Exam Notes
Electrostatics Class 12 Physics Revision — Grandmaster Guide
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Last Updated: 2026-06-01
- E = kq/r²
- V = kq/r
- F = kq₁q₂/r²
- U = kq₁q₂/(2r)
- C = 4πε₀R
- q = CV
- ε₀ = 8.85 × 10⁻¹² F/m
- εᵣ = ε₀ε
- D = ε₀E
- P = ε₀(E₁ - E₂)
- σ = q/A
- E = σ/(2ε₀)
- V = Ed
- R = ρ(L/A)
- C = ε₀A/d
🪤 The 5 Mistakes That Cost Marks
- Assuming electric field is always ∈ the direction of force
- Forgetting to consider the sign of charge when calculating electric field or potential
- Not using the correct value of ε₀
- Not considering the effect of dielectric constant on capacitance
- Assuming that electric field inside a conductor is always zero
✏️ 3 Solved PYQs
- A point charge of 5 μC is placed at the center of a sphere of radius 10 cm. Find the electric flux through the sphere. Step 1: Calculate the electric field at the surface of the sphere using E = kq/r² Step 2: Calculate the electric flux through the sphere using Φ = EA Answer: Φ = 1.13 × 10⁵ Nm²C⁻¹
- Two charges of 2 μC and -3 μC are placed 10 cm apart. Find the electric potential at a point midway between the charges. Step 1: Calculate the electric potential due to each charge using V = kq/r Step 2: Calculate the total electric potential at the midpoint using V = V₁ + V₂ Answer: V = 1.8 × 10⁵ V
- A capacitor of capacitance 100 μF is connected to a 12 V battery. Find the charge on the capacitor. Step 1: Calculate the charge on the capacitor using q = CV Answer: q = 1.2 × 10⁻³ C
🧠 The One Thing Most Students Get Wrong
- Most students get wrong the concept of electric field and potential inside a conductor. They assume that electric field inside a conductor is always zero, which is not true. The electric field inside a conductor is zero only when the conductor is ∈ equilibrium. If the conductor is not ∈ equilibrium, there can be an electric field inside the conductor.
👁️ Ayush's Note
- To solve problems ∈ electrostatics, first identify the type of problem. Is it a problem involving electric field, potential, or capacitance?
- Use the correct formulas and equations to solve the problem.
- Always consider the sign of charge and the direction of electric field.
- Use the concept of symmetry to simplify the problem.
- Practice, practice, practice. The more you practice, the better you will become at solving problems ∈ electrostatics.
🔁 Last 5 Minutes Box
- Revision of key concepts: electric field, potential, capacitance
- Revision of key formulas: E = kq/r², V = kq/r, C = 4πε₀R
- Revision of key equations: q = CV, ε₀ = 8.85 × 10⁻¹² F/m
- Practice of quick problems: finding electric field and potential due to a point charge, finding capacitance of a capacitor
📝 Practice MCQs
1. What is the electric field at a point 10 cm from a point charge of 5 μC?
A) 2.25 × 10⁵ N/C
B) 1.13 × 10⁵ N/C
C) 5.6 × 10⁵ N/C
D) 1.8 × 10⁵ N/C
Answer: B) 1.13 × 10⁵ N/C. Use the formula E = kq/r² to find the electric field.
2. What is the electric potential at a point 10 cm from a point charge of 5 μC?
A) 1.8 × 10⁵ V
B) 2.25 × 10⁵ V
C) 5.6 × 10⁵ V
D) 1.13 × 10⁵ V
Answer: A) 1.8 × 10⁵ V. Use the formula V = kq/r to find the electric potential.
3. What is the capacitance of a capacitor with plates of area 100 cm² and separation 1 mm?
A) 8.85 × 10⁻¹² F
B) 8.85 × 10⁻¹¹ F
C) 8.85 × 10⁻¹⁰ F
D) 8.85 × 10⁻⁹ F
Answer: C) 8.85 × 10⁻¹⁰ F. Use the formula C = ε₀A/d to find the capacitance.
4. What is the charge on a capacitor of capacitance 100 μF connected to a 12 V battery?
A) 1.2 × 10⁻³ C
B) 1.2 × 10⁻⁴ C
C) 1.2 × 10⁻⁵ C
D) 1.2 × 10⁻⁶ C
Answer: A) 1.2 × 10⁻³ C. Use the formula q = CV to find the charge.
5. What is the electric flux through a sphere of radius 10 cm with a point charge of 5 μC at its center?
A) 2.25 × 10⁵ Nm²C⁻¹
B) 1.13 × 10⁵ Nm²C⁻¹
C) 5.6 × 10⁵ Nm²C⁻¹
D) 1.8 × 10⁵ Nm²C⁻¹
Answer: B) 1.13 × 10⁵ Nm²C⁻¹. Use the formula Φ = EA to find the electric flux.
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📚 Academic References
Content verified against peer-reviewed research:
- El Agente: An autonomous agent for quantum chemistry — Matter (2025) 🔓 — DOI ↗
🔓 = Open Access article
This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.
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