Class 12 Physics Test Paper Chapter 1 AND 2 | CBSE Board | Detailed Solutions & Marking Scheme
VEDANT SKILL ASSESSMENT SERIES - ACADEMIC YEAR 2026-27
CLASS: XII | SUBJECT: PHYSICS (042)
CHAPTERS: Ch-1: Electric Charges & Fields, Ch-2: Electric Potential & Capacitance
TIME ALLOWED: 1 Hour | MAXIMUM MARKS: 30
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 (7 Marks)
Q1. A highly pessimistic student combs his hair on a dry day and notices that the comb attracts small pieces of paper. If the comb acquired a negative charge of 3.2 × 10⁻⁹ C, the number of excess electrons transferred to the comb from his hair is: [1]
A. 2.0 × 10¹⁰
B. 3.2 × 10⁹
C. 2.0 × 10⁹
D. 6.25 × 10¹⁸
Q2. Two point charges +q and -4q are placed at a distance L apart on a straight line. The precise position of a third point charge Q on the line where it will experience absolutely zero net electrostatic force is: [1]
A. At a distance L to the left of charge +q
B. At a distance L/2 to the right of charge -4q
C. Exactly mid-way between the two given charges
D. At a distance 2L to the left of charge +q
Q3. An electric dipole consisting of charges +q and -q separated by distance 2a is placed inside a perfectly sealed, hollow cube of side L (L > 2a). The total net outward electric flux passing through all six faces of this cube combined is: [1]
A. q / ε₀
B. 2q / ε₀
C. Zero
D. q / 6ε₀
Q4. Four identical large metallic plates, each of surface area A, are kept parallel to each other at equal separation d. Alternate plates are cross-connected. The equivalent capacitance between the final terminals is: [1]
A. ε₀A / d
B. 2ε₀A / d
C. 3ε₀A / d
D. 4ε₀A / d
Q5. A parallel plate air capacitor is fully charged by a battery and then isolated. If a dielectric slab of dielectric constant K is now carefully slid into the gap, how do the stored energy U and potential difference V change? [1]
A. Both U and V increase by K times
B. Both U and V decrease by K times
C. U increases by K times, V remains constant
D. U decreases by K times, V remains constant
For Questions 6 and 7, select the correct option from the following:
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, but R is NOT the correct explanation of A.
C. A is true, but R is false.
D. A is false, but R is true.
Q6. Assertion (A): An electron moves spontaneously from a region of lower electrostatic potential to a region of higher electrostatic potential.
Reason (R): An electron carries a negative charge, and every free charged particle always experiences a net force pushing it towards a region of lower potential energy. [1]
Q7. Assertion (A): The electric field inside a hollow charged conductor is always zero, but the electrostatic potential inside it can vary from point to point depending on the geometry.
Reason (R): The relationship between electric field and potential is given by E = -dV/dr, so if E = 0, then V must be constant throughout the volume. [1]
SECTION B (8 Marks)
Q8. An old, leaky water tap drips water droplets. Two identical metallic spheres hanging from thin insulating threads are touched simultaneously by a single charged plastic rod. They repel each other and stand at an angle. If a student coats one sphere with a thin layer of conducting soot and leaves the other clean, explain with logical steps what happens to the separation and the force when they are brought into contact again and released. [2]
Q9. Two concentric thin metallic spherical shells of radii R and 2R carry charges Q and 3Q respectively. Calculate the potential difference (Vᵣ - V₂ᴿ) between the inner and outer shells. If the outer shell is now connected to the earth via a wire, what will be the new potential of the inner shell? [2]
Q10. A laboratory technician mistakenly connects a parallel plate capacitor to a high-voltage supply until it is fully charged. He then carefully disconnects it. Without discharging the capacitor, he doubles the distance between the plates using insulated tools. Deduce the ratio of the final electric field intensity to the initial electric field intensity between the plates. What happens to the energy stored? [2]
Q11. Draw a clean, clear schematic graph showing the variation of: [2]
(i) Stored electrostatic potential energy U vs charge Q accumulated on a given capacitor.
(ii) Electric potential V vs distance r measured from the center of a hollow charged sphere of radius R.
SECTION C (6 Marks)
Q12.
(a) Derive the expression for the net torque acting on an electric dipole of dipole moment p placed in a uniform external electric field E and express it in vector notation.
(b) If this dipole is rotated from its orientation of stable equilibrium to unstable equilibrium, calculate the total work done by the external agency. Clearly mention the orientation angles in both states. [3]
Q13. State Gauss's Law in electrostatics. Using this law, derive a neat mathematical expression for the electric field intensity at a distance r from an infinitely long straight wire having a uniform linear charge density λ. Clearly describe the shape of the Gaussian surface chosen and explain why the field at the flat end caps does not contribute to the net flux. [3]
SECTION D (5 Marks)
Q14. Answer the following questions regarding a parallel plate capacitor:
(a) Derive the standard expression for the capacitance of a parallel plate capacitor with plate area A and separation d when the medium between the plates is air/vacuum.
(b) A dielectric slab of thickness t (t < d) and dielectric constant K is introduced between the plates. Derive the modified expression for its capacitance.
(c) A network contains three identical capacitors each of capacitance 6 μF connected in series across a 12 V DC battery. If one of the capacitors is completely filled with a dielectric slab of K = 3, calculate the new charge on each capacitor and the potential difference across the filled capacitor. [5]
SECTION E (4 Marks - Case Study)
Q15. CASE STUDY: Van de Graaff Generator Principles & Real-world Electrostatics
In industrial processes and high-energy physics experiments, large electric potentials are generated using static charge transfer mechanisms. Consider a large hollow spherical conducting shell of radius R carrying a total charge Q. If a small conducting sphere of radius r carrying a small charge q is placed exactly at the center of this large shell, a remarkable electrostatic property emerges. Even though the large shell may already be at a very high potential, if we connect the inner small sphere to the outer large shell by a conducting wire, the charge q flows entirely and spontaneously from the inner sphere to the outer shell. This happens because the potential difference between the inner sphere and outer shell depends uniquely on the inner charge and geometry, ensuring that the inner sphere is always at a higher absolute potential than the outer shell, regardless of how massive Q is. This principle allows scientists to continuously build up massive voltages on the outer dome, bounded only by the dielectric breakdown strength of the surrounding air.
Based on the text above, answer the following sub-questions:
(i) Write the exact mathematical formula for the potential difference V(r) - V(R) between the inner small sphere and the outer large shell before they are connected by a wire. [1]
(ii) Explain why the charge q is driven to flow exclusively from the inside to the outside dome when a connection is made. What physical quantity dictates this unidirectional behavior? [1]
(iii) If the maximum electric field that air can withstand without breakdown (sparking) is 3.0 × 10⁶ V/m, calculate the maximum potential that can be safely stored on an outer isolated spherical dome of radius 2.0 meters. What happens if the charge added exceeds this threshold value? [2]
VEDANT SKILL ASSESSMENT SERIES – ANSWER KEY WITH MARKING SCHEME
CLASS XII – PHYSICS (042)
CHAPTERS: Electric Charges & Fields, Electric Potential & Capacitance
TOTAL MARKS: 30
SECTION A (7 × 1 = 7 Marks)
Q1. Correct Option: A
Number of electrons,
n = Q/e
= (3.2 × 10⁻⁹)/(1.6 × 10⁻¹⁹)
= 2.0 × 10¹⁰
Answer: A (2.0 × 10¹⁰)
Marks: 1
Q2. Correct Option: A
Let the null point be at distance x to the left of +q.
For zero force,
kq/x² = k(4q)/(x + L)²
1/x² = 4/(x + L)²
x + L = 2x
x = L
Answer: A
Marks: 1
Q3. Correct Option: C
According to Gauss's Law,
Φ = Qenclosed/ε₀
The dipole contains +q and –q.
Net enclosed charge = 0
Therefore,
Φ = 0
Answer: C
Marks: 1
Q4. Correct Option: C
Three capacitors are formed between adjacent plates.
Each capacitance:
C = ε₀A/d
All three are connected in parallel.
Ceq = C + C + C
= 3ε₀A/d
Answer: C
Marks: 1
Q5. Correct Option: B
Battery disconnected ⇒ Charge remains constant.
After dielectric insertion:
C' = KC
V' = Q/C' = V/K
U' = Q²/(2C')
= U/K
Both decrease by factor K.
Answer: B
Marks: 1
Q6. Correct Option: A
Assertion: True
Electron moves naturally toward higher potential.
Reason: True
A charge always tends toward lower potential energy.
For electron,
U = qV
Since q is negative, higher V means lower U.
Answer: A
Marks: 1
Q7. Correct Option: D
Assertion: False
Potential inside conductor is constant.
Reason: True
Since E = –dV/dr and E = 0,
dV/dr = 0
Hence V remains constant.
Answer: D
Marks: 1
SECTION B (4 × 2 = 8 Marks)
Q8. Metallic Spheres and Charge Redistribution
Solution:
When the spheres are brought into contact, charges redistribute equally because both are conductors.
After separation:
• Each sphere acquires equal charge.
• Both spheres again carry like charges.
• Repulsive force exists between them.
• Separation remains non-zero.
Marking Scheme:
Equal redistribution of charge – 1 mark
Effect on force/separation explained – 1 mark
Total: 2 Marks
Q9. Concentric Spherical Shells
Potential at inner shell:
VR = kQ/R + k(3Q)/(2R)
VR = kQ/R + 3kQ/(2R)
Potential at outer shell:
V2R = k(Q + 3Q)/(2R)
= 2kQ/R
Potential difference:
VR – V2R
= [kQ/R + 3kQ/(2R)] – 2kQ/R
= kQ/(2R)
Answer:
VR – V2R = kQ/(2R)
When outer shell is earthed:
Potential of outer shell = 0
Potential of inner shell becomes
V = kQ/R
Marking Scheme:
Potential difference derivation – 1 mark
New inner-shell potential – 1 mark
Total: 2 Marks
Q10. Capacitor Isolated and Plate Separation Doubled
Solution:
Since battery disconnected,
Q = constant
Electric field between plates:
E = σ/ε₀
Since σ unchanged,
Efinal = Einitial
Therefore,
Efinal/Einitial = 1
Capacitance:
C' = C/2
Energy:
U = Q²/(2C)
U' = Q²/(2C')
= 2U
Stored energy doubles.
Answer:
Efinal/Einitial = 1
Energy becomes 2U
Marking Scheme:
Electric field ratio – 1 mark
Energy change – 1 mark
Total: 2 Marks
Q11. Graphs
(i) U vs Q
U = Q²/(2C)
Parabola opening upward passing through origin.
(ii) V vs r for charged conducting sphere
Inside sphere (r ≤ R):
V = constant
Outside sphere (r > R):
V = kQ/r
Hyperbolic decrease.
Marking Scheme:
Correct graph (i) – 1 mark
Correct graph (ii) – 1 mark
Total: 2 Marks
SECTION C (2 × 3 = 6 Marks)
Q12. Torque on Electric Dipole
(a)
Force on charges:
F = qE
These equal and opposite forces form a couple.
Torque:
τ = F(2a)sinθ
τ = qE(2a)sinθ
τ = pE sinθ
Vector form:
τ⃗ = p⃗ × E⃗
(b)
Potential energy:
U = –pE cosθ
Stable equilibrium:
θ = 0°
U₁ = –pE
Unstable equilibrium:
θ = 180°
U₂ = +pE
Work done:
W = U₂ – U₁
= pE – (–pE)
= 2pE
Answer:
τ = pE sinθ
τ⃗ = p⃗ × E⃗
Work done = 2pE
Marking Scheme:
Torque derivation – 1 mark
Vector form – 1 mark
Work done calculation – 1 mark
Total: 3 Marks
Q13. Gauss's Law and Infinite Line Charge
Statement:
The total electric flux through a closed surface equals enclosed charge divided by ε₀.
∮E·dA = Qenc/ε₀
Choose cylindrical Gaussian surface:
Radius = r
Length = l
Electric field is radial and constant over curved surface.
Flux through curved surface:
Φ = E(2πrl)
Flux through flat ends = 0
because electric field is parallel to end surfaces.
Enclosed charge:
Qenc = λl
Applying Gauss's Law:
E(2πrl) = λl/ε₀
E = λ/(2π ε₀ r)
Answer:
E = λ/(2π ε₀ r)
Marking Scheme:
Statement of law – 1 mark
Gaussian surface and flux calculation – 1 mark
Final expression – 1 mark
Total: 3 Marks
SECTION D (5 Marks)
Q14.
(a) Parallel Plate Capacitor
Electric field:
E = σ/ε₀
Potential difference:
V = Ed
= σd/ε₀
Capacitance:
C = Q/V
= σA/(σd/ε₀)
Therefore,
C = ε₀A/d
(b) Dielectric Slab Introduced
Effective separation:
deff = d – t + t/K
Therefore,
C = ε₀A/[d – t + (t/K)]
(c) Numerical
Capacitors:
6 μF, 6 μF, 18 μF
Series combination:
1/Ceq
= 1/6 + 1/6 + 1/18
= 7/18
Ceq = 18/7 μF
Charge:
Q = CeqV
= (18/7)(12)
= 216/7 μC
≈ 30.86 μC
Same charge flows through all series capacitors.
Charge on each capacitor:
30.86 μC
Voltage across dielectric-filled capacitor:
V = Q/C
= 30.86/18
≈ 1.71 V
Answer:
Charge = 30.86 μC
Voltage across filled capacitor = 1.71 V
Marking Scheme:
Part (a) derivation – 1 mark
Part (b) derivation – 2 marks
Equivalent capacitance – 1 mark
Charge and voltage calculation – 1 mark
Total: 5 Marks
SECTION E (CASE STUDY – 4 Marks)
Q15.
(i)
Potential due to inner sphere:
V(r) = kq/r + kQ/R
Potential at outer shell:
V(R) = kq/R + kQ/R
Therefore,
V(r) – V(R)
= kq(1/r – 1/R)
Answer:
V(r) – V(R) = (1/4πε₀)q(1/r – 1/R)
Marks: 1
(ii)
Charge flows from higher potential to lower potential through conductor.
Inner sphere always remains at higher potential.
The driving quantity is:
Potential Difference (Voltage)
Marks: 1
(iii)
Maximum field at surface:
E = V/R
Therefore,
Vmax = ER
= (3 × 10⁶)(2)
= 6 × 10⁶ V
Answer:
Maximum safe potential = 6 MV
If charge exceeds this value, dielectric breakdown of air occurs and sparking/discharge takes place.
Marking Scheme:
Calculation – 1 mark
Breakdown explanation – 1 mark
Total: 2 Marks
TOTAL MARKS = 30
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