Class 12 Physics Test Paper Chapter 1 Electric Charges and Fields | CBSE Board | Detailed Solutions & Marking Scheme
VEDANT IGNITE TEST SERIES
Class: 12 | Subject: Physics
Chapter: Electric Charges and Fields
Maximum Marks: 30
(Competency-based questions aligned with CBSE pattern and PYQ trends)
SECTION A (1 × 7 = 7 Marks)
MCQs (5 Questions)
Q1. Two identical small metal spheres A and B carry charges +6 μC and –2 μC respectively. They are brought into contact and separated. What will be the final charge on each sphere?
(a) +2 μC each (b) +4 μC each (c) –2 μC each (d) +1 μC each
Q2. A point charge +q is placed at the centre of a cube. The electric flux through one face of the cube is:
(a) q/ε₀ (b) q/6ε₀ (c) q/12ε₀ (d) zero
Q3. Two charges +q and –q are placed at a small distance forming an electric dipole. At a point on the axial line far away from the dipole, the electric field varies as:
(a) 1/r² (b) 1/r³ (c) 1/r (d) constant
Q4. A charge of 2 μC is placed in an electric field of magnitude 4 × 10⁵ N/C. The force experienced by the charge is:
(a) 0.2 N (b) 0.4 N (c) 0.8 N (d) 0.08 N
Q5. Which of the following statements about electric field lines is correct?
(a) They start from negative charges and end on positive charges
(b) They never intersect each other
(c) They form closed loops
(d) Their direction is opposite to electric field
Assertion–Reason (2 Questions)
Q6.
Assertion (A): Electric field inside a conductor in electrostatic equilibrium is zero.
Reason (R): Free charges inside the conductor redistribute themselves until the net electric field becomes zero.
Options:
(a) Both A and R are true and R is correct explanation of A
(b) Both A and R are true but R is not correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Q7.
Assertion (A): Electric field lines are perpendicular to the surface of a charged conductor.
Reason (R): If the field lines were not perpendicular, charges would move along the surface.
Options:
(a) Both A and R are true and R is correct explanation of A
(b) Both A and R are true but R is not correct explanation
(c) A is true but R is false
(d) A is false but R is true
SECTION B (2 × 4 = 8 Marks)
Q8. Two point charges +3 μC and +12 μC are separated by 2 m in air.
Calculate the magnitude of electrostatic force between them. State whether the force is attractive or repulsive.
Q9. Define electric field. A charge of 5 μC experiences a force of 0.1 N in an electric field.
Calculate the magnitude of the electric field.
Q10. Draw the electric field lines for:
(a) an isolated positive point charge
(b) a dipole
State one property of electric field lines.
Q11. State Gauss’s law in electrostatics.
Explain briefly why electric flux through a closed surface depends only on the enclosed charge.
SECTION C (3 × 2 = 6 Marks)
Q12. Derive the expression for electric field intensity at a point on the axial line of an electric dipole.
Also state the direction of the field.
Q13. A uniformly charged thin spherical shell has charge Q and radius R.
Using Gauss’s law, determine the electric field:
(a) outside the shell
(b) inside the shell.
SECTION D (5 × 1 = 5 Marks)
Q14.
Three point charges are placed at the corners of an equilateral triangle of side 0.5 m. Two charges are +2 μC each and the third charge is –2 μC.
(a) Draw the configuration of charges.
(b) Determine the net electrostatic force on the –2 μC charge due to the other two charges.
(c) State the principle used in solving this problem.
SECTION E – CASE STUDY (4 Marks)
Q15. Electric Field Around Charged Objects
During an experiment, a teacher places a positively charged metal sphere on an insulating stand. Students observe the electric field pattern around it using small test charges and field mapping techniques.
Based on the observation, answer the following questions:
(a) What is the direction of electric field around a positively charged sphere? (1)
(b) If the distance from the charge is doubled, how does the electric field change? (1)
(c) Two charges +q and +4q are placed at a distance r apart. Find the point on the line joining them where the electric field becomes zero. (2)
VEDANT IGNITE TEST SERIES
Class 12 – Physics
Chapter: Electric Charges and Fields
DETAILED SOLVED ANSWER KEY (With Stepwise Marking Scheme – CBSE Style)
✅ SECTION A (1 × 7 = 7 Marks)
Q1.
Total charge = +6 μC − 2 μC = +4 μC
Since spheres are identical, charge is equally shared:
Charge on each = 4/2 = +2 μC
✔ Correct Option: (a) (1 mark)
Q2.
By Gauss Law:
Total flux through cube = q/ε₀
Cube has 6 faces, so flux per face = (q/ε₀)/6
= q/6ε₀
✔ Correct Option: (b) (1 mark)
Q3.
Electric field due to dipole at far axial point:
E ∝ 1/r³
✔ Correct Option: (b) (1 mark)
Q4.
Given:
q = 2 × 10⁻⁶ C
E = 4 × 10⁵ N/C
Formula:
F = qE
Substitute:
F = (2 × 10⁻⁶)(4 × 10⁵)
F = 8 × 10⁻¹
F = 0.8 N
✔ Correct Option: (c) (1 mark)
Q5.
Electric field lines never intersect; otherwise direction becomes undefined.
✔ Correct Option: (b) (1 mark)
Q6.
Assertion: True
Reason: True
Explanation: Charges redistribute until internal field becomes zero.
✔ Correct Option: (a) (1 mark)
Q7.
Assertion: True
Reason: True
Explanation: Tangential component would cause motion of charges.
✔ Correct Option: (a) (1 mark)
✅ SECTION B (2 × 4 = 8 Marks)
Q8. Electrostatic Force
Step 1: Formula (½ mark)
F = k q₁q₂ / r²
Step 2: Substitution (1 mark)
= 9×10⁹ × (3×10⁻⁶ × 12×10⁻⁶) / (2)²
Step 3: Simplification (1 mark)
= 9×10⁹ × 36×10⁻¹² / 4
= 9×10⁹ × 9×10⁻¹²
Step 4: Final Answer (1 mark)
F = 0.081 N
Step 5: Nature (½ mark)
Both charges positive → Repulsive force
Q9. Electric Field
Step 1: Definition (1 mark)
Electric field is force per unit positive test charge.
Step 2: Formula (½ mark)
E = F/q
Step 3: Substitution (1 mark)
= 0.1 / (5×10⁻⁶)
Step 4: Calculation (½ mark)
E = 2 × 10⁴ N/C
✔ Final Answer: 2 × 10⁴ N/C
Q10. Electric Field Lines
(a) Positive charge (½ mark)
Field lines radiate outward symmetrically
(b) Dipole (½ mark)
Lines originate from + and terminate at −
Diagram (1 mark)
(Any correct diagram acceptable)
Property (1 mark)
Field lines never intersect / represent direction of field
Q11. Gauss Law
Statement (1 mark)
Electric flux through closed surface = q_enc / ε₀
Explanation (1 mark)
External charges produce equal inward and outward flux → cancel
Only enclosed charge contributes to net flux
✅ SECTION C (3 × 2 = 6 Marks)
Q12. Axial Field of Dipole
Step 1: Dipole moment (½ mark)
p = q(2a)
Step 2: Formula (1 mark)
E = (1/4πɛ₀)(2p/r³)
Step 3: Direction (½ mark)
Along axis from + to − charge
Q13. Spherical Shell
Using Gauss Law
(a) Outside shell (1 mark)
Gaussian surface encloses total charge Q
E(4πr²) = Q/ε₀
E = (1/4πɛ₀)(Q/r²)
(b) Inside shell (1 mark)
Enclosed charge = 0
E = 0
✅ SECTION D (5 Marks)
Q14. Force on –2 μC
Step 1: Formula (1 mark)
F = k q₁q₂ / r²
Step 2: Calculation (1 mark)
= 9×10⁹ × (2×10⁻⁶ × 2×10⁻⁶) / (0.5)²
Step 3: Simplification (1 mark)
= 9×10⁹ × 4×10⁻¹² / 0.25
= 0.144 N
Step 4: Resultant Force (1 mark)
Two equal forces at 60°
R = 2F cos30°
= 2 × 0.144 × 0.866
R ≈ 0.25 N
Step 5: Direction & Principle (1 mark)
Direction: toward midpoint between charges
Principle: Superposition principle
✅ SECTION E (CASE STUDY – 4 Marks)
Q15
(a) Direction (1 mark)
Electric field is directed away from positive charge
(b) Variation (1 mark)
E ∝ 1/r²
If distance doubles:
E_new = E/4
(c) Zero Field Position (2 marks)
Step 1: Equation (1 mark)
kq/x² = 4kq/(r−x)²
Step 2: Simplification (½ mark)
(r−x)² = 4x²
Step 3: Solve (½ mark)
r − x = 2x
r = 3x
x = r/3
✔ Final Answer: Point lies at distance r/3 from +q
✅ OVERALL EXAMINER NOTES (VERY IMPORTANT FOR YOUR TEST SERIES)
✔ Award full marks for correct formula even if minor calculation mistake
✔ Units must be correct for full marks
✔ Diagrams: give marks if conceptually correct (no need perfect drawing)
✔ Step marking strictly followed (CBSE pattern)
✔ In numericals:
Formula → ½ to 1 mark
Substitution → 1 mark
Final answer → ½ to 1 mark
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