Class 10 Physics — Chapter 15: Electrostatics

Quran Education Point
Subject: Physics
Class: 10th
Board: Punjab

A. Multiple Choice Questions (MCQs)

15.1 Electrical breakdown is:

(a) The process of ionizing a gas due to a strong electric field
(b) The process of neutralizing a charged object
(c) The process of charging an insulator
(d) The process of discharging a conductor

Correct Answer: (a) The process of ionizing a gas due to a strong electric field

15.2 Which of the following is a visible example of electrical breakdown?

(a) Corona discharge
(b) Electric current
(c) Magnetic field
(d) Electric potential

Correct Answer: (a) Corona discharge

15.3 What happens when two unlike charges are brought close to each other?

(a) They attract
(b) They repel
(c) They neutralize each other
(d) Nothing happens

Correct Answer: (a) They attract

15.4 According to Coulomb’s law, what happens to the attraction between two oppositely charged objects as their distance of separation increases?

(a) Decreases
(b) Increases
(c) Remains unchanged
(d) Cannot be determined

Correct Answer: (a) Decreases

15.5 Which device works on the principle of electrostatics?

(a) Electric bell
(b) Photocopier
(c) Electric heater
(d) Transformer

Correct Answer: (b) Photocopier

B. Short Answer Questions

15.1 State Coulomb’s law of electrostatics and write its mathematical form.

Answer:

Statement: The force of attraction or repulsion between two point electric charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centers.

Mathematical Form:

\[ F=k\frac{|q_1q_2|}{r^2} \]

Where:

\(F\) = Electrostatic force
\(q_1\) and \(q_2\) = Magnitudes of the charges
\(r\) = Distance between the charges
\(k\) = Coulomb’s constant

\[ k=\frac{1}{4\pi\varepsilon_0} \approx 9\times10^9\ \text{N m}^2\text{C}^{-2} \]

15.2 What is an electric field?

Answer:

An electric field is a region or space around an electric charge in which another charge experiences an electrostatic force of attraction or repulsion.

15.3 Give two examples each of electrical conductors and electrical insulators.

Answer:

Conductors: Copper and aluminium.
Insulators: Rubber and glass.

15.4 List some applications of electrostatics in daily life.

Answer:

Photocopiers (Xerography): Use electrostatic charges to attract toner powder onto paper.
Electrostatic Paint Spraying: Ensures an even coat of paint on metallic surfaces.
Electrostatic Precipitators: Used in factory chimneys to remove smoke and dust particles.

15.5 What happens when like charges and unlike charges are brought close to each other?

Answer:

Like charges repel each other. For example, positive and positive charges or negative and negative charges.
Unlike charges attract each other. For example, positive and negative charges.

15.6 What is meant by electrical breakdown, and when does it occur?

Answer:

Definition: Electrical breakdown is the process in which an insulating material, such as air, becomes an electrical conductor due to ionization.

Condition: It occurs when the applied electric field intensity exceeds the dielectric strength of the insulating material.

15.7 What are the dangers of static electricity?

Answer:

Sparks and Fire Hazards: Static discharge near flammable gases or fuel vapours can trigger explosions, particularly at gas stations and chemical plants.
Damage to Electronics: Sensitive electronic components can be damaged by sudden electrostatic discharges (ESD).
Lightning: Lightning is a massive static discharge between clouds or between clouds and the ground. It can cause serious property damage and fatal injuries.

C. Constructed Response Questions

15.1 Why do some materials gain electrons while others lose electrons when rubbed together? What property of the materials determines this behaviour?

Answer:

Reason: Different materials have different electron affinities and triboelectric properties. A material with more weakly bound outer electrons tends to lose electrons, while a material with a stronger tendency to attract electrons gains them.

Determining Property: Electron affinity and atomic structure, particularly how tightly the nucleus holds the valence electrons, determine this behaviour.

15.2 A charged comb attracts small pieces of paper, but after some time, the paper pieces fall off. Why does this happen?

Answer:

Initial Attraction: The charged comb induces a redistribution of charges in the neutral paper pieces. The side nearer the comb becomes effectively oppositely charged, producing an attractive electrostatic force.

Falling Off: When the paper touches the comb, electrons may transfer between the comb and the paper. The paper can acquire the same type of charge as the comb, causing repulsion and making the pieces fall off.

15.3 Why do dust particles often get attracted to television screens or computer monitors after they are turned ON?

Answer:

When a screen becomes electrically charged, its electric field induces charge separation in nearby neutral dust particles. The resulting electrostatic attraction draws the dust particles towards the screen.

15.4 If we rub a balloon on our dry hair and place it near a thin stream of water, the water bends towards the balloon. What does this tell us about the nature of water molecules?

Answer:

This indicates that water molecules are polar molecules. They have permanent electric dipoles with partially positive and partially negative ends. The static charge on the balloon attracts the oppositely charged side of the water molecules, causing the stream of water to bend towards the balloon.

15.5 A conductor and an insulator are both placed in an electric field. How will their behaviour differ, and why?

Answer:

Conductor: A conductor contains free electrons that can move in response to an electric field. In electrostatic equilibrium, the charges redistribute so that the internal electric field becomes zero. Excess charge resides on the conductor’s outer surface.
Insulator: An insulator does not contain freely moving electrons in the same way. Its bound charges shift slightly, causing dielectric polarization without free charge moving throughout the material.

D. Comprehensive Questions

15.1 Discuss the working of a gold leaf electroscope. How can it be used to determine the nature and magnitude of charge on a body?

Answer:

Structure and Working:

A gold leaf electroscope consists of a brass rod passing through an insulating stopper into a glass jar. Two thin gold leaves are attached to the lower end of the rod.

When a charged object touches the brass disc at the top, charge is transferred to the rod and gold leaves. The leaves acquire like charges and repel each other, causing them to diverge.

Determining the Magnitude of Charge:

The degree of divergence of the gold leaves gives an indication of the amount of charge present. A greater divergence generally indicates a greater charge, provided other conditions remain the same.

Determining the Nature of Charge:

Charge the electroscope with a known charge, such as a positive charge.
Bring the unknown charged body near the disc.
If the leaf divergence increases, the body has the same type of charge.
If the leaf divergence decreases, the body has the opposite type of charge.

15.2 How does Coulomb’s law help in understanding the interaction between charged particles? What factors affect the force between them?

Answer:

Understanding the Interaction:

Coulomb’s law quantifies the electrostatic force between stationary point charges. The forces are equal in magnitude and opposite in direction, acting along the line joining the centres of the charges.

Factors Affecting the Force:

1. Magnitude of Charges \((q_1,q_2)\): The force is directly proportional to the product of the magnitudes of the charges.

\[ F\propto |q_1q_2| \]

2. Distance of Separation \((r)\): The force is inversely proportional to the square of the distance.

\[ F\propto\frac{1}{r^2} \]

3. Medium: The permittivity of the medium affects the electrostatic force. A dielectric medium generally reduces the force compared with the force in a vacuum.

\[ k=\frac{1}{4\pi\varepsilon} \]

Here, \(\varepsilon\) is the permittivity of the medium.

15.3 Explain the concept of electric field intensity. How can the electric field of a point charge and a parallel plate capacitor be represented using field lines?

Answer:

Electric Field Intensity \((E)\):

Electric field intensity is the force experienced per unit positive test charge placed at a point in an electric field.

\[ E=\frac{F}{q_0} \]

Where:

\(E\) = Electric field intensity
\(F\) = Electric force
\(q_0\) = Positive test charge

It is a vector quantity measured in newtons per coulomb \((\text{N C}^{-1})\) or volts per metre \((\text{V m}^{-1})\).

Representation Using Field Lines:

Point Charge: Electric field lines point radially outward from a positive charge and radially inward towards a negative charge.
Parallel Plate Capacitor: The field is represented by uniform, parallel, equally spaced lines directed from the positive plate towards the negative plate, away from the edges.

15.4 Explain how and why electrical conductors and insulators behave differently in terms of charge transfer and movement of electrons.

Answer:

Conductors:

Conductors contain loosely bound valence electrons, also called free electrons. When charge is added at one point, it spreads rapidly across the surface because electrons can move freely through the metallic structure.

Insulators:

In insulators, electrons are tightly bound to individual atoms or molecules. Charges generally remain localized where they are placed because they cannot move freely throughout the material.

15.5 Discuss how electrostatics plays a role in real-life hazards, such as dust explosions in factories and static shocks from synthetic clothing. How can these hazards be minimized?

Answer:

Hazards:

Dust Explosions: Fine combustible particles, such as flour or coal dust, can accumulate static charge due to friction. A static spark may ignite the suspended dust cloud.
Static Shocks: Friction between synthetic clothing and the skin can build up static charge. When a person touches a grounded metal object, the charge may discharge, producing a shock.

Methods to Minimize These Hazards:

Grounding/Earthing: Ground machinery, pipes and storage containers to discharge accumulated static electricity safely.
Maintaining Humidity: Keeping indoor environments from becoming excessively dry can reduce static charge buildup.
Anti-Static Equipment: Use anti-static wrist straps, conductive flooring and anti-static sprays where appropriate.

E. Numerical Problems

Problem 15.1

Given:

Test charge, \(q=25\,\mu\text{C}=25\times10^{-6}\,\text{C}\)
Force, \(F=0.500\,\text{N}\)

To Find:

Electric field intensity, \(E=?\)

Formula:

\[ E=\frac{F}{q} \]

Calculation:

\[ E=\frac{0.500}{25\times10^{-6}} \]

\[ E=\frac{0.500}{2.5\times10^{-5}} \]

\[ E=2.0\times10^4\,\text{N C}^{-1} \]

Answer:

Electric field intensity = \(2.0\times10^4\,\text{N C}^{-1}\)

Problem 15.2

Given:

Charge 1, \(q_1=8\,\mu\text{C}=8\times10^{-6}\,\text{C}\)
Charge 2, \(q_2=4\,\mu\text{C}=4\times10^{-6}\,\text{C}\)
Distance, \(r=120\,\text{cm}=1.2\,\text{m}\)
Coulomb’s constant, \(k=9\times10^9\,\text{N m}^2\text{C}^{-2}\)

To Find:

Coulomb’s force, \(F=?\)
Nature of force

Formula:

\[ F=k\frac{|q_1q_2|}{r^2} \]

Calculation:

\[ F=\frac{(9\times10^9)(8\times10^{-6})(4\times10^{-6})}{(1.2)^2} \]

\[ F=\frac{0.288}{1.44} \]

\[ F=0.2\,\text{N} \]

Answer:

Magnitude of force = \(0.2\,\text{N}\)

Nature of force = Repulsive, since both charges are positive.

Problem 15.3

Given:

Initial distance, \(r_1=9\,\text{cm}=0.09\,\text{m}\)
Initial force, \(F_1=0.2\,\text{N}\)
New distance, \(r_2=3\,\text{cm}=0.03\,\text{m}\)

To Find:

New force, \(F_2=?\)

Formula:

According to Coulomb’s law,

\[ F\propto\frac{1}{r^2} \]

Therefore,

\[ F_1r_1^2=F_2r_2^2 \]

\[ F_2=F_1\left(\frac{r_1}{r_2}\right)^2 \]

Calculation:

\[ F_2=0.2\left(\frac{9}{3}\right)^2 \]

\[ F_2=0.2(3)^2 \]

\[ F_2=0.2\times9=1.8\,\text{N} \]

Answer:

New force = \(1.8\,\text{N}\)

Problem 15.4

Given:

Point charge, \(q=10\,\mu\text{C}=10\times10^{-6}\,\text{C}=10^{-5}\,\text{C}\)
Distance, \(r=0.5\,\text{m}\)
Coulomb’s constant, \(k=9\times10^9\,\text{N m}^2\text{C}^{-2}\)

To Find:

Electric field strength, \(E=?\)

Formula:

\[ E=k\frac{|q|}{r^2} \]

Calculation:

\[ E=\frac{(9\times10^9)(10^{-5})}{(0.5)^2} \]

\[ E=\frac{9\times10^4}{0.25} \]

\[ E=3.6\times10^5\,\text{N C}^{-1} \]

Answer:

Electric field strength = \(3.6\times10^5\,\text{N C}^{-1}\)

Problem 15.5

Given:

Charge on each sphere, \(q_1=q_2=q=3\,\mu\text{C}=3\times10^{-6}\,\text{C}\)
Distance, \(r=2.0\,\text{cm}=0.02\,\text{m}\)
Coulomb’s constant, \(k=9\times10^9\,\text{N m}^2\text{C}^{-2}\)

To Find:

Repulsive force, \(F=?\)

Formula:

\[ F=k\frac{q^2}{r^2} \]

Calculation:

\[ F=\frac{(9\times10^9)(3\times10^{-6})^2}{(0.02)^2} \]

\[ F=\frac{(9\times10^9)(9\times10^{-12})}{0.0004} \]

\[ F=\frac{0.081}{0.0004} \]

\[ F=202.5\,\text{N} \]

Answer:

Repulsive force = \(202.5\,\text{N}\)

Problem 15.6

Given:

Repulsive force, \(F=0.5\,\text{N}\)
Distance, \(r=3\,\text{cm}=0.03\,\text{m}\)
Equal charges, \(q_1=q_2=q\)
Coulomb’s constant, \(k=9\times10^9\,\text{N m}^2\text{C}^{-2}\)

To Find:

Charge on each sphere, \(q=?\)

Formula:

\[ F=k\frac{q^2}{r^2} \]

Rearranging,

\[ q^2=\frac{Fr^2}{k} \]

Calculation:

\[ q^2=\frac{0.5(0.03)^2}{9\times10^9} \]

\[ q^2=\frac{0.5\times0.0009}{9\times10^9} \]

\[ q^2=\frac{0.00045}{9\times10^9} \]

\[ q^2=5\times10^{-14}\,\text{C}^2 \]

Taking the square root,

\[ q=\sqrt{5\times10^{-14}} \]

\[ q\approx2.236\times10^{-7}\,\text{C} \]

Converting to microcoulombs,

\[ q\approx0.224\,\mu\text{C} \]

Answer:

Charge on each sphere = \(0.224\,\mu\text{C}\), or \(2.24\times10^{-7}\,\text{C}\).

Note: The supplied textbook key lists \(224\,\mu\text{C}\), but the calculation using the given SI values gives \(0.224\,\mu\text{C}\). The textbook key appears to contain a unit-conversion error.

Problem 15.7

Given:

Force, \(F=85\,\text{N}\)
Distance, \(r=3.2\,\text{cm}=0.032\,\text{m}\)
First charge, \(q_1=25\,\mu\text{C}=25\times10^{-6}\,\text{C}\)
Coulomb’s constant, \(k=9\times10^9\,\text{N m}^2\text{C}^{-2}\)

To Find:

Magnitude of the second charge, \(q_2=?\)

Formula:

\[ F=k\frac{|q_1q_2|}{r^2} \]

Rearranging,

\[ |q_2|=\frac{Fr^2}{k|q_1|} \]

Calculation:

\[ |q_2|=\frac{85(0.032)^2}{(9\times10^9)(25\times10^{-6})} \]

\[ |q_2|=\frac{85\times0.001024}{2.25\times10^5} \]

\[ |q_2|=\frac{0.08704}{2.25\times10^5} \]

\[ |q_2|\approx3.868\times10^{-7}\,\text{C} \]

Converting to microcoulombs,

\[ |q_2|\approx0.387\,\mu\text{C} \]

Answer:

Magnitude of the second charge = \(0.387\,\mu\text{C}\), or \(3.87\times10^{-7}\,\text{C}\).

Note: The supplied calculation’s final conversion to \(3.87\,\mu\text{C}\) is incorrect. Since \(1\,\mu\text{C}=10^{-6}\,\text{C}\), the correct result is approximately \(0.387\,\mu\text{C}\).

Notes by Sir Dilawar

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