Chapter 16: Electric Charges and Fields

Introduction

Key Points

• Static electricity arises from the discharge of electric charges through the body.
• Common experience: electric shock when touching car door or bus iron bar after sliding.
• Electrostatics studies forces, fields, and potentials from static charges.

Solved Examples

Ex 1: Why do you feel a shock when touching a car door after sliding?

• Sliding causes charge buildup (static electricity).
• Touching door discharges the static charge through the body.

Practice Set

Q1: What causes the sensation of electric shock when touching a metal object after sliding?

• A. Current flow from battery
• B. Discharge of static electricity
• C. Magnetic field induction
• D. Heat conduction

Q2: Electrostatics primarily studies:

• A. Moving charges and currents
• B. Forces and potentials from static charges
• C. Magnetic fields from currents
• D. Heat generated by electric currents

Q3: Which of the following is NOT related to electrostatics?

• A. Electric field due to a charged sphere
• B. Current in a wire
• C. Potential difference between charged plates
• D. Force between two charged particles at rest

Answer Key

• Q1: B - Static electricity discharge causes the shock sensation.
• Q2: B - Electrostatics studies forces and potentials from static charges.
• Q3: B - Current in a wire involves moving charges, not static charges.

Memory Aids

• Static charge buildup = friction + insulation.
• Electrostatics = study of charges at rest.
• Shock = sudden discharge of static electricity.

Electric Charge

Key Points

• Neutral body becomes charged by gaining or losing electrons.
• Two types of charges: positive and negative.
• Loss of electrons → positive charge; gain of electrons → negative charge.
• Like charges repel; unlike charges attract.
• Glass rod or cat’s fur rubbed with silk → positive charge.
• Plastic rod or silk rubbed → negative charge.
• Charged object is electrified; neutral object has no net charge.
• Gold-leaf electroscope detects presence and type of charge.

Solved Examples

Ex 1: What charge does a glass rod acquire when rubbed with silk?

• Glass rod loses electrons → becomes positively charged.

Ex 2: What happens when a plastic rod is rubbed with silk?

• Plastic rod gains electrons → becomes negatively charged.

Ex 3: What force acts between two positively charged rods?

• Like charges repel → rods repel each other.

Practice Set

Q1: When an object loses electrons, it becomes:

• A. Negatively charged
• B. Positively charged
• C. Neutral
• D. Electrically neutral

Q2: Two rods repel each other. What can be concluded about their charges?

• A. Both are neutral
• B. Both have opposite charges
• C. Both have like charges
• D. One is charged, other is neutral

Q3: A gold-leaf electroscope shows divergence of leaves when a charged rod is brought near. What does this indicate?

• A. The rod is neutral
• B. The rod is charged
• C. The electroscope is faulty
• D. No charge on rod or electroscope

Answer Key

• Q1: B - Loss of electrons causes positive charge.
• Q2: C - Like charges repel, so rods have same type of charge.
• Q3: B - Divergence indicates presence of charge on rod.

Memory Aids

• Loss of electrons = positive charge.
• Gain of electrons = negative charge.
• Like charges repel; unlike charges attract.
• Gold-leaf electroscope detects charge by leaf divergence.

Conductors and Insulators

Properties of Conductors

• Allow electric current to pass easily due to free electrons.
• Examples: Metals (copper, aluminum), human and animal bodies, earth.
• Charge distributes uniformly over the surface when transferred.

Properties of Insulators

• Resist electric current; electrons are tightly bound.
• Examples: Glass, porcelain, plastic, nylon, wood.
• Charge remains localized where it is placed.

Electric Wiring in Houses

• Three wires: live, neutral, and earth.
• Live and neutral carry current from power station.
• Earth wire connected to ground; carries extra charge safely.
• Metallic bodies of appliances connected to earth wire for safety.

Solved Examples

• Ex 1: Charge distribution on a metal sphere.
  • Given charge Q on conductor.
  • Charge spreads uniformly on surface.
  • Surface charge density \( \sigma = \frac{Q}{4\pi r^2} \).
• Ex 2: Identifying conductor or insulator.
  • Material allows current easily → conductor.
  • Material resists current → insulator.

Practice Set

• Q1: Which of the following is a good conductor?
  • A) Glass
  • B) Copper
  • C) Plastic
  • D) Wood
• Q2: When charge is placed on an insulator, it:
  • A) Spreads uniformly
  • B) Moves to earth wire
  • C) Remains localized
  • D) Flows as current
• Q3: In household wiring, the earth wire is used to:
  • A) Carry current to appliances
  • B) Connect live and neutral wires
  • C) Safely discharge extra charge to ground
  • D) Increase voltage supply

Answer Key

• Q1: B) Copper — Metals have free electrons allowing easy current flow.
• Q2: C) Remains localized — Insulators do not allow free movement of charge.
• Q3: C) Safely discharge extra charge to ground — Earth wire prevents electric shocks.

Memory Aids

• Conductors: "Free electrons flow" → metals, body, earth.
• Insulators: "Charge stays put" → glass, plastic, wood.
• Earth wire = safety path for extra charge.

Basic Properties of Electric Charge

Addition of Charges

• Total charge in a system with charges \(q_1, q_2, ..., q_n\) is \(q = q_1 + q_2 + ... + q_n\).

Conservation of Charge

• Total charge in an isolated system remains constant.
• Charge cannot be created or destroyed, only transferred.

Quantisation of Charge

• Charge is always an integral multiple of the elementary charge \(e = 1.6 \times 10^{-19}\) C.
• Mathematically, \(Q = \pm ne\), where \(n = 1, 2, 3, ...\)
• 1 Coulomb contains approximately \(6 \times 10^{18}\) electrons.

Solved Examples

• Ex 1: Find total charge if \(q_1 = 2 \times 10^{-6}\) C and \(q_2 = -3 \times 10^{-6}\) C.
  • Total charge \(q = q_1 + q_2 = 2 \times 10^{-6} - 3 \times 10^{-6} = -1 \times 10^{-6}\) C.
• Ex 2: Calculate number of electrons in 3 C charge.
  • Number of electrons \(n = \frac{Q}{e} = \frac{3}{1.6 \times 10^{-19}} \approx 1.875 \times 10^{19}\).

Practice Set

• Q1: Total charge of charges \(2 \times 10^{-6}\) C and \(5 \times 10^{-6}\) C is?
  • A) \(7 \times 10^{-6}\) C
  • B) \(3 \times 10^{-6}\) C
  • C) \(10 \times 10^{-6}\) C
  • D) \(1 \times 10^{-6}\) C
• Q2: Which statement is true?
  • A) Charge can be created
  • B) Charge is conserved
  • C) Charge can be destroyed
  • D) Charge is not quantised
• Q3: Number of electrons in 0.32 C charge is approximately?
  • A) \(2 \times 10^{18}\)
  • B) \(5 \times 10^{18}\)
  • C) \(1 \times 10^{18}\)
  • D) \(3 \times 10^{18}\)

Answer Key

• Q1: A; Sum of charges \(2 \times 10^{-6} + 5 \times 10^{-6} = 7 \times 10^{-6}\) C.
• Q2: B; Charge is conserved in isolated systems.
• Q3: A; Number of electrons \(= \frac{0.32}{1.6 \times 10^{-19}} = 2 \times 10^{18}\).

Memory Aids

• Charge addition: sum all charges algebraically.
• Conservation: total charge constant in isolated system.
• Quantisation: charge = integer × elementary charge.
• Elementary charge \(e = 1.6 \times 10^{-19}\) C.

Coulomb's Law

Fundamental Formula

• Force between two point charges \( q_1 \) and \( q_2 \) separated by distance \( r \) in vacuum:

  \[ F = k \frac{|q_1 q_2|}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_1 q_2|}{r^2} \]

• Permittivity of free space: \( \varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2 \text{N}^{-1} \text{m}^{-2} \)
• Coulomb constant: \( k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{C}^{-2} \)

Vector Form

• Force on \( q_2 \) due to \( q_1 \):

  \[ \vec{F}_{21} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{|\vec{r}_2 - \vec{r}_1|^3} (\vec{r}_2 - \vec{r}_1) = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{|\vec{r}_{21}|^2} \hat{r}_{21} = - \vec{F}_{12} \]

• Obeys Newton's third law: forces are equal in magnitude and opposite in direction.

Force Direction

• Like charges (\( q_1 q_2 > 0 \)) repel: force vectors point away from each other.
• Unlike charges (\( q_1 q_2 < 0 \)) attract: force vectors point towards each other.

Effect of Dielectric Medium

• In a medium with dielectric constant \( K \), force reduces by factor \( K \):

  \[ F = \frac{F_0}{K} = \frac{1}{4 \pi \varepsilon_0 K} \frac{q_1 q_2}{r^2} \]

• \( F_0 \) is force in vacuum.

Solved Examples

Ex 1

• Two charges \( q_1 = 2 \mu C \), \( q_2 = -3 \mu C \) are 0.5 m apart in vacuum.
• Calculate magnitude of force \( F \):

  \[ F = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{(0.5)^2} = 0.216 \text{ N} \]

• Force is attractive due to opposite signs.

Practice Set

Q1

Two point charges each of \( 1 \mu C \) are placed 1 m apart in vacuum. What is the magnitude of the force between them?

• A) 9 N
• B) 1 N
• C) 0.9 N
• D) 0.09 N

Q2

Two charges \( q_1 = 4 \mu C \) and \( q_2 = -2 \mu C \) are 0.2 m apart in a medium with dielectric constant \( K = 5 \). What is the force magnitude?

• A) 1.8 N
• B) 18 N
• C) 3.6 N
• D) 0.36 N

Q3

Two charges \( q_1 = 3 \mu C \) and \( q_2 = 3 \mu C \) are 0.1 m apart in vacuum. What is the force magnitude?

• A) 8.1 N
• B) 81 N
• C) 0.81 N
• D) 0.081 N

Answer Key

Q1

Option A. \( F = 9 \times 10^9 \times \frac{(1 \times 10^{-6})^2}{1^2} = 9 \text{ N} \)

Q2

Option D. Force in vacuum: \( F_0 = 9 \times 10^9 \times \frac{4 \times 10^{-6} \times 2 \times 10^{-6}}{0.2^2} = 1.8 \text{ N} \). In medium: \( F = \frac{1.8}{5} = 0.36 \text{ N} \)

Q3

Option C. \( F = 9 \times 10^9 \times \frac{3 \times 10^{-6} \times 3 \times 10^{-6}}{0.1^2} = 8.1 \text{ N} \)

Memory Aids

• Force \( F \) varies inversely with square of distance \( r^2 \).
• Force \( F \) varies directly with product of charges \( q_1 q_2 \).
• Dielectric reduces force by factor \( K \).
• Like charges repel, unlike charges attract.
• Newton's third law: \( \vec{F}_{12} = - \vec{F}_{21} \).

Forces between Multiple Charges

Principle of Superposition

• Force on a charge due to multiple charges is the vector sum of individual forces.
• Each force acts independently; presence of other charges does not alter individual forces.
• Mathematically, \( \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} + \cdots + \vec{F}_{1n} \).
• Each force \( \vec{F}_{1i} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_i}{r_{1i}^2} \hat{r}_{1i} \), where \( \hat{r}_{1i} \) is the unit vector from \( q_1 \) to \( q_i \).

Vector Addition of Forces

• Position vectors \( \vec{r}_1, \vec{r}_2, \vec{r}_3 \) locate charges in space.
• Vectors \( \vec{r}_{12} = \vec{r}_2 - \vec{r}_1 \), \( \vec{r}_{13} = \vec{r}_3 - \vec{r}_1 \) define directions between charges.
• Unit vectors \( \hat{r}_{12}, \hat{r}_{13} \) indicate force directions.
• Net force on \( q_1 \) is \( \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} \).

Electric Field from Multiple Charges

• Total electric field at a point is vector sum of fields from all charges.
• Electric field due to charge \( q_i \) at distance \( r_i \) is \( \vec{E}_i = \frac{1}{4\pi\epsilon_0} \frac{q_i}{r_i^2} \hat{r}_i \).
• Net electric field \( \vec{E} = \sum_i \vec{E}_i \).
• Useful for calculating forces: \( \vec{F} = q \vec{E} \).

Solved Examples

Ex 1

• Given charges \( q_1 = 2\mu C \), \( q_2 = 3\mu C \), distance \( r_{12} = 0.5m \).
• Calculate force \( \vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{r}_{12} \).
• Use \( k = 9 \times 10^9 \) Nm²/C²: \( F = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{0.5^2} = 0.216 N \).

Ex 2

• Three charges \( q_1, q_2, q_3 \) at known positions.
• Calculate \( \vec{F}_{12} \) and \( \vec{F}_{13} \) using Coulomb's law.
• Find net force \( \vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} \) by vector addition.

Practice Set

Q1

Two charges \( +2\mu C \) and \( +3\mu C \) are 1 m apart. What is the force between them?

• A) 0.054 N
• B) 0.36 N
• C) 0.12 N
• D) 0.18 N

Q2

Charge \( q_1 = 1\mu C \) experiences forces \( \vec{F}_{12} = 3 \hat{i} \) N and \( \vec{F}_{13} = 4 \hat{j} \) N. What is magnitude of net force?

• A) 5 N
• B) 7 N
• C) 1 N
• D) 12 N

Q3

Three charges \( q_1, q_2, q_3 \) are placed at vertices of an equilateral triangle of side 0.5 m. Each charge is \( 2\mu C \). What is net force on \( q_1 \)? (Use \( k=9\times10^9 \) Nm²/C²)

• A) 0 N
• B) 0.216 N
• C) 0.374 N
• D) 0.432 N

Answer Key

Q1

Option B. \( F = k \frac{q_1 q_2}{r^2} = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{1^2} = 0.054 N \) (Check calculation: 9e9*6e-12=0.054 N, so correct is A)

Q2

Option A. Net force magnitude = \( \sqrt{3^2 + 4^2} = 5 \) N.

Q3

Option C. Each force magnitude \( F = k \frac{q^2}{r^2} = 9 \times 10^9 \times \frac{(2 \times 10^{-6})^2}{0.5^2} = 0.144 N \). Net force is vector sum of two forces at 60°, \( F_{net} = \sqrt{F^2 + F^2 + 2FF\cos 60^\circ} = 0.374 N \).

Memory Aids

• Superposition: Add forces vectorially, independent of other charges.
• Coulomb's law: \( F = k \frac{q_1 q_2}{r^2} \), \( k = 9 \times 10^9 \) Nm²/C².
• Use unit vectors to find force directions.
• Net force = vector sum of all individual forces.

Electric Field

Definition and Formula

• Electric field is the region around a charge where its force is experienced by a unit positive test charge.
• Electric field \( \vec{E} \) due to a point charge \( q \) at position \( \vec{r} \):
  \[ \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r} = \frac{\vec{F}}{q_0} \]
• \( q_0 \) is the positive test charge; \( \hat{r} \) is the unit vector from charge to point.

Electric Field from Multiple Charges

• Net force on test charge \( q_0 \) due to \( n \) charges \( q_1, q_2, ..., q_n \):
  \[ \vec{F} = \vec{F}_1 + \vec{F}_2 + ... + \vec{F}_n \]
• Net electric field at test charge position:
  \[ \vec{E} = \frac{\vec{F}_1}{q_0} + \frac{\vec{F}_2}{q_0} + ... + \frac{\vec{F}_n}{q_0} = \vec{E}_1 + \vec{E}_2 + ... + \vec{E}_n \]
• General formula:
  \[ \vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^n \frac{q_i}{r_i^2} \hat{r}_{p_i} \]
• \( \hat{r}_{p_i} \) points from the \( i^{th} \) charge to point \( P \).

Physical Significance

• Electric field is a property of the charge system, independent of the test charge.
• It is a vector quantity, like force.
• Changes in electric field propagate at the speed of light due to electromagnetic waves.
• Electric field is detected by forces on charges and can transport energy.

Solved Examples

Ex 1

• Find electric field at 0.5 m from a point charge \( q = 2 \mu C \).
• Use formula:
  \[ E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \]
• Calculate:
  \( E = 9 \times 10^9 \times \frac{2 \times 10^{-6}}{(0.5)^2} = 7.2 \times 10^4 \ \text{N/C} \)

Ex 2

• Calculate net electric field at point P due to two charges \( q_1 = 3 \mu C \) and \( q_2 = -3 \mu C \) placed 1 m apart, P at midpoint.
• Distance from each charge to P is 0.5 m.
• Calculate fields:
  \( E_1 = E_2 = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{(0.5)^2} = 1.08 \times 10^5 \ \text{N/C} \)
• Directions opposite, fields add up:
  \( E_{net} = 2 \times 1.08 \times 10^5 = 2.16 \times 10^5 \ \text{N/C} \)

Practice Set

• Q1: Electric field due to a point charge decreases with distance as:
  A) \( 1/r \)   B) \( 1/r^2 \)   C) \( r^2 \)   D) \( r \)
• Q2: Two charges \( +q \) and \( -q \) are placed 2 m apart. Electric field at midpoint is:
  A) Zero   B) Directed from +q to -q   C) Directed from -q to +q   D) Infinite
• Q3: Three charges \( q, q, -2q \) are placed at vertices of an equilateral triangle of side \( a \). Net electric field at the vertex with charge \( -2q \) is:
  A) Zero   B) \( \frac{1}{4\pi\epsilon_0} \frac{3q}{a^2} \) towards \( -2q \)   C) \( \frac{1}{4\pi\epsilon_0} \frac{3q}{a^2} \) away from \( -2q \)   D) \( \frac{1}{4\pi\epsilon_0} \frac{q}{a^2} \)

Answer Key

• Q1: B
  Electric field varies inversely with square of distance.
• Q2: B
  Fields from +q and -q add at midpoint, direction from positive to negative charge.
• Q3: B
  Net field magnitude is sum of fields from two \( q \) charges, directed towards \( -2q \) due to attraction.

Memory Aids

• Electric field \( E \) ~ \( q/r^2 \) (Inverse square law)
• Field direction: away from positive, towards negative charges
• Net field = vector sum of individual fields
• Electric field independent of test charge magnitude

Electric Field Lines

Definition and Direction

• Electric field lines represent the direction of the electric field around charges.
• At any point, the tangent to the field line shows the direction of the net electric field.
• Lines start from positive charges and end at negative charges; for isolated charges, they may start or end at infinity.

Field Line Patterns

• Positive point charge: lines radiate outward.
• Negative point charge: lines point inward.
• Two like charges: lines repel and do not cross.
• Dipole (positive and negative): lines go from positive to negative.

Properties

• Field lines are continuous and do not break in charge-free regions.
• Tangent at any point gives the electric field direction.
• Field lines never intersect; intersection implies ambiguous field direction.
• Electrostatic field lines do not form closed loops (conservative field).
• Field strength is proportional to the density of field lines.
• Field lines are perpendicular (normal) to conductor surfaces where they start or end.

Dependence on Distance

• Electric field strength decreases with distance from the charge.
• Closer points have denser field lines indicating stronger field.
• Farther points have sparse field lines indicating weaker field.
• Field lines spread over solid angle \(\Delta \Omega\), showing field distribution.

Solved Examples

Ex 1

• Find the direction of the electric field at a point near a positive charge.
• Solution: Field lines radiate outward; direction is away from the positive charge.

Ex 2

• Determine the relative strength of the electric field at two points \(P_1\) and \(P_2\) at distances \(r_1\) and \(r_2\) from a point charge \(q\), where \(r_2 = 2r_1\).
• Solution: \(E \propto \frac{1}{r^2}\), so \(E_2 = \frac{E_1}{4}\). Field at \(P_2\) is one-fourth that at \(P_1\).

Practice Set

Q1

Which statement about electric field lines is correct?

• A. They start from negative charges and end at positive charges.
• B. They can intersect each other.
• C. They are always perpendicular to conductor surfaces.
• D. They form closed loops around charges.

Q2

At a distance \(r\) from a point charge \(q\), the electric field strength is \(E\). What is the field strength at distance \(2r\)?

• A. \(E/2\)
• B. \(E/4\)
• C. \(2E\)
• D. \(4E\)

Q3

Two positive charges are placed close to each other. Which of the following is true about their electric field lines?

• A. Field lines start from one charge and end on the other.
• B. Field lines cross each other between the charges.
• C. Field lines repel and do not cross, showing repulsion.
• D. Field lines form closed loops around both charges.

Answer Key

• Q1: C. Field lines are always perpendicular to conductor surfaces where they start or end.
• Q2: B. Electric field strength varies as \(1/r^2\), so at \(2r\), it is \(E/4\).
• Q3: C. Like charges repel; their field lines repel and do not cross.

Memory Aids

• Field lines start at +ve, end at -ve or infinity.
• Density of lines ∝ field strength.
• Lines never cross (unique field direction).
• Field strength \(E = k \frac{q}{r^2}\).
• Lines perpendicular to conductor surfaces.

Electric Flux

Definition and Formula

• Electric flux (\(\Phi\)) is the number of electric field lines passing through a surface.
• Mathematically, \(\Phi = \int_S \vec{E} \cdot d\vec{A}\).
• For uniform electric field over surface \(S\), \(\Phi = \vec{E} \cdot \vec{A} = EA \cos \theta\), where \(\theta\) is angle between \(\vec{E}\) and area vector \(\vec{A}\).
• Area vector \(\vec{A}\) is perpendicular to the surface.
• If \(\theta = 90^\circ\), flux \(\Phi = 0\) (field lines parallel to surface).
• If \(\theta = 0^\circ\), flux is maximum: \(\Phi_{max} = EA\).
• Electric flux is a scalar quantity measured in volt-meter (V·m) or newton meter squared per coulomb (N·m²/C).
• Electric field can be non-zero even if net flux is zero, but zero flux implies no net electric field through the surface.

Solved Examples

Ex 1

• Given: \(E = 5 \times 10^3\,\mathrm{N/C}\), \(A = 2\,\mathrm{m}^2\), \(\theta = 30^\circ\).
• Calculate electric flux: \(\Phi = EA \cos \theta = 5 \times 10^3 \times 2 \times \cos 30^\circ = 5 \times 10^3 \times 2 \times 0.866 = 8660\,\mathrm{N\cdot m^2/C}\).

Ex 2

• Electric field \(E = 1000\,\mathrm{N/C}\) is parallel to surface (\(\theta = 90^\circ\)), area \(A = 1\,\mathrm{m}^2\).
• Flux \(\Phi = EA \cos 90^\circ = 1000 \times 1 \times 0 = 0\).

Practice Set

Q1

Electric field \(E = 2000\,\mathrm{N/C}\) acts at \(60^\circ\) to a surface of area \(0.5\,\mathrm{m}^2\). Find the electric flux.

• A) 500 \(\mathrm{N\cdot m^2/C}\)
• B) 1000 \(\mathrm{N\cdot m^2/C}\)
• C) 2000 \(\mathrm{N\cdot m^2/C}\)
• D) 1732 \(\mathrm{N\cdot m^2/C}\)

Q2

A uniform electric field of \(3000\,\mathrm{N/C}\) is perpendicular to a surface of area \(0.2\,\mathrm{m}^2\). Calculate the flux.

• A) 600 \(\mathrm{N\cdot m^2/C}\)
• B) 0 \(\mathrm{N\cdot m^2/C}\)
• C) 1500 \(\mathrm{N\cdot m^2/C}\)
• D) 3000 \(\mathrm{N\cdot m^2/C}\)

Q3

Electric field \(E = 4000\,\mathrm{N/C}\) makes an angle \(45^\circ\) with a surface of area \(0.1\,\mathrm{m}^2\). What is the electric flux?

• A) 282.8 \(\mathrm{N\cdot m^2/C}\)
• B) 400 \(\mathrm{N\cdot m^2/C}\)
• C) 141.4 \(\mathrm{N\cdot m^2/C}\)
• D) 0 \(\mathrm{N\cdot m^2/C}\)

Answer Key

• Q1: D; \(\Phi = EA \cos 60^\circ = 2000 \times 0.5 \times 0.5 = 500\) is incorrect, correct is \(2000 \times 0.5 \times 0.5 = 500\) but options show 1732, correct calculation: \(2000 \times 0.5 \times \cos 60^\circ = 2000 \times 0.5 \times 0.5 = 500\). So correct option is A. Correction: A is correct.
• Q2: A; Flux \(= 3000 \times 0.2 \times \cos 0^\circ = 600\).
• Q3: A; Flux \(= 4000 \times 0.1 \times \cos 45^\circ = 400 \times 0.707 = 282.8\).

Memory Aids

• Electric flux \(\Phi = EA \cos \theta\).
• \(\theta = 0^\circ\) → max flux; \(\theta = 90^\circ\) → zero flux.
• Flux is scalar, unit: V·m or N·m²/C.
• Flux zero does not imply zero electric field.

Electric Dipole

Definition and Dipole Moment

• An electric dipole consists of two equal and opposite charges (+q and -q) separated by a small distance 2a.
• Dipole moment \( \vec{p} = q \times 2\vec{a} \), a vector from negative to positive charge.
• Dipole moment is a measure of the strength and orientation of the dipole.

Electric Field on Axial Line (End-on Position)

• Electric field at point on axis: \( \vec{E} = \frac{q \hat{p}}{4\pi\epsilon_0} \left[ \frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} \right] \)
• Equivalent form: \( \vec{E} = \frac{2 \vec{p} r}{4\pi\epsilon_0 (r^2 - a^2)^2} \)
• At large distance \( (r \gg a) \), \( \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{2 \vec{p}}{r^3} \)

Electric Field on Equatorial Line (Broadside-on Position)

• Electric field at point on equatorial line: \( \vec{E} = -\frac{\vec{p}}{4\pi\epsilon_0 (r^2 + a^2)^{3/2}} \)
• At large distance \( (r \gg a) \), \( \vec{E} = -\frac{\vec{p}}{4\pi\epsilon_0 r^3} \)

Physical Significance

• Nonpolar molecules (e.g., CO₂, CH₄) have coinciding centers of positive and negative charges; no permanent dipole moment.
• Polar molecules (e.g., H₂O, NH₃) have permanent dipole moments due to non-coinciding charge centers.
• External electric fields can induce dipole moments in nonpolar molecules.

Electric Field at a General Point

• Dipole moment \( \vec{p} \) resolved into components: \( p \cos\theta \) (axial) and \( p \sin\theta \) (equatorial).
• Electric field components at point P:
  • Axial component: \( E_1 = \frac{2k p \cos\theta}{r^3} \)
  • Equatorial component: \( E_2 = \frac{k p \sin\theta}{r^3} \)
• Resultant electric field magnitude: \[ E = \frac{k p}{r^3} \sqrt{1 + 3 \cos^2 \theta} \]
• Angle \( \alpha \) between \( \vec{E} \) and \( \vec{r} \): \[ \tan \alpha = \frac{E_2}{E_1} = \frac{\tan \theta}{2} \Rightarrow \alpha = \tan^{-1} \left( \frac{\tan \theta}{2} \right) \]

Solved Examples

Ex 1

• Find electric field on axial line at distance \( r \gg a \) for dipole moment \( p = 4 \times 10^{-30} \) C·m.
• Use formula: \( E = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} \).
• Substitute values to calculate \( E \).

Ex 2

• Calculate electric field on equatorial line at large distance \( r \) for dipole moment \( p \).
• Use formula: \( E = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} \).
• Plug in values to find \( E \).

Practice Set

Q1

What is the direction of the electric dipole moment vector?

• A. From positive to negative charge
• B. From negative to positive charge
• C. Perpendicular to the line joining charges
• D. Along the equatorial line

Q2

At a point on the axial line far from the dipole, the electric field varies as:

• A. \( 1/r^2 \)
• B. \( 1/r^3 \)
• C. \( 1/r \)
• D. \( 1/r^4 \)

Q3

A dipole has moment \( 3 \times 10^{-30} \) C·m. Calculate the electric field at a point 0.1 m away on the axial line. (Use \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \) Nm²/C²)

• A. 5.4 × 10³ N/C
• B. 1.8 × 10⁴ N/C
• C. 5.4 × 10⁴ N/C
• D. 1.8 × 10³ N/C

Answer Key

• Q1: B. Dipole moment points from negative to positive charge.
• Q2: B. Electric field on axial line varies as \( 1/r^3 \) at large distances.
• Q3: C. \( E = \frac{2kp}{r^3} = \frac{2 \times 9 \times 10^9 \times 3 \times 10^{-30}}{(0.1)^3} = 5.4 \times 10^4 \) N/C.

Memory Aids

• Dipole moment \( \vec{p} \) points from negative to positive charge.
• Axial field \( \propto 2p/r^3 \), equatorial field \( \propto p/r^3 \).
• At large distances, axial field is twice the equatorial field.
• Resultant field magnitude: \( E = kp/r^3 \sqrt{1 + 3 \cos^2 \theta} \).

Dipole in a Uniform External Field

Concepts and Formulas

• An electric dipole consists of two equal and opposite charges (+q and -q) separated by distance 2a.
• Placed in uniform electric field \(\vec{E}\), forces on charges are equal in magnitude and opposite in direction: \(+q\vec{E}\) and \(-q\vec{E}\).
• Net force on dipole in uniform field is zero, but forces create a torque \(\tau\).
• Torque magnitude: \(\tau = qE \times 2a \sin \theta = pE \sin \theta\), where \(p = q \times 2a\) is dipole moment and \(\theta\) is angle between \(\vec{p}\) and \(\vec{E}\).
• Vector form: \(\vec{\tau} = \vec{p} \times \vec{E}\).
• Torque is zero when \(\theta = 0^\circ\) or \(180^\circ\), maximum at \(90^\circ\).
• Torque tends to align dipole along the electric field direction.

Solved Examples

• Ex 1: Calculate torque on dipole with \(p=4 \times 10^{-29} \text{ C m}\) in field \(E=5 \times 10^{4} \text{ N/C}\) at \(30^\circ\).
  • \(\tau = pE \sin \theta = 4 \times 10^{-29} \times 5 \times 10^{4} \times \sin 30^\circ = 1 \times 10^{-24} \text{ Nm}\)

Practice Set

• Q1: What is the net force on an electric dipole in a uniform electric field?
  • A) Zero
  • B) \(pE\)
  • C) \(qE\)
  • D) Depends on \(\theta\)
• Q2: Torque on a dipole is maximum when \(\theta\) is:
  • A) 0°
  • B) 45°
  • C) 90°
  • D) 180°
• Q3: A dipole with moment \(2 \times 10^{-29} \text{ C m}\) is in a field of \(10^{5} \text{ N/C}\) at 60°. Calculate torque.
  • A) \(1.73 \times 10^{-24} \text{ Nm}\)
  • B) \(2 \times 10^{-24} \text{ Nm}\)
  • C) \(3.46 \times 10^{-24} \text{ Nm}\)
  • D) \(0\)

Answer Key

• Q1: A) Zero — Forces on charges cancel out, net force is zero in uniform field.
• Q2: C) 90° — Torque \(\tau = pE \sin \theta\) is maximum at \(\theta=90^\circ\).
• Q3: A) \(1.73 \times 10^{-24} \text{ Nm}\) — \(\tau = pE \sin 60^\circ = 2 \times 10^{-29} \times 10^{5} \times 0.866 = 1.73 \times 10^{-24}\).

Memory Aids

• Torque \(\tau = pE \sin \theta\) aligns dipole with field.
• Net force zero in uniform field; torque rotates dipole.
• Dipole moment \(\vec{p}\) points from -q to +q.

Dipole in a Non-Uniform Electric Field

Concepts and Formulas

• In non-uniform field, forces on +q and -q differ in magnitude and direction.
• Net force on dipole is non-zero, causing translation and torque.
• Force direction depends on dipole orientation relative to field gradient.
• If \(\vec{E}\) not parallel or antiparallel to \(\vec{p}\), dipole experiences both net force and torque.
• Oscillating dipole in uniform field has time period \(T = 2\pi \sqrt{\frac{I}{pE}}\), where \(I\) is moment of inertia.

Solved Examples

• Ex 1: Dipole in increasing field experiences net force. If force on +q is \(F_+ = 5 \times 10^{-6} \text{ N}\) and on -q is \(F_- = 3 \times 10^{-6} \text{ N}\), net force is:
  • \(F_{net} = F_+ - F_- = 2 \times 10^{-6} \text{ N}\)

Practice Set

• Q1: What happens to a dipole in a non-uniform electric field?
  • A) No net force, only torque
  • B) Net force and torque
  • C) Only net force
  • D) No force or torque
• Q2: If dipole moment \(p=3 \times 10^{-29} \text{ C m}\), field \(E=4 \times 10^{4} \text{ N/C}\), and moment of inertia \(I=2 \times 10^{-46} \text{ kg m}^2\), find oscillation period \(T\).
  • A) \(2.88 \times 10^{-3} \text{ s}\)
  • B) \(3.14 \times 10^{-3} \text{ s}\)
  • C) \(1.57 \times 10^{-3} \text{ s}\)
  • D) \(4.44 \times 10^{-3} \text{ s}\)
• Q3: In a non-uniform field, if force on +q is \(6 \times 10^{-6} \text{ N}\) right and on -q is \(4 \times 10^{-6} \text{ N}\) left, net force is:
  • A) \(2 \times 10^{-6} \text{ N}\) right
  • B) \(10 \times 10^{-6} \text{ N}\) right
  • C) \(2 \times 10^{-6} \text{ N}\) left
  • D) Zero

Answer Key

• Q1: B) Net force and torque — Non-uniform field causes both translation and rotation.
• Q2: A) \(2.88 \times 10^{-3} \text{ s}\) — \(T=2\pi \sqrt{I/(pE)}=2\pi \sqrt{2\times10^{-46}/(3\times10^{-29}\times4\times10^{4})}\).
• Q3: A) \(2 \times 10^{-6} \text{ N}\) right — Net force is difference, direction of larger force.

Memory Aids

• Non-uniform field: dipole moves and rotates.
• Oscillation period \(T=2\pi \sqrt{I/(pE)}\) for dipole in uniform field.
• Net force = difference of forces on charges in non-uniform field.

Continuous Charge Distribution

Charge Densities

• Electric charge is treated as continuous at macroscopic scale due to high electron concentration.
• Line charge density (\(\lambda\)): Charge per unit length, \(\lambda = \frac{\Delta Q}{\Delta l}\), units: C/m.
• Surface charge density (\(\sigma\)): Charge per unit area, \(\sigma = \frac{\Delta Q}{\Delta S}\), units: C/m².
• Volume charge density (\(\rho\)): Charge per unit volume, \(\rho = \frac{\Delta Q}{\Delta V}\), units: C/m³.

Electric Field from Continuous Charges

• Electric field \(\vec{E}\) is sum of contributions from small charge elements.
• For line charge: \(\vec{E}_{\lambda} = \frac{1}{4\pi\varepsilon_0} \sum \frac{\lambda \Delta l}{r'^2} \hat{r}'\)
• For surface charge: \(\vec{E}_{\sigma} = \frac{1}{4\pi\varepsilon_0} \sum \frac{\sigma \Delta S}{r'^2} \hat{r}'\)
• For volume charge: \(\vec{E}_{\rho} = \frac{1}{4\pi\varepsilon_0} \sum \frac{\rho \Delta V}{r'^2} \hat{r}'\)
• \(\varepsilon_0\) is permittivity of free space, \(r'\) is distance from charge element to point.

Solved Examples

Ex 1

• Find linear charge density if \(\Delta Q = 5 \times 10^{-6} C\) over \(\Delta l = 0.1 m\).
• \(\lambda = \frac{5 \times 10^{-6}}{0.1} = 5 \times 10^{-5} C/m\).

Practice Set

• Q1: What is the unit of surface charge density?
  A) C/m
  B) C/m²
  C) C/m³
  D) C
• Q2: Electric field due to a small charge element \(dq\) at distance \(r\) is proportional to:
  A) \(\frac{1}{r}\)
  B) \(\frac{1}{r^2}\)
  C) \(r^2\)
  D) \(r\)
• Q3: For a volume charge density \(\rho = 2 \times 10^{-6} C/m^3\), charge in volume \(0.5 m^3\) is:
  A) \(1 \times 10^{-6} C\)
  B) \(1 \times 10^{-5} C\)
  C) \(1 \times 10^{-7} C\)
  D) \(1 \times 10^{-4} C\)

Answer Key

• Q1: B) C/m² - Surface charge density is charge per unit area.
• Q2: B) \(\frac{1}{r^2}\) - Electric field varies inversely with square of distance.
• Q3: A) \(1 \times 10^{-6} C\) - Charge = \(\rho \times V = 2 \times 10^{-6} \times 0.5 = 1 \times 10^{-6} C\).

Memory Aids

• \(\lambda\) for line (length), \(\sigma\) for surface (area), \(\rho\) for volume.
• Electric field \(\propto \frac{1}{r^2}\) from point charge elements.
• Sum small charge contributions to get total field.

Electric Field Linear Charge Distributions

Types and Formulas

• Finite wire: \(E_P = \frac{k \lambda}{r} (\sin \alpha_1 + \sin \alpha_2)\)
• Infinite wire: \(E_P = \frac{2k \lambda}{r}\)
• Charged arc: \(E_P = \frac{k \lambda}{r} (2 \sin \alpha)\)
• \(k = \frac{1}{4\pi\varepsilon_0}\), \(r\) is distance to point P, \(\alpha\) are angles subtended.

Solved Examples

Ex 1

• Calculate \(E_P\) at 0.5 m from infinite wire with \(\lambda = 4 \times 10^{-6} C/m\).
• \(E_P = \frac{2k \lambda}{r} = \frac{2 \times 9 \times 10^9 \times 4 \times 10^{-6}}{0.5} = 1.44 \times 10^5 N/C\).

Practice Set

• Q1: Electric field at point P due to finite wire depends on:
  A) Length only
  B) Distance and angles subtended
  C) Charge only
  D) None
• Q2: For charged arc, electric field magnitude is:
  A) \(\frac{k \lambda}{r} (2 \sin \alpha)\)
  B) \(\frac{k \lambda}{r} (\cos \alpha)\)
  C) \(\frac{2k \lambda}{r}\)
  D) \(\frac{k \lambda}{r} (\alpha)\)
• Q3: Electric field at 1 m from infinite wire with \(\lambda = 1 \times 10^{-6} C/m\) is:
  A) \(1.8 \times 10^4 N/C\)
  B) \(3.6 \times 10^4 N/C\)
  C) \(9 \times 10^3 N/C\)
  D) \(4.5 \times 10^4 N/C\)

Answer Key

• Q1: B) Distance and angles subtended - Formula includes \(r\) and \(\sin \alpha\).
• Q2: A) \(\frac{k \lambda}{r} (2 \sin \alpha)\) - Standard formula for charged arc.
• Q3: B) \(3.6 \times 10^4 N/C\) - \(E = \frac{2k \lambda}{r} = \frac{2 \times 9 \times 10^9 \times 1 \times 10^{-6}}{1} = 3.6 \times 10^4\).

Memory Aids

• Infinite wire field is twice that of half the finite wire at same distance.
• Arc field depends on \(2 \sin \alpha\) factor.
• Use symmetry and angles to simplify calculations.

Electric Field Charged Ring

Key Concepts and Formula

• Ring radius: \(a\), total charge: \(Q\), point P on axis at distance \(x\).
• Electric field at P: \(E = \frac{k Q x}{(a^2 + x^2)^{3/2}}\).
• Field direction along axis OP due to symmetry.
• At center (\(x=0\)), \(E=0\) due to cancellation.
• Maximum field occurs at \(x = \frac{a}{\sqrt{2}}\).

Solved Examples

Ex 1

• Find electric field at 0.1 m on axis of ring with \(a=0.05 m\), \(Q=2 \times 10^{-6} C\).
• \(E = \frac{9 \times 10^9 \times 2 \times 10^{-6} \times 0.1}{(0.05^2 + 0.1^2)^{3/2}} = 5.76 \times 10^4 N/C\).

Practice Set

• Q1: Electric field at center of charged ring is:
  A) Maximum
  B) Zero
  C) Depends on charge
  D) Infinite
• Q2: Formula for electric field at distance \(x\) on axis is:
  A) \(\frac{k Q}{x^2}\)
  B) \(\frac{k Q x}{(a^2 + x^2)^{3/2}}\)
  C) \(\frac{k Q a}{(a^2 + x^2)^{3/2}}\)
  D) \(\frac{k Q}{(a^2 + x^2)}\)
• Q3: For ring radius 0.1 m, at what \(x\) is field maximum?
  A) 0.1 m
  B) 0.05 m
  C) 0.07 m
  D) 0.071 m

Answer Key

• Q1: B) Zero - Symmetry causes field cancellation at center.
• Q2: B) \(\frac{k Q x}{(a^2 + x^2)^{3/2}}\) - Standard formula for ring axis field.
• Q3: D) 0.071 m - Maximum at \(x = \frac{a}{\sqrt{2}} = 0.1/1.414 = 0.071 m\).

Memory Aids

• Field zero at ring center due to symmetry.
• Field formula includes \(x\) in numerator and \((a^2 + x^2)^{3/2}\) in denominator.
• Maximum field at \(x = a/\sqrt{2}\).

Gauss's Law

Fundamental Concept

• Gauss's law: Total electric flux \( \Phi \) through a closed surface \( S \) is \( \frac{1}{\varepsilon_0} \) times the total charge \( q_{\text{enclosed}} \) inside the surface.
• Mathematically: \( \Phi = \oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{q_{\text{enclosed}}}{\varepsilon_0} \)
• Electric field \( \mathbf{E} \) due to a point charge \( Q \) at distance \( R \): \( E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2} \)
• For a spherical Gaussian surface of radius \( R \), flux \( \Phi = E \times 4\pi R^2 = \frac{Q}{\varepsilon_0} \)
• Gauss's law holds for any closed surface, regardless of shape or size.
• Only charges inside the Gaussian surface contribute to \( q_{\text{enclosed}} \).
• Electric field at the surface is due to all charges (inside and outside), but flux depends only on enclosed charge.
• Gaussian surface must not pass through discrete charges; it can pass through continuous charge distributions.
• Useful for calculating electric fields in symmetric charge distributions.
• Based on Coulomb's inverse square law; violation implies deviation from inverse square dependence.

Solved Examples

• Ex 1: Calculate electric flux through a spherical surface of radius 0.5 m enclosing a charge of 2 \( \mu C \).
  • Given: \( Q = 2 \times 10^{-6} C \), \( \varepsilon_0 = 8.85 \times 10^{-12} F/m \)
  • Flux: \( \Phi = \frac{Q}{\varepsilon_0} = \frac{2 \times 10^{-6}}{8.85 \times 10^{-12}} = 2.26 \times 10^{5} \; Nm^2/C \)
• Ex 2: Find electric field on the surface of a sphere of radius 0.3 m enclosing charge 5 \( \mu C \).
  • Electric field: \( E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2} = 9 \times 10^{9} \times \frac{5 \times 10^{-6}}{(0.3)^2} = 5 \times 10^{5} \; N/C \)

Practice Set

• Q1: Gauss's law relates electric flux through a closed surface to:
  • A) Total charge outside the surface
  • B) Total charge inside the surface
  • C) Electric field at a point
  • D) Potential difference across the surface
• Q2: A charge of 3 \( \mu C \) is enclosed in a spherical surface of radius 0.4 m. What is the electric flux through the surface? (\( \varepsilon_0 = 8.85 \times 10^{-12} F/m \))
  • A) \( 3.39 \times 10^{5} \; Nm^2/C \)
  • B) \( 1.07 \times 10^{5} \; Nm^2/C \)
  • C) \( 2.69 \times 10^{5} \; Nm^2/C \)
  • D) \( 5.39 \times 10^{5} \; Nm^2/C \)
• Q3: The electric field at the surface of a sphere of radius 0.2 m enclosing charge \( 4 \times 10^{-6} C \) is:
  • A) \( 9 \times 10^{5} \; N/C \)
  • B) \( 4.5 \times 10^{5} \; N/C \)
  • C) \( 1.8 \times 10^{6} \; N/C \)
  • D) \( 3.6 \times 10^{5} \; N/C \)

Answer Key

• Q1: B — Gauss's law states flux depends on total charge enclosed inside the surface.
• Q2: A — \( \Phi = \frac{Q}{\varepsilon_0} = \frac{3 \times 10^{-6}}{8.85 \times 10^{-12}} = 3.39 \times 10^{5} \; Nm^2/C \)
• Q3: C — \( E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2} = 9 \times 10^{9} \times \frac{4 \times 10^{-6}}{(0.2)^2} = 1.8 \times 10^{6} \; N/C \)

Memory Aids

• Flux \( \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} \) — remember flux depends only on enclosed charge.
• Electric field of point charge: \( E = k \frac{Q}{R^2} \), where \( k = \frac{1}{4\pi\varepsilon_0} \).
• Gaussian surface: closed surface enclosing charge; avoid passing through discrete charges.
• Gauss's law confirms Coulomb's inverse square law.

Applications of Gauss's Law

Electric Field of an Infinitely Long Line Charge

• Charge distributed uniformly along a line with linear charge density \(\lambda\) (C/m).
• Electric field magnitude: \(E = \frac{\lambda}{2\pi\epsilon_0 r}\), where \(r\) is radial distance (m).
• Field direction: radially outward from the line charge.

Electric Field of an Infinitely Large Non-Conducting Thin Sheet

• Surface charge density \(\sigma\) (C/m²) uniformly spread.
• Electric field magnitude: \(E = \frac{\sigma}{2\epsilon_0}\), constant and independent of distance.
• Field direction: perpendicular to the sheet on both sides.

Electric Field of an Infinitely Large Charged Conducting Sheet

• Charges free to move, uniformly distributed on surface.
• Electric field magnitude: \(E = \frac{\sigma}{\epsilon_0}\), twice that of non-conducting sheet.
• Field direction: perpendicular to the surface, only on one side.

Electric Field of a Uniformly Charged Spherical Shell or Solid Conducting Sphere

• Total charge \(Q\) on surface.
• Outside (\(r \geq R\)): \(E = \frac{kQ}{r^2}\), behaves like point charge.
• Inside (\(r < R\)): \(E = 0\), no electric field inside shell.

Electric Field of a Uniformly Charged Solid Non-Conducting Sphere

• Charge distributed uniformly with volume charge density \(\rho\) (C/m³).
• Outside (\(r \geq R\)): \(E = \frac{kQ}{r^2}\), point charge behavior.
• Inside (\(r < R\)): \(E = \frac{kQr}{R^3}\), increases linearly with \(r\).

Solved Examples

Ex 1

Find the electric field at 0.1 m from an infinitely long line charge with \(\lambda = 5 \times 10^{-6} \text{ C/m} \).

• Use \(E = \frac{\lambda}{2\pi\epsilon_0 r}\).
• \(\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \).
• Calculate \(E = \frac{5 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 0.1} \approx 8.99 \times 10^{4} \text{ N/C} \).

Ex 2

Calculate the electric field just outside a charged conducting sheet with surface charge density \(\sigma = 1 \times 10^{-6} \text{ C/m}^2\).

• Use \(E = \frac{\sigma}{\epsilon_0}\).
• Calculate \(E = \frac{1 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 1.13 \times 10^{5} \text{ N/C} \).

Practice Set

Q1

What is the electric field at 0.2 m from an infinitely long line charge with \(\lambda = 2 \times 10^{-6} \text{ C/m} \)?

• A) \(1.8 \times 10^{4} \text{ N/C}\)
• B) \(1.8 \times 10^{5} \text{ N/C}\)
• C) \(3.6 \times 10^{4} \text{ N/C}\)
• D) \(3.6 \times 10^{5} \text{ N/C}\)

Q2

Electric field due to an infinitely large non-conducting sheet with surface charge density \(\sigma = 4 \times 10^{-6} \text{ C/m}^2\) is:

• A) \(2.26 \times 10^{5} \text{ N/C}\)
• B) \(4.52 \times 10^{5} \text{ N/C}\)
• C) \(1.13 \times 10^{5} \text{ N/C}\)
• D) \(9.04 \times 10^{5} \text{ N/C}\)

Q3

Inside a uniformly charged solid sphere of radius 0.1 m and total charge \(Q = 1 \times 10^{-6} \text{ C}\), what is the electric field at \(r = 0.05 \text{ m}\)?

• A) \(4.5 \times 10^{4} \text{ N/C}\)
• B) \(9.0 \times 10^{4} \text{ N/C}\)
• C) \(1.8 \times 10^{5} \text{ N/C}\)
• D) \(0 \text{ N/C}\)

Answer Key

Q1

Answer: A) \(1.8 \times 10^{4} \text{ N/C}\)
Calculation: \(E = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 0.2} \approx 1.8 \times 10^{4} \text{ N/C}\).

Q2

Answer: A) \(2.26 \times 10^{5} \text{ N/C}\)
Calculation: \(E = \frac{\sigma}{2\epsilon_0} = \frac{4 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \approx 2.26 \times 10^{5} \text{ N/C}\).

Q3

Answer: A) \(4.5 \times 10^{4} \text{ N/C}\)
Calculation: \(E = \frac{kQr}{R^3} = \frac{9 \times 10^{9} \times 1 \times 10^{-6} \times 0.05}{(0.1)^3} = 4.5 \times 10^{4} \text{ N/C}\).

Memory Aids

• Line charge field \(\propto 1/r\).
• Non-conducting sheet field constant: \(E = \sigma/(2\epsilon_0)\).
• Conducting sheet field double non-conducting: \(E = \sigma/\epsilon_0\).
• Spherical shell: zero inside, \(\propto 1/r^2\) outside.
• Solid sphere inside field \(\propto r\), outside \(\propto 1/r^2\).