Electrostatic Potential and Capacitance — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Electrostatic Potential and Capacitance" — 8 important questions with detailed answers for CBSE board exam preparation.
✓ 100% Free
✓ No Login Needed
✓ NCERT / CBSE Aligned
✓ Download as PDF
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Electrostatic Potential and Capacitance" — 8 important questions with detailed answers…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
🤖 Stuck on any question? Ask Syllab's free AI Tutor for a step-by-step explanation — instant, unlimited, no login.
Key Questions Covered:
- Define electric potential. Write expression for potential due to point charge.
- What is potential difference? A charge of 2 C is moved from point A to point …
- Define capacitance. Write expression for capacitance of parallel plate capaci…
- Two capacitors of 2 μF and 4 μF are connected in series across 600 V supply. …
- Derive expression for energy stored in a capacitor.
- A parallel plate capacitor with plate area 100 cm² and separation 2 mm is cha…
- + 2 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Define electric potential. Write expression for potential… | ✓ Solved |
| What is potential difference? A charge of 2 C is moved fr… | ✓ Solved |
| Define capacitance. Write expression for capacitance of p… | ✓ Solved |
| Two capacitors of 2 μF and 4 μF are connected in series a… | ✓ Solved |
| Derive expression for energy stored in a capacitor. | ✓ Solved |
| A parallel plate capacitor with plate area 100 cm² and se… | ✓ Solved |
Showing 6 of 8 questions
Q1: Define electric potential. Write expression for potential due to point charge.
Electric Potential (V): Work done per unit positive charge in bringing it from infinity to that point in electric field.
V = W/q₀ (potential is scalar quantity)
SI Unit: Volt (V) or J/C
Potential due to point charge q at distance r:
V = kq/r = q/(4πε₀r)
where k = 9 × 10⁹ N⋅m²/C²
ε₀ = 8.85 × 10⁻¹² F/m
Note: V = 0 at infinity (reference point)
Q2: What is potential difference? A charge of 2 C is moved from point A to point B, and 20 J of work is done. Find potential difference between A and B.
Potential Difference (V_AB): Difference in electric potential between two points.
V_AB = V_A - V_B = W_AB/q
where W_AB = work done by external agent to move charge q from B to A
Given: q = 2 C
W = 20 J
Step 1: V_AB = W/q = 20/2 = 10 V
Final Answer: Potential difference = 10 V
This means point A is at 10 V higher potential than point B. Work done against electric field = 20 J.
Q3: Define capacitance. Write expression for capacitance of parallel plate capacitor.
Capacitance (C): Ability of conductor to store charge for given potential difference.
C = Q/V (charge per unit potential difference)
SI Unit: Farad (F) = C/V
For Parallel Plate Capacitor:
C = ε₀εᵣA/d = ε₀A/d (in vacuum/air)
where ε₀ = 8.85 × 10⁻¹² F/m (permittivity of free space)
εᵣ = relative permittivity of dielectric
A = area of each plate
d = separation between plates
Note: Capacitance is independent of charge and voltage, depends only on geometry and material.
Q4: Two capacitors of 2 μF and 4 μF are connected in series across 600 V supply. Find charge on each capacitor and voltage across each.
Given: C₁ = 2 μF, C₂ = 4 μF, V = 600 V (series connection)
In series connection: Same charge on both capacitors
Step 1: Find equivalent capacitance
1/C_eq = 1/C₁ + 1/C₂ = 1/2 + 1/4 = 3/4
C_eq = 4/3 μF
Step 2: Total charge Q = C_eq × V = (4/3) × 600 = 800 μC
Step 3: Charge on each capacitor = 800 μC
(Q₁ = Q₂ = 800 μC since in series)
Step 4: Voltage across C₁
V₁ = Q/C₁ = 800/2 = 400 V
Step 5: Voltage across C₂
V₂ = Q/C₂ = 800/4 = 200 V
Verification: V₁ + V₂ = 400 + 200 = 600 V ✓
Final Ans...
Q5: Derive expression for energy stored in a capacitor.
Consider charging capacitor from uncharged state to final charge Q.
At any instant, if charge is q, potential difference = q/C
Work done to add dq: dW = V⋅dq = (q/C)dq
Total work done:
W = ∫₀^Q (q/C)dq = (1/C)∫₀^Q q⋅dq = (1/C)[q²/2]₀^Q = Q²/(2C)
This work is stored as electrostatic potential energy.
Energy stored: U = Q²/(2C)
Alternative forms:
U = (1/2)QV (since Q = CV)
U = (1/2)CV² (substituting Q = CV)
where Q = charge on capacitor
V = final potential difference
C = capacitance
Energy...
Q6: A parallel plate capacitor with plate area 100 cm² and separation 2 mm is charged to 100 V. Calculate: (a) Capacitance (b) Charge stored (c) Energy stored
Given: A = 100 cm² = 100 × 10⁻⁴ m² = 0.01 m²
d = 2 mm = 2 × 10⁻³ m
V = 100 V
ε₀ = 8.85 × 10⁻¹² F/m
(a) Capacitance:
C = ε₀A/d = (8.85 × 10⁻¹²) × (0.01) / (2 × 10⁻³)
C = 8.85 × 10⁻¹⁴ / (2 × 10⁻³)
C = 4.425 × 10⁻¹¹ F
C ≈ 44.25 pF
(b) Charge stored:
Q = CV = 4.425 × 10⁻¹¹ × 100
Q = 4.425 × 10⁻⁹ C
Q ≈ 4.425 nC
(c) Energy stored:
U = (1/2)CV² = (1/2) × 4.425 × 10⁻¹¹ × (100)²
U = 0.5 × 4.425 × 10⁻¹¹ × 10⁴
U = 2.2125 × 10⁻⁷ J
U ≈ 221.25 nJ
Final Answer: (a) C ≈ 44.25 pF (b) Q ≈ 4.425 nC (c) U ≈ 221...
Showing 6 of 8 questions. Visit the full page for complete solutions.
More Class 12 Physics NCERT Solutions
- Electric Charges and Fields — Class 12 Physics NCERT Solutions
- Current Electricity — Class 12 Physics NCERT Solutions
- Moving Charges and Magnetism — Class 12 Physics NCERT Solutions
- Magnetism and Matter — Class 12 Physics NCERT Solutions
- Electromagnetic Induction — Class 12 Physics NCERT Solutions
- Alternating Current — Class 12 Physics NCERT Solutions
- Electromagnetic Waves — Class 12 Physics NCERT Solutions
- Ray Optics and Optical Instruments — Class 12 Physics NCERT Solutions
- Wave Optics — Class 12 Physics NCERT Solutions
- Dual Nature of Radiation and Matter — Class 12 Physics NCERT Solutions
- Atoms — Class 12 Physics NCERT Solutions
- Nuclei — Class 12 Physics NCERT Solutions
- Semiconductor Electronics — Class 12 Physics NCERT Solutions
- Current Electricity Exemplar — Class 12 Physics NCERT Solutions