting the direction: for a positive Q, E is radially outward; for a negative Q, E is radially inward.
E(r) = (1/4πε₀) * Σ (qᵢ / rᵢP²) * r̂ᵢP
Reads as: Electric field at point P equals one over four pi epsilon-naught times the sum over all charges of the charge divided by the square of the distance from that charge to P, times the unit vector from that charge to P.
Variables: E(r) = net electric field at point P (N/C), qᵢ = i-th source charge (C), rᵢP = distance from qᵢ to P (m), r̂ᵢP = unit vector from qᵢ to P.
Use when: Calculating the net electric field at a point due to a system of multiple point charges using the superposition principle.
⚠ Performing a scalar sum instead of a vector sum; the directions of individual field vectors must be considered before adding them.
How It Works
Charging by Rubbing (Friction)
  1. Step 1: Two different insulating materials (e.g., glass rod and silk) are rubbed together.
  2. Step 2: Due to friction, electrons are transferred from one material to the other. The material that loses electrons becomes positively charged, and the material that gains electrons becomes negatively charged.
  3. Step 3: The total charge of the system (rod + silk) remains zero, demonstrating the conservation of charge.
💡 Only electrons move during charging by rubbing; protons do not move. The material with a higher electron affinity gains electrons.
Using a Gold-leaf Electroscope
  1. Step 1: A charged object is brought into contact with the metal knob at the top of the electroscope.
  2. Step 2: The charge flows down the metal rod to the two thin gold leaves at the bottom.
  3. Step 3: Since both leaves acquire the same type of charge (both positive or both negative), they repel each other and diverge.
  4. Step 4: The degree of divergence is an indicator of the amount of charge on the object.
💡 The electroscope detects charge, not its sign. To determine the sign, a known charge is brought near the knob.
Coulomb's Torsion Balance Experiment
  1. Step 1: Coulomb used a torsion balance to measure the force between two charged spheres.
  2. Step 2: He charged a fixed sphere and a movable sphere. The force between them caused the movable sphere to twist a fiber.
  3. Step 3: He varied the distance between the spheres and measured the force (by the angle of twist).
  4. Step 4: To vary the charge, he used a method of 'charging by contact': touching a charged sphere to an identical uncharged sphere, which splits the charge equally (q, q/2, q/4, etc.).
  5. Step 5: By systematically varying charge and distance, he established the inverse-square law.
💡 Coulomb did not know the absolute value of charge; he worked with relative charges (q, q/2, q/4).
Calculating Net Force using Superposition Principle
  1. Step 1: Identify all charges in the system and the charge on which the net force is to be calculated.
  2. Step 2: Calculate the force on the chosen charge due to each other charge individually using Coulomb's law in vector form.
  3. Step 3: Represent each force as a vector with the correct direction (repulsive for like charges, attractive for unlike charges).
  4. Step 4: Add all these force vectors together using vector addition (parallelogram law or component method) to find the net force.
💡 The presence of other charges does not affect the pair-wise force between any two charges. Each force is calculated independently.
Mapping an Electric Field
  1. Step 1: Place a very small positive test charge (q₀) at a point in space around a source charge system.
  2. Step 2: Measure the electrostatic force (F) experienced by the test charge.
  3. Step 3: The electric field at that point is defined as E = F / q₀.
  4. Step 4: Repeat this process at different points to map the electric field vector at each point in space.
💡 The test charge must be negligibly small so that it does not disturb the original charge distribution of the source.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Charge Transfer with Creation
✗ When a glass rod is rubbed with silk, positive charge is created on the rod and negative charge is created on the silk.
→ Charge is never created or destroyed; it is only transferred. The rod loses electrons to become positive, and the silk gains those electrons to become negative.
✓ When a glass rod is rubbed with silk, electrons are transferred from the glass rod to the silk. The rod becomes positively charged due to a deficit of electrons, and the silk becomes negatively charged due to an excess of electrons.
⚠ Misapplying Quantisation to Macroscopic Charges
✗ A charge of 2.5 × 10⁻⁶ C is not possible because it is not an integral multiple of e.
→ While the statement is true for microscopic charges, at the macroscopic level (e.g., microcoulombs), the charge is an enormous multiple of e (~10¹³ times). The quantisation is not practically observable, and such a charge is perfectly possible.
✓ At the macroscopic level, the quantisation of charge has no practical consequence and can be ignored. A charge of 2.5 × 10⁻⁶ C is possible because it contains an extremely large number of elementary charges.
⚠ Incorrect Direction of Electric Field
✗ The electric field due to a negative point charge points radially outward.
→ The electric field direction is defined as the direction of force on a positive test charge. A negative source charge attracts a positive test charge, so the force (and thus the field) is radially inward.
✓ The electric field due to a negative point charge points radially inward, towards the charge.
⚠ Forgetting Vector Nature of Superposition
✗ The net electric field at a point is the sum of the magnitudes of the fields due to individual charges.
→ Electric field is a vector quantity. The net field is the vector sum, not the scalar sum. Directions must be considered.
✓ The net electric field at a point due to a system of charges is the vector sum of the electric fields due to the individual charges. For example, if two fields are in opposite directions, they may cancel out partially or completely.
⚠ Misinterpreting Coulomb's Law Constant
✗ The value of k in Coulomb's law is 9 × 10⁹ N m²/C², and it is a universal constant.
→ The value 9 × 10⁹ is an approximation for 1/4πε₀ in vacuum. The constant 'k' is not universal; it depends on the medium. In a medium, the permittivity changes, and the force (and thus the effective 'k') changes.
✓ In vacuum, the constant in Coulomb's law is k = 1/4πε₀ ≈ 9 × 10⁹ N m²/C². This value changes if the charges are placed in a different medium.
Compare & Contrast
Nature of Force
Coulomb's Law (Electrostatic Force)
Newton's Law of Gravitation
Coulomb's law can be either attractive or repulsive depending on the signs of the charges, whereas Newton's law of gravitation is always attractive.
Movement of Charge
Conductors
Insulators
In conductors, charge readily distributes over the entire surface because free electrons can move. In insulators, charge stays at the same place where it is deposited because electrons are tightly bound.
Dependence on Test Charge
Electric Field (E)
Electrostatic Force (F)
The electric field E at a point is a property of the source charges and is independent of the test charge. The electrostatic force F on a charge placed at that point depends on both the field and the magnitude of the charge (F = qE).
Exam Focus
Marks pattern: 1 MCQ (1 mark) on properties of charge or Coulomb's law, 1 SA-I (2 marks) on a simple numerical or definition, 1 SA-II (3 marks) on superposition principle or electric field calculation, 1 LA (5 marks) on a complex numerical involving forces/fields from multiple charges.
Most asked: Properties of charge (additivity, conservation, quantisation) · Coulomb's law and its vector form · Principle of superposition for forces and fields · Electric field due to a point charge and a system of charges · Comparison of electrostatic and gravitational forces · Numericals on finding net force or net electric field
Don't skip: Numerical problems on finding net force or net electric field at a point due to a system of 3 or more charges (e.g., at the centroid of a triangle).; Derivation or application of the vector form of Coulomb's law.; Problems involving the quantisation of charge (q = ne).
One-Line Revision (read aloud before the exam)
  1. Electric charge is an intrinsic property of matter that exists in two types: positive and negative.
  2. Like charges repel each other, and unlike charges attract each other.
  3. Charge is a scalar quantity and is additive, conserved, and quantised.
  4. The total charge of an isolated system always remains constant.
  5. All observable charges are integral multiples of the elementary charge e = 1.6 × 10⁻¹⁹ C.
  6. Coulomb's law states that the electrostatic force between two point charges is inversely proportional to the square of the distance between them.
  7. The SI unit of charge is the coulomb (C).
  8. The force between two charges is a vector, acting along the line joining them.
  9. The principle of superposition allows us to calculate the net force on a charge due to multiple other charges by vector addition.
  10. The electric field E at a point is the force per unit positive test charge placed at that point.
  11. The electric field due to a positive point charge is radially outward, and due to a negative point charge is radially inward.
  12. Conductors have free electrons, while insulators do not.
  13. The gold-leaf electroscope is used to detect the presence of charge.
  14. Coulomb's law is analogous to Newton's law of gravitation but with both attractive and repulsive forces.
  15. The value of 1/4πε₀ is approximately 9 × 10⁹ N m² C⁻².

MCQs — 10 Questions with Answers

Q1. An electric dipole is formed by charges +q and −q separated by distance 2a, giving dipole moment p = q × 2a. If the charge is doubled to 2q and the separation is halved to a, what happens to the dipole moment?

  • A. It doubles to 2p
  • B. It remains unchanged at p ✓
  • C. It becomes half, p/2
  • D. It becomes four times, 4p

Answer: B — New dipole moment = (2q) × a = 2qa = q × 2a = p. Doubling the charge and halving the separation cancel out exactly, so the dipole moment is unchanged.

Q2. A neutral copper rod is rubbed with a cloth and acquires a charge of +5 μC. How much charge does the cloth acquire?

  • A. +5 μC
  • B. −5 μC ✓
  • C. 0 μC
  • D. +10 μC

Answer: B — By conservation of charge, if the rod loses 5 μC of electrons (becomes +5 μC), the cloth gains those electrons (−5 μC).

Q3. A point charge Q is enclosed inside a spherical Gaussian surface, and the total electric flux through it is φ. An identical charge Q is now placed just outside the sphere. What is the new total electric flux through the same spherical surface?

  • A. 2φ, because the total charge nearby has doubled
  • B. φ/2, because the outside charge partially cancels the flux
  • C. φ, because only the charge enclosed within the surface determines the flux ✓
  • D. 0, because the two charges form a dipole and cancel each other

Answer: C — Gauss's law states that total electric flux = q_enclosed / ε₀. Only the charge inside the Gaussian surface counts; the charge placed outside contributes zero net flux through the closed surface.

Q4. A plastic rod is rubbed with cat fur and becomes negatively charged. What type of charge does the cat fur acquire?

  • A. Negative charge
  • B. Positive charge ✓
  • C. No charge (remains neutral)
  • D. Both positive and negative charges

Answer: B — Unlike charges are created when dissimilar materials are rubbed; if plastic becomes negative, fur becomes positive by conservation of charge.

Q5. Two charged pith balls, each with charge q, are separated by distance r. If the separation is reduced to r/2, how does the repulsive force change?

  • A. Becomes 1/4 of original
  • B. Becomes 1/2 of original
  • C. Becomes 2 times original
  • D. Becomes 4 times original ✓

Answer: D — Since force is inversely proportional to distance squared (F ∝ 1/r²), reducing distance to r/2 increases force by factor of (r/(r/2))² = 4.

Q6. Which statement is NOT correct about electric charge?

  • A. Charge is quantized in units of electron charge
  • B. Charge can be created by rubbing two insulators ✓
  • C. Like charges always repel each other
  • D. Total charge in an isolated system is conserved

Answer: B — Rubbing does not create charge; it only transfers electrons from one material to another — total charge before and after remains unchanged.

Q7. A gold-leaf electroscope shows maximum divergence when a charged object is brought near it. What does this indicate? (Assertion–Reason style) Assertion (A): The electroscope has detected a large amount of charge. Reason (R): The degree of leaf divergence is proportional to the magnitude of charge.

  • A. Both A and R are true, and R is the correct explanation of A ✓
  • B. Both A and R are true, but R is not the explanation of A
  • C. A is true but R is false
  • D. Both A and R are false

Answer: A — Maximum divergence directly indicates large charge because the repulsive force between leaves increases with charge magnitude, causing greater separation.

Q8. A system contains three point charges: q₁ = +2 C, q₂ = −3 C, and q₃ = +5 C. What is the net charge of the system?

  • A. +4 C ✓
  • B. +10 C
  • C. −4 C
  • D. +2 C

Answer: A — By additivity principle, net charge = q₁ + q₂ + q₃ = (+2) + (−3) + (+5) = +4 C.

Q9. A metal rod with a wooden handle is rubbed vigorously. Why does the metal rod get charged while a metal spoon placed on a table does not get charged when rubbed?

  • A. Metal is a conductor; the wooden handle insulates it from ground ✓
  • B. Metal spoon is too large to acquire charge
  • C. Friction force is not applied to the metal spoon
  • D. The metal rod has more electrons than the metal spoon

Answer: A — The wooden handle breaks electrical contact with Earth; charge cannot leak away, so charge accumulates on the metal rod — but the spoon on table contacts ground and leaks all charge.

Q10. Two identical conducting spheres, one with charge +6 μC and the other with charge −4 μC, are touched together and then separated. What is the final charge on each sphere? (HOTS: requires understanding charge redistribution)

  • A. +6 μC and −4 μC (unchanged)
  • B. +1 μC and +1 μC ✓
  • C. +2 μC and −2 μC
  • D. +3 μC and −3 μC

Answer: B — Total charge = +6 + (−4) = +2 μC; when identical conductors touch, charge distributes equally: each gets +2/2 = +1 μC by charge redistribution principle.

Flashcards

What happens when two glass rods rubbed with silk are brought close together?

They repel each other because like charges repel.

Define an electrified body.

A body that has acquired excess positive or negative charge by losing or gaining electrons.

Why does a plastic comb get charged when rubbed but a metal spoon does not?

Plastic is an insulator so charge stays on it, while metal is a conductor so charge leaks to the ground through our body.

What is created when a glass rod is rubbed with silk?

No new charge is created; electrons from the glass transfer to silk, making glass positive and silk negative.

What is the principle of additivity of charges?

Total charge in a system equals the algebraic sum of all individual charges, like adding real numbers.

How does a gold-leaf electroscope detect charge?

When a charged object touches the metal knob, charge flows to the gold leaves and they diverge by a degree indicating charge amount.

Distinguish between a conductor and an insulator.

Conductors allow electricity to pass easily because electrons are free to move; insulators resist electricity because electrons are tightly bound.

What is the historical origin of the term 'electricity'?

It comes from the Greek word 'elektron' meaning amber, as Thales of Miletus observed amber attracts light objects when rubbed.

When a charged glass rod touches uncharged silk, what happens?

Unlike charges neutralize each other and both become electrically neutral with no charge remaining.

Define a point charge.

A charged body whose size is negligible compared to distances between it and other charges, so all charge is assumed concentrated at one point.

Important Board Questions

Define electric charge. State whether charge is created during rubbing or just transferred between objects. Give one example. [2 marks]

Charge is the property causing force in electric field; explain conservation of charge with glass-silk rubbing example showing electron transfer, not creation.

Explain why a plastic comb gets electrified when rubbed with dry hair, but a metal spoon does not show any signs of charging. How would you charge a metal rod if it is normally insulated by the Earth? [5 marks]

Distinguish conductor (metal) from insulator (plastic); explain why charge leaks through body-Earth connection for metal spoon; show wooden handle breaks this path, allowing charge accumulation on metal rod.

State and explain the law of additivity of electric charges. Two point charges +3 μC and −2 μC are placed in a system. Calculate the total charge and explain what happens when these charges are brought into contact with each other. Show all steps. [6 marks]

Define additivity: total charge = algebraic sum of individual charges (Q = q₁ + q₂); calculate total = +3 − 2 = +1 μC; explain neutralization: when brought into contact, unlike charges partially cancel, leaving net +1 μC; clarify charge conservation throughout.

Assertion-Reason Questions for Electric Charges and Fields

Q1. Assertion (A):

When a glass rod is rubbed with silk, the glass rod acquires positive charge and the silk acquires negative charge.

Reason (R):

During rubbing, electrons are transferred from glass to silk because electrons in silk are more tightly bound to their atoms than in glass.

Answer: C — A is true but R is false — The assertion is true—this is the established convention. However, the reason is false. Electrons transfer from glass TO silk (making glass positive), which means electrons in glass are LESS tightly bound, not more. The correct reason would be that electrons in glass are weakly bound and easily transferred to silk, which is more electronegative.

Q2. Assertion (A):

A metal comb does not retain charge when rubbed with hair, but a plastic comb does.

Reason (R):

Metals are conductors and charges leak through the human body to the ground, while plastics are insulators and charges remain localized on the surface.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the assertion is true. Conductivity and the path to ground (through body) determine charge retention. Insulators block charge movement, so charge remains where it was deposited.

Q3. Assertion (A):

The total electric charge in an isolated system always remains constant.

Reason (R):

This is because electric charge can be created or destroyed only by conversion to or from electromagnetic radiation.

Answer: C — A is true but R is false — The assertion is true—charge conservation is a fundamental law. However, the reason is false. Charge is not created or destroyed by converting to/from radiation; it is simply conserved. The reason describes energy conversion, not charge conservation. Charge cannot be created or destroyed under any known physical process.

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Case-Based Questions (CBQ) for Electric Charges and Fields

Case Study:

During a physics demonstration in class, a teacher uses a Van de Graaff generator to study electrostatic phenomena. The generator charges a metal sphere to +15 μC. A small neutral pith ball is suspended by a silk thread near the charged sphere. Initially, the pith ball is attracted to the sphere and touches it. The teacher then records that after 2 seconds of contact, when the pith ball is separated from the sphere, both objects repel each other strongly. The teacher explains that charge redistribution occurs differently in this scenario compared to two conductors of equal size.

Q1. Why was the pith ball initially attracted to the positive sphere before making contact?

  • A. The pith ball was negatively pre-charged
  • B. The positive sphere induced a negative charge on the near side of the pith ball through charge induction ✓
  • C. The silk thread attracted the sphere
  • D. The pith ball has an inherent negative charge

Answer: B

Q2. After contact and separation, both objects repel each other. This indicates:

  • A. The pith ball became positively charged by receiving electrons from the sphere
  • B. The pith ball became positively charged by losing electrons to the sphere ✓
  • C. The pith ball became negatively charged by receiving electrons from the sphere
  • D. Charge was neutralized and both objects became uncharged

Answer: B

Q3. The teacher notes that unlike two identical metal spheres where charge divides equally (7.5 μC each), the charge distribution here is unequal. This is most likely because:

  • A. The pith ball is smaller than the metal sphere, so it receives less charge ✓
  • B. The pith ball is an insulator and charge distributes differently in insulators
  • C. The silk thread prevents equal charge distribution
  • D. The pith ball's irregular shape affects charge distribution

Answer: A

Q4. If the same experiment is repeated with the pith ball touching the sphere for 5 seconds instead of 2 seconds, the repulsive force between them after separation will:

  • A. Increase because more charge transfers with longer contact time ✓
  • B. Decrease because charge leaks away over time
  • C. Remain the same because charge distribution is instantaneous
  • D. Become zero because they reach equilibrium

Answer: A

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Previous Year Questions (PYQ) for Electric Charges and Fields

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 55/1/1 · 1 mark

A conducting wire connects two charged metallic spheres A and B of radii r₁ and r₂ respectively. The distance between the spheres is very large compared to their radii. The ratio of electric fields, (E_A/E_B) at the surfaces of spheres A and B will be

  • r₁/r₂
  • r₂/r₁
  • r₁²/r₂²
  • r₂²/r₁²

CBSE 2026 · Set 55/1/1 · 3 marks

Using Gauss's law, deduce an expression for electric field at a point due to a uniformly charged infinite plane thin sheet.

CBSE 2026 · Set 55/1/1 · 5 marks

An electric dipole consists of two point charges q and −q separated by a distance 2a. Derive an expression for the electric field E⃗ due to this dipole at a point distant r from the centre of the dipole on the equatorial plane. Write the expression for the electric field at a far off point, i.e. r >> a. A dipole is placed in x-y plane such that charges q and −q are located at x = a and x = b respectively. There exists an electric field E⃗ = 2î N/C in the region. Calculate the force F⃗ and torque τ⃗ experienced by the dipole.

CBSE 2026 · Set 55/1/2 · 5 marks

An electric dipole consists of two point charges q and −q separated by a distance 2a. Derive an expression for the electric field E⃗ due to this dipole at a point distant r from the centre of the dipole on the equatorial plane. Write the expression for the electric field at a far off point, i.e. r >> a.

CBSE 2026 · Set 55/1/3 · 3 marks

Using Gauss's law, deduce an expression for electric field at a point due to a uniformly charged infinite plane thin sheet. (b) Two large thin plane sheets, each having surface charge density σ, are held close and parallel to each other in air. What is the net electric field at a point (i) inside and (ii) outside, the sheets?

CBSE 2026 · Set 55/1/3 · 5 marks

(a) An electric dipole consists of two point charges q and −q separated by a distance 2a. Derive an expression for the electric field E⃗ due to this dipole at a point distant r from the centre of the dipole on the equatorial plane. Write the expression for the electric field at a far off point, i.e. r >> a. (b) A dipole is placed in x-y plane such that charges q and −q are located at x = a and x = b respectively. There exists an electric field E⃗ = 2î N/C in the region. Calculate the force F⃗ and torque τ⃗ experienced by the dipole.

CBSE 2026 · Set 55/2/1 · 1 mark

Three point charges 2q, −2q and q are kept at the vertices of an equilateral triangle of side l. The potential energy of the system is

  • zero
  • −2q²/(πε₀l)
  • q²/(2πε₀l)
  • −q²/(πε₀l)

CBSE 2026 · Set 55/2/1 · 3 marks

(a) An electric field E⃗ = E₀î exists in a region of space. Draw three equipotential surfaces in the region. (b) Two point charges −q and +q are located at points (−a, 0, 0) and (a, 0, 0) respectively. Find the electrostatic potential at point (x, 0, 0) where x >> a.

CBSE 2026 · Set 55/2/2 · 1 mark

Three point charges +q, −2q and +q are placed along x-axis at points x = −1 m, x = 0 m and x = 2 m respectively. The potential energy of the system is

  • −q²/(πε₀)
  • −2q²/(3πε₀)
  • q²/(4πε₀)
  • −5q²/(4πε₀)

CBSE 2026 · Set 55/2/2 · 3 marks

(i) Draw electric field lines and equipotential surfaces for a system of two equal and opposite point charges separated by some distance. (ii) Why electric field E⃗ at a point on an equipotential surface must be perpendicular to the surface at that point?

CBSE 2026 · Set 55/2/3 · 1 mark

A point charge of 5 μC is located at (1 cm, 1 cm) in x-y plane. A second charge of 6 μC is brought from infinity to point (−3 cm, 4 cm). The work done in the process is

  • 0.54 J
  • 5.4 J
  • −0.54 J
  • 10.8 J

CBSE 2026 · Set 55/2/3 · 3 marks

(a) Draw electric field lines and equipotential surfaces for a system of two equal positive charges separated by some distance.

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Class 12 Physics Formula Sheet

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Class 12 Physics Derivations

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