Class 12 Physics QP Worksheet Chapter 1 Electric Charges and Fields | Detailed Solutions & Marking Scheme
VEDANT WORKSHEET SERIES
Class: XII
Subject: Physics
Chapter 1: Electric Charges and Fields
Time: 1 Hour
Maximum Marks: 30
Section A – (1 × 7 = 7 Marks)
1. Two point charges +2 μC and –2 μC are separated by a distance of 10 cm in vacuum. The force between them will be:
(A) Attractive and equal to (3.6,N)
(B) Repulsive and equal to (3.6,N)
(C) Attractive and equal to (0.36,N)
(D) Repulsive and equal to (0.36,N)
2. According to the principle of superposition, the net force on a charge due to multiple charges is:
(A) Product of individual forces
(B) Vector sum of individual forces
(C) Scalar sum of individual forces
(D) Average of all forces
3. The electric field at a point due to a point charge depends on:
(A) Only magnitude of the charge
(B) Only distance from the charge
(C) Magnitude of charge and distance from it
(D) Nature of surrounding medium only
4. The number of electric field lines passing normally through a surface is called:
(A) Electric field intensity
(B) Electric flux
(C) Electric potential
(D) Electric charge density
5. The electric field inside a uniformly charged thin spherical shell is:
(A) Zero everywhere
(B) Same as outside field
(C) Maximum at centre
(D) Depends on radius
6. Assertion (A): The electric field lines never intersect each other.
Reason (R): At any point, the electric field has a unique direction.
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
7. Assertion (A): The electric field at the centre of an electric dipole is zero.
Reason (R): Electric fields due to equal and opposite charges cancel at the midpoint.
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Section B – (2 × 4 = 8 Marks)
8. State the law of conservation of electric charge. Give one example demonstrating this principle in electrostatics.
9. Write the expression for Coulomb’s law in vector form. Mention the factors on which the electrostatic force between two charges depends.
10. Define electric dipole moment. Write its SI unit and explain its direction.
11. Explain the concept of electric field lines. Write any two important properties of electric field lines.
Section C – (3 × 2 = 6 Marks)
12. Using Gauss’s theorem, obtain the expression for the electric field due to an infinitely long straight uniformly charged wire. State the assumptions made.
13. A small electric dipole of dipole moment (p) is placed in a uniform electric field (E).
(a) Write the expression for the torque acting on the dipole.
(b) When will the torque be maximum?
(c) What happens when the dipole is aligned parallel to the field?
Section D – (5 × 1 = 5 Marks)
14. Derive the expression for the electric field due to a uniformly charged thin spherical shell at a point:
(a) Outside the shell
(b) Inside the shell
Use Gauss’s theorem in your derivation and explain the physical significance of the result.
Section E – Case Study Based Question (4 Marks)
A photocopy machine and laser printer use electrostatic principles for printing. In these machines, a charged drum attracts toner particles due to electrostatic force. These particles are then transferred to paper where they form images. The principle involves electric charges, electric fields, and electrostatic attraction.
15. Answer the following questions based on the above case:
(a) What type of force is responsible for attracting toner particles to the charged drum? (1 mark)
(b) If the electric field near the drum increases, what will happen to the force acting on charged toner particles? (1 mark)
(c) Explain how the principle of superposition of electric fields may influence the movement of multiple charged toner particles in the printing process. (2 marks)
Here is the Detailed Answer Key with Marking Scheme in easy-to-copy plain maths format (CBSE examiner style):
VEDANT WORKSHEET SERIES – ANSWER KEY (DETAILED)
Class 12 Physics – Chapter 1: Electric Charges and Fields
Maximum Marks: 30
Section A (1 × 7 = 7 Marks)
1.
F = k q1 q2 / r²
= (9 × 10^9 × 2 × 10^-6 × 2 × 10^-6) / (0.1)^2
= (9 × 4 × 10^-12) / 0.01
= 36 × 10^-12 / 0.01
= 3.6 N
Charges are opposite → force is attractive
Answer: (A)
Marks: 1
2. Net force = vector sum of individual forces
Answer: (B)
Marks: 1
3. Electric field E = kQ / r²
Depends on charge and distance
Answer: (C)
Marks: 1
4. Electric flux = number of field lines passing normally through a surface
Answer: (B)
Marks: 1
5. Inside spherical shell → E = 0
Answer: (A)
Marks: 1
6. Assertion-Reason
Field lines do not intersect because direction of field at a point is unique
Answer: (A)
Marks: 1
7. Assertion-Reason
At centre of dipole, fields cancel
Answer: (A)
Marks: 1
Section B (2 × 4 = 8 Marks)
8. Law of Conservation of Charge
Charge can neither be created nor destroyed, only transferred.
Example:
Rubbing glass rod with silk → charges appear but total charge remains zero.
Marking Scheme:
• Statement: 1 mark
• Example: 1 mark
9. Coulomb’s Law (Vector Form)
F12 = (1 / 4πϵ0) × (q1 q2 / r²) r̂
Depends on:
• Magnitude of charges
• Distance between charges
• Medium (permittivity)
Marking Scheme:
• Formula: 1 mark
• Factors (any two): 1 mark
10. Electric Dipole Moment
p = q × 2a
SI unit: Coulomb meter (C m)
Direction: from negative to positive charge
Marking Scheme:
• Definition/formula: 1 mark
• Unit + direction: 1 mark
11. Electric Field Lines
Imaginary lines representing direction of electric field.
Properties (any two):
• Start from positive and end at negative charges
• Never intersect
• Tangent gives direction of field
Marking Scheme:
• Definition: 1 mark
• Properties (any two): 1 mark
Section C (3 × 2 = 6 Marks)
12. Electric Field due to Infinite Line Charge
Using Gauss Law:
Φ = E × (2πrL)
Charge enclosed = λL
So,
E × (2πrL) = λL / ϵ0
E = λ / (2πϵ0 r)
Assumptions:
• Infinite length
• Uniform charge distribution
• Cylindrical symmetry
Marking Scheme:
• Gaussian surface & setup: 1 mark
• Final formula: 1 mark
• Assumptions: 1 mark
13. Dipole in Electric Field
(a) Torque:
τ = pE sinθ
(b) Maximum when sinθ = 1 → θ = 90°
(c) When parallel (θ = 0):
τ = 0 → stable equilibrium
Marking Scheme:
• Formula: 1 mark
• Condition of max torque: 1 mark
• Explanation: 1 mark
Section D (5 × 1 = 5 Marks)
14. Electric Field due to Spherical Shell
Using Gauss Law:
Φ = E × 4πr²
Charge enclosed = Q
So,
E × 4πr² = Q / ϵ0
E = (1 / 4πϵ0) × Q / r²
(a) Outside shell:
Acts like point charge
E = kQ / r²
(b) Inside shell:
Charge enclosed = 0
So, E = 0
Physical Significance:
• Field inside conductor is zero
• Charges reside on surface
Marking Scheme:
• Gaussian law application: 2 marks
• Outside result: 1 mark
• Inside result: 1 mark
• Significance: 1 mark
Section E – Case Study (4 Marks)
15.
(a) Electrostatic force
Marks: 1
(b) F = qE → Force increases
Marks: 1
(c) Superposition Principle:
Net field = vector sum of fields due to all charges
In printer:
• Multiple toner particles experience combined fields
• Their motion depends on resultant field direction
Marking Scheme:
• Statement of principle: 1 mark
• Application explanation: 1 mark
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