Class 12 Physics Case Study Worksheet Chapter 1 Electric Charges and Fields | Detailed Solutions & Marking Scheme
VEDANT WORKSHEET SERIES
Class XII – Physics
Chapter 1: Electric Charges and Fields
Time: 1 Hour
Maximum Marks: 20
SECTION A – CASE STUDY BASED QUESTIONS (5 × 4 = 20 Marks)
Case Study 1: Electrostatic Charges in Daily Life
During winter, Riya notices that when she removes her woollen sweater, it crackles and sometimes small sparks are seen in the dark. This happens due to transfer of electrons between the sweater and her body because of friction. The body acquires electrostatic charge, which may later discharge through nearby conductors. Such phenomena demonstrate the conservation of electric charge and the behaviour of charges on insulating materials.
Answer the following questions:
a) What type of charging occurs when two objects are rubbed together? (1 Mark)
b) State the law of conservation of charge. (1 Mark)
c) If an object initially has zero charge and gains (3 \times 10^{12}) electrons during rubbing, calculate the net charge acquired by it.
(Charge of electron = (1.6 \times 10^{-19}) C) (2 Marks)
Case Study 2: Coulomb’s Law in Electrostatic Interactions
Two small metallic spheres are placed on insulating stands in a laboratory. They are given charges and allowed to interact. The force between them depends on the magnitude of their charges and the distance between them. This relationship is described by Coulomb’s Law, which is fundamental in electrostatics and explains forces between stationary charges.
Answer the following questions:
a) Write the mathematical expression for Coulomb’s law. (1 Mark)
b) If the distance between two charges is doubled, how does the electrostatic force change? (1 Mark)
c) Two charges (q_1 = 2 \times 10^{-6}) C and (q_2 = 4 \times 10^{-6}) C are separated by 3 m in air. Calculate the electrostatic force between them.
((k = 9 \times 10^9 , Nm^2C^{-2})) (2 Marks)
Case Study 3: Electric Field Around Charges
Electric charges produce an electric field in the space surrounding them. This field can be visualized using electric field lines, which show the direction and strength of the electric field. Field lines originate from positive charges and terminate at negative charges. The density of field lines represents the magnitude of the field.
Answer the following questions:
a) Define electric field intensity at a point. (1 Mark)
b) In which direction do electric field lines originate and terminate? (1 Mark)
c) Calculate the electric field at a point 0.2 m away from a point charge of (5 \times 10^{-6}) C in vacuum.
((k = 9 \times 10^9 , Nm^2C^{-2})) (2 Marks)
Case Study 4: Electric Dipole in an Electric Field
An electric dipole consists of two equal and opposite charges separated by a small distance. Dipoles are commonly found in molecules such as water. When a dipole is placed in a uniform electric field, it experiences a torque that tends to align the dipole with the direction of the field. This principle is used in several devices and molecular physics applications.
Answer the following questions:
a) Define an electric dipole moment. (1 Mark)
b) What happens to a dipole when placed in a uniform electric field? (1 Mark)
c) A dipole with charges (+q) and (-q) separated by distance (2a) is placed in a uniform electric field (E). Write the expression for the torque acting on it. (2 Marks)
Case Study 5: Applications of Gauss’s Theorem
Gauss’s theorem provides a powerful method to calculate electric fields for highly symmetrical charge distributions. It states that the total electric flux through a closed surface is proportional to the charge enclosed within the surface. This theorem helps in calculating the electric field due to an infinitely long straight wire, infinite plane sheet, and spherical shell.
Answer the following questions:
a) State Gauss’s theorem in electrostatics. (1 Mark)
b) What is the electric field inside a uniformly charged thin spherical shell? (1 Mark)
c) Using Gauss’s law, write the expression for electric field due to an infinitely long straight wire with linear charge density (λ) at a distance ( r ). (2 Marks)
DETAILED ANSWER KEY WITH MARKING SCHEME
Class XII – Physics | Chapter 1: Electric Charges and Fields
Worksheet: Vedant Worksheet Series (2026–27)
Case Study 1: Electrostatic Charges in Daily Life
a) Answer:
Charging by friction (also called triboelectric charging).
Marks: 1
b) Answer:
The law of conservation of charge states that total electric charge of an isolated system remains constant; charge can neither be created nor destroyed, only transferred.
Marks: 1
c) Solution:
Number of electrons gained, n = 3 × 10^12
Charge of one electron, e = 1.6 × 10^-19 C
Net charge,
Q = - n × e
Q = - (3 × 10^12) × (1.6 × 10^-19)
Q = - 4.8 × 10^-7 C
Final Answer: -4.8 × 10^-7 C
Marking Scheme:
Formula (Q = ne) → 1 mark
Substitution & calculation → 1 mark
Case Study 2: Coulomb’s Law
a) Answer:
F = k q1 q2 / r^2
(where k = 1 / 4πε₀)
Marks: 1
b) Answer:
If distance is doubled (r → 2r):
F' = F / 4
So, force becomes one-fourth.
Marks: 1
c) Solution:
Given:
q1 = 2 × 10^-6 C
q2 = 4 × 10^-6 C
r = 3 m
k = 9 × 10^9
F = k q1 q2 / r^2
F = (9 × 10^9 × 2 × 10^-6 × 4 × 10^-6) / (3)^2
F = (9 × 10^9 × 8 × 10^-12) / 9
F = 8 × 10^-3 N
Final Answer: 8 × 10^-3 N
Marking Scheme:
Formula → 1 mark
Substitution & simplification → 1 mark
Case Study 3: Electric Field
a) Answer:
Electric field intensity at a point is defined as the force experienced by a unit positive test charge placed at that point.
E = F / q
Marks: 1
b) Answer:
Electric field lines originate from positive charges and terminate on negative charges.
Marks: 1
c) Solution:
Given:
q = 5 × 10^-6 C
r = 0.2 m
k = 9 × 10^9
E = kq / r^2
E = (9 × 10^9 × 5 × 10^-6) / (0.2)^2
E = (45 × 10^3) / 0.04
E = 1.125 × 10^6 N/C
Final Answer: 1.125 × 10^6 N/C
Marking Scheme:
Formula → 1 mark
Calculation → 1 mark
Case Study 4: Electric Dipole
a) Answer:
Electric dipole moment is defined as the product of magnitude of one charge and separation between charges.
p = q × 2a
(Direction: from -ve to +ve)
Marks: 1
b) Answer:
A dipole placed in a uniform electric field experiences a torque that tends to align it along the direction of the electric field.
Marks: 1
c) Answer:
Torque on dipole:
τ = p E sinθ
(where θ is angle between dipole moment and electric field)
Marks: 2
(Formula → 1, explanation of terms → 1)
Case Study 5: Gauss’s Theorem
a) Answer:
Gauss’s theorem states that the total electric flux through a closed surface is equal to 1/ε₀ times the net charge enclosed by the surface.
Φ = Q / ε₀
Marks: 1
b) Answer:
Electric field inside a uniformly charged thin spherical shell is zero.
Marks: 1
c) Answer:
Electric field due to infinitely long straight wire:
E = λ / (2π ε₀ r)
Marks: 2
(Formula → 1, correct variables → 1)
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