Class 12 Physics Test Paper Chapter 1 AND 2 | GSEB Board | Detailed Solutions & Marking Scheme
VEDANT SKILL ASSESSMENT SERIES
ACADEMIC YEAR 2026-27
CLASS: XII (GSEB) | TIME: 1 Hour | MAX. MARKS: 30 SUBJECT: PHYSICS (Chapter 1: Electric Charge and Field & Chapter 2: Electrostatic Potential and Capacitance)
GENERAL INSTRUCTIONS:
Read all questions carefully before answering. — Misreading the question and then blaming the paper will not increase your marks.
Show proper steps wherever required. — “Sir, answer toh yahi aana tha” is not an accepted mathematical method.
Write neatly and clearly. — If your handwriting requires a decoder machine, checking may become an adventure.
Manage your time wisely. — Spending 45 minutes on one question and calling the rest “optional” is not a strategy.
If you do not know the answer, you may cry silently. — Loud crying, emotional speeches, and negotiations for hints are strictly prohibited.
SECTION A (Multiple Choice Questions)
[14 × 1 = 14 Marks]
Q1. A glass rod rubbed with silk acquires a charge of +4.8 × 10^-19 C. The number of electrons lost by the glass rod is: A. 3 B. 4 C. 6 D. zero
Q2. Two point charges placed at a distance r in air exert a force F on each other. If they are immersed in a dielectric medium of dielectric constant K = 4 without changing the distance, the new force becomes: A. 4F B. 2F C. F / 2 D. F / 4
Q3. What is the SI unit of electric flux? A. N / C B. N . m^2 / C C. N . C / m^2 D. C^2 / (N . m^2)
Q4. An electric dipole consists of two equal and opposite charges separated by a small distance 2a. The total net charge of this system is: A. 2q B. -2q C. zero D. q / 2
Q5. The dimensional formula for electric potential difference is: A. M^1 L^2 T^-3 A^-1 B. M^1 L^2 T^-2 A^-1 C. M^1 L^1 T^-3 A^-1 D. M^1 L^2 T^-3 A^-2
Q6. The electric potential at any point on the equatorial plane of an electric dipole is always: A. maximum B. minimum but non-zero C. zero D. dependent on the distance from the center
Q7. Two capacitors of capacitances 3 uF and 6 uF are connected in series across a voltage source. The ratio of the charges stored on them (q1 : q2) is: A. 1 : 2 B. 2 : 1 C. 1 : 1 D. 1 : 4
Q8. If the distance between the plates of a parallel plate air capacitor is doubled, its capacitance: A. becomes four times B. is doubled C. is halved D. remains unchanged
Q9. A total work of 40 J is performed to move a charge of 2 C between two specific points inside an electric field. The potential difference between these points is: A. 80 V B. 20 V C. 10 V D. 5 V
Q10. The electric field lines are always ________ to an equipotential surface at every point. A. parallel B. perpendicular C. inclined at 45° D. tangential
Q11. A dielectric slab of dielectric constant K is completely filled in the gap of a capacitor while it remains connected to a battery. The energy stored in the capacitor: A. decreases by a factor of K B. increases by a factor of K C. increases by a factor of K^2 D. stays completely constant
Q12. The electric field due to an infinitely long straight uniformly charged wire varies with the distance r as: A. E ∝ 1/r B. E ∝ 1/r^2 C. E ∝ 1/r^3 D. E ∝ r
Q13. Total energy density in the medium between the plates of a parallel plate capacitor is given by: A. 0.5 . e0 . E B. 0.5 . e0 . E^2 C. e0 . E^2 D. 0.5 . (e0 / E^2)
Q14. The angle between the electric dipole moment vector p and the electric field vector E when the dipole is in stable equilibrium is: A. 0° B. 90° C. 180° D. 45°
SECTION B (Short Answer Questions — Answer any 3 out of 4)
[3 × 2 = 6 Marks]
Q15. Explain the concept of quantization of electric charge and state its fundamental formula. [2]
Q16. State Coulomb's law in scalar form. Write down the value and units of the electrostatic constant k and permittivity of free space e0. [2]
Q17. Define an equipotential surface. Sketch the shape of the equipotential surfaces for a single isolated positive point charge. [2]
Q18. Two point charges q1 = +2 uC and q2 = -2 uC are separated by a distance of 6 × 10^-3 m in air. Calculate the magnitude of the electric dipole moment of this system. [2]
SECTION C (Medium Answer Questions — Answer any 2 out of 3)
[2 × 3 = 6 Marks]
Q19. Derive the formula for the torque acting on an electric dipole placed inside a uniform external electric field. State the conditions under which it experiences maximum and minimum torque. [3]
Q20. Using Gauss's law, derive the expression for the electric field intensity at any point outside a uniformly charged thin spherical shell of radius R and total charge Q. [3]
Q21. An electric point charge of +5 uC is located at the center of a cubic Gaussian surface of edge length 10 cm. Calculate the net electric flux passing through the entire cube and the flux passing through any one individual face of the cube. (Take e0 = 8.85 × 10^-12 C^2 / N.m^2) [3]
SECTION D (Long Answer Questions — Answer any 1 out of 2)
[1 × 4 = 4 Marks]
Q22. Derive the mathematical expression for the capacitance of a parallel plate capacitor containing air as the medium between its plates. Explain comprehensively how its capacitance changes when a dielectric medium of dielectric constant K completely fills the gap. [4]
Q23. Answer both sub-questions: (a) Derive an expression for the total electrostatic potential energy of a system consisting of two point charges q1 and q2 separated by a distance r12 in the complete absence of an external electric field. [2] (b) Three capacitors of capacitances 2 uF, 3 uF, and 6 uF are connected in a series configuration across a 120 V DC power supply line. Determine the equivalent capacitance of the combination and the total electrical energy stored within the system. [2]
SOLVED ANSWER KEY WITH MARKING SCHEME
VEDANT SKILL ASSESSMENT SERIES
ACADEMIC YEAR 2026–27
CLASS: XII (GSEB) | SUBJECT: PHYSICS
Chapter 1: Electric Charge and Field & Chapter 2: Electrostatic Potential and Capacitance
Max. Marks: 30
SECTION A – MULTIPLE CHOICE QUESTIONS
[14 × 1 = 14 Marks]
Q1.
Given:
Charge = +4.8 × 10^-19 C
Charge on one electron:
e = 1.6 × 10^-19 C
Number of electrons lost:
n = q/e
n = (4.8 × 10^-19)/(1.6 × 10^-19)
n = 3
Correct Answer: A. 3
Marking Scheme: 1 Mark
Q2.
Force in dielectric medium:
F' = F/K
Given:
K = 4
F' = F/4
Correct Answer: D. F / 4
Marking Scheme: 1 Mark
Q3.
SI unit of electric flux:
Electric flux = Electric field × Area
= (N/C) × m²
= N.m²/C
Correct Answer: B. N.m²/C
Marking Scheme: 1 Mark
Q4.
Net charge of electric dipole:
(+q) + (−q) = 0
Correct Answer: C. zero
Marking Scheme: 1 Mark
Q5.
Electric potential difference:
V = Work/Charge
Dimensional formula:
[M¹L²T^-2] / [AT]
= M¹L²T^-3A^-1
Correct Answer: A. M¹L²T^-3A^-1
Marking Scheme: 1 Mark
Q6.
Potential at equatorial line of dipole:
Potential due to +q and −q cancel each other.
Hence,
V = 0
Correct Answer: C. zero
Marking Scheme: 1 Mark
Q7.
In series connection:
Charge remains same on all capacitors.
q₁ = q₂
Ratio:
q₁ : q₂ = 1 : 1
Correct Answer: C. 1 : 1
Marking Scheme: 1 Mark
Q8.
Capacitance:
C = ε₀A/d
If distance doubled:
C becomes half.
Correct Answer: C. is halved
Marking Scheme: 1 Mark
Q9.
Potential difference:
V = W/q
V = 40/2
V = 20 V
Correct Answer: B. 20 V
Marking Scheme: 1 Mark
Q10.
Electric field lines are always perpendicular to equipotential surface.
Correct Answer: B. perpendicular
Marking Scheme: 1 Mark
Q11.
Capacitor connected to battery:
Voltage remains constant.
Energy:
U = (1/2)CV²
New capacitance:
C' = KC
Hence,
U' = KU
Energy increases by factor K.
Correct Answer: B. increases by a factor of K
Marking Scheme: 1 Mark
Q12.
Electric field due to infinite line charge:
E ∝ 1/r
Correct Answer: A. E ∝ 1/r
Marking Scheme: 1 Mark
Q13.
Energy density:
u = (1/2)ε₀E²
Correct Answer: B. 0.5 ε₀E²
Marking Scheme: 1 Mark
Q14.
Stable equilibrium occurs when dipole aligns with electric field.
θ = 0°
Correct Answer: A. 0°
Marking Scheme: 1 Mark
SECTION B – SHORT ANSWER QUESTIONS
[Answer any 3 out of 4]
[3 × 2 = 6 Marks]
Q15. Quantization of Electric Charge [2]
Definition:
Electric charge exists in discrete packets and not continuously.
Charge on any body is always an integral multiple of elementary charge.
Formula:
q = ±ne
Where:
n = integer (1, 2, 3, ...)
e = 1.6 × 10^-19 C
Marking Scheme:
Explanation = 1 Mark
Formula = 1 Mark
Q16. Coulomb’s Law [2]
Statement:
The electrostatic force between two point charges is directly proportional to product of charges and inversely proportional to square of distance between them.
Scalar form:
F = k(q₁q₂/r²)
Where,
k = 9 × 10^9 N.m²/C²
Permittivity of free space:
ε₀ = 8.85 × 10^-12 C²/N.m²
Marking Scheme:
Statement = 1 Mark
Formula and constants = 1 Mark
Q17. Equipotential Surface [2]
Definition:
A surface on which electric potential remains same at every point is called equipotential surface.
For an isolated positive charge:
Equipotential surfaces are concentric spherical surfaces around the charge.
Marking Scheme:
Definition = 1 Mark
Correct sketch/shape = 1 Mark
Q18. Electric Dipole Moment [2]
Given:
q = 2 μC
Distance between charges:
2a = 6 × 10^-3 m
Dipole moment:
p = q × d
= (2 × 10^-6)(6 × 10^-3)
= 12 × 10^-9
= 1.2 × 10^-8 C m
Final Answer:
p = 1.2 × 10^-8 C m
Marking Scheme:
Formula = 1 Mark
Correct answer = 1 Mark
SECTION C – MEDIUM ANSWER QUESTIONS
[Answer any 2 out of 3]
[2 × 3 = 6 Marks]
Q19. Torque on Electric Dipole [3]
When dipole placed in uniform electric field:
Equal and opposite forces act on charges.
Net force = 0
But a torque acts.
Torque:
τ = Force × perpendicular distance
τ = qE × (2a sinθ)
Since,
p = q(2a)
Therefore:
τ = pE sinθ
Final Formula:
τ = pE sinθ
Maximum torque:
sinθ = 1
θ = 90°
τ(max) = pE
Minimum torque:
sinθ = 0
θ = 0° or 180°
τ = 0
Marking Scheme:
Derivation = 2 Marks
Maximum and minimum condition = 1 Mark
Q20. Electric Field due to Charged Spherical Shell [3]
Using Gauss law.
Gaussian surface radius = r
(r > R)
Electric flux:
Φ = EA
= E(4πr²)
Charge enclosed:
Q
By Gauss law:
E(4πr²) = Q/ε₀
Therefore:
E = Q/(4πε₀r²)
Final Answer:
E = (1/4πε₀)(Q/r²)
Acts radially outward.
Marking Scheme:
Gaussian surface = 1 Mark
Derivation = 1 Mark
Final expression = 1 Mark
Q21. Electric Flux Through Cube [3]
Given:
q = 5 μC
= 5 × 10^-6 C
By Gauss law:
Φ = q/ε₀
Φ = (5 × 10^-6)/(8.85 × 10^-12)
Φ = 5.65 × 10^5 N.m²/C
Since charge at center,
Flux equally distributed through 6 faces.
Flux through one face:
Φ(face) = Φ/6
= (5.65 × 10^5)/6
= 9.42 × 10^4 N.m²/C
Final Answer:
Total flux = 5.65 × 10^5 N.m²/C
Flux through one face = 9.42 × 10^4 N.m²/C
Marking Scheme:
Total flux formula = 1 Mark
Numerical calculation = 1 Mark
Flux through one face = 1 Mark
SECTION D – LONG ANSWER QUESTIONS
[Answer any 1 out of 2]
[1 × 4 = 4 Marks]
Q22. Capacitance of Parallel Plate Capacitor [4]
Capacitance:
C = Q/V
Electric field:
E = σ/ε₀
Potential difference:
V = Ed
V = (σd)/ε₀
Since,
σ = Q/A
V = Qd/(Aε₀)
Capacitance:
C = Q/V
C = ε₀A/d
Final Formula:
C = ε₀A/d
Effect of dielectric medium:
If dielectric constant = K
New capacitance:
C' = Kε₀A/d
C' = KC
Thus, capacitance increases K times.
Marking Scheme:
Derivation = 2 Marks
Final formula = 1 Mark
Effect of dielectric = 1 Mark
Q23.
(a) Electrostatic Potential Energy [2]
Potential energy of system:
U = (1/4πε₀)(q₁q₂/r₁₂)
Marking Scheme:
Derivation/concept = 1 Mark
Formula = 1 Mark
(b) Capacitors in Series [2]
Given:
C₁ = 2 μF
C₂ = 3 μF
C₃ = 6 μF
Series combination:
1/Ceq = 1/2 + 1/3 + 1/6
1/Ceq = 1
Ceq = 1 μF
Energy stored:
U = (1/2)CV²
= (1/2)(1 × 10^-6)(120)²
= 0.5 × 10^-6 × 14400
= 7.2 × 10^-3 J
Final Answer:
Equivalent capacitance = 1 μF
Energy stored = 7.2 × 10^-3 J
Marking Scheme:
Equivalent capacitance = 1 Mark
Energy calculation = 1 Mark
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