Class 12 Physics Test Paper Chapter 1 Electric Charges and Fields | GSEB Board | Detailed Solutions & Marking Scheme
VEDANT IGNITE TEST SERIES
Class: 12
Subject: Physics
Board: GSEB
Chapter: 1 – Electric Charges and Fields
Time: 1 Hour
Marks: 30
SECTION A
(7 × 1 = 7 Marks)
Numerical Based MCQs
(Write the correct option)
Two charges (+2 \mu C) and (+3 \mu C) are separated by a distance of 1 m in vacuum. The electrostatic force between them is:
(a) 0.054 N (b) 0.036 N (c) 0.018 N (d) 0.027 NTwo identical charges each of (5 \times 10^{-6} C) are placed 2 m apart in vacuum. The force between them is:
(a) 0.1125 N (b) 0.056 N (c) 0.225 N (d) 0.45 NThe electric field at a distance of 0.3 m from a charge (2 \times 10^{-6} C) is:
(a) (2 \times 10^5) N/C (b) (4 \times 10^5) N/C (c) (1 \times 10^5) N/C (d) (6 \times 10^5) N/CA charge of (4 \times 10^{-9} C) experiences a force of (8 \times 10^{-5} N) in an electric field. The magnitude of the electric field is:
(a) (1 \times 10^4) N/C (b) (2 \times 10^4) N/C (c) (3 \times 10^4) N/C (d) (4 \times 10^4) N/CThe electric field due to a point charge (5 \mu C) at a distance of 2 m is:
(a) (1.125 \times 10^4) N/C (b) (2.25 \times 10^4) N/C (c) (4.5 \times 10^4) N/C (d) (9 \times 10^4) N/CTwo charges (+Q) and (-Q) are separated by a distance (2a). The electric field at the midpoint is:
(a) Zero (b) Along the line joining charges from +Q to −Q (c) Perpendicular to the line joining charges (d) InfiniteIf the distance between two charges is doubled, the electrostatic force becomes:
(a) Half (b) Double (c) One-fourth (d) Four times
SECTION B
(5 Questions – Write Any 3)
(3 × 2 = 6 Marks)
State Coulomb’s Law in electrostatics. Write its mathematical expression and explain the meaning of each term.
Define electric field. Write the expression for electric field due to a point charge.
Write the principle of superposition of electric forces.
Define electric field lines. Write any two properties of electric field lines.
Define electric flux and write its SI unit.
SECTION C
(5 Questions – Write Any 3)
(3 × 3 = 9 Marks)
Derive the expression for electric field due to a point charge at a distance (r).
Derive the expression for electrostatic force between two point charges (Coulomb’s Law) in vector form.
Derive the expression for electric field on the axial line of an electric dipole.
Explain electric field lines and draw diagrams showing field lines for:
(a) Positive charge
(b) Negative charge
(c) Electric dipoleDefine electric dipole. Derive the expression for electric dipole moment.
SECTION D
(3 Questions – Write Any 2)
(2 × 4 = 8 Marks)
Derive the expression for electric field on the equatorial line of an electric dipole.
State Gauss’s Law in electrostatics and derive its mathematical expression.
Using Gauss’s law, derive the expression for electric field due to an infinitely long straight charged wire.
VEDANT IGNITE TEST SERIES – ANSWER KEY
Class 12 | Physics | GSEB
Chapter: Electric Charges and Fields
Marks: 30
✅ SECTION A – MCQs (1 Mark Each)
(Answer + Key Step)
Q1
F = k q₁q₂ / r²
= (9×10⁹ × 2×10⁻⁶ × 3×10⁻⁶) / 1²
= 54×10⁻³ = 0.054 N
✔ Ans: (a)
Q2
F = (9×10⁹ × 5×10⁻⁶ × 5×10⁻⁶) / (2²)
= (9×10⁹ × 25×10⁻¹²) / 4
= 56.25×10⁻³ ≈ 0.056 N
✔ Ans: (b)
Q3
E = kQ / r²
= (9×10⁹ × 2×10⁻⁶) / (0.3)²
= (18×10³) / 0.09
= 2×10⁵ N/C
✔ Ans: (a)
Q4
E = F/q
= (8×10⁻⁵) / (4×10⁻⁹)
= 2×10⁴ N/C
✔ Ans: (b)
Q5
E = (9×10⁹ × 5×10⁻⁶) / (2²)
= (45×10³)/4
= 1.125×10⁴ N/C
✔ Ans: (a)
Q6
At midpoint, fields due to +Q and −Q are in same direction (from + to −).
✔ Ans: (b)
Q7
F ∝ 1/r²
If r → 2r → F → F/4
✔ Ans: (c)
✅ SECTION B (2 Marks Each)
(Attempt any 3 — Marking Scheme included)
Q1 Coulomb’s Law (2 Marks)
Statement (1M):
Force between two charges is directly proportional to product of charges and inversely proportional to square of distance.
Formula (1M):
F = k q₁q₂ / r²
Q2 Electric Field (2 Marks)
Definition (1M):
Electric field = Force per unit positive charge
E = F/q
Formula for point charge (1M):
E = kQ / r²
Q3 Superposition Principle (2 Marks)
Statement (2M):
Net force on a charge = vector sum of forces due to all other charges
F = F₁ + F₂ + F₃
Q4 Electric Field Lines (2 Marks)
Definition (1M):
Imaginary lines showing direction of electric field
Any 2 properties (1M):
• Start from + and end at −
• Never intersect
• Density shows strength
Q5 Electric Flux (2 Marks)
Definition (1M):
Number of field lines passing through a surface
Formula + Unit (1M):
Φ = E·A
Unit = N m²/C
✅ SECTION C (3 Marks Each)
(Attempt any 3 — Detailed + Marking Scheme)
Q1 Electric Field due to Point Charge (3 Marks)
Force on test charge:
F = kQq / r² (1M)
Electric field:
E = F/q (1M)
Substitute:
E = (kQq / r²)/q
E = kQ / r² (1M)
Q2 Coulomb’s Law (Vector Form) (3 Marks)
Magnitude:
F = k q₁q₂ / r² (1M)
Vector form:
F₁₂ = (1/4πε₀)(q₁q₂ / r²) r̂ (1M)
Explanation of r̂ (1M)
Q3 Field on Axial Line of Dipole (3 Marks)
Dipole moment:
p = q(2a) (1M)
Using derivation (steps):
E = (1/4πε₀)(2p / r³) (2M)
Q4 Electric Field Lines (3 Marks)
Positive charge → outward lines (1M)
Negative charge → inward lines (1M)
Dipole → from + to − (1M)
Q5 Electric Dipole (3 Marks)
Definition (1M):
Two equal & opposite charges separated by small distance
Dipole moment:
p = qd (1M)
Unit: C·m (1M)
✅ SECTION D (4 Marks Each)
(Attempt any 2 — Full Derivations + Marking Scheme)
Q1 Equatorial Line of Dipole (4 Marks)
Expression:
E = (1/4πε₀)(p / r³)
Steps:
• Resolve fields (1M)
• Cancel components (1M)
• Final derivation (1M)
• Direction opposite to dipole moment (1M)
Q2 Gauss Law (4 Marks)
Statement (1M):
Total flux through closed surface = Q/ε₀
Formula (1M):
Φ = ∮E·dA
Derivation (2M):
Φ = EA
⇒ EA = Q/ε₀
Q3 Infinite Line Charge (4 Marks)
Gaussian surface: cylinder (1M)
Flux:
Φ = EA = E(2πrL) (1M)
Charge enclosed:
Q = λL (1M)
Using Gauss law:
E(2πrL) = λL/ε₀
E = λ / (2π ε₀ r) (1M)
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