Electrostatics & Current Electricity
Coulomb's law, electric field and potential, capacitors, Ohm's law, and circuits
Key Concepts
- Coulomb's law gives the force between point charges
- Gauss's law: total flux through a closed surface = Q_enc/ε₀
- Electric field inside a conductor is zero; surface is equipotential
- Ohm's law (V = IR) holds for ohmic conductors at constant temperature
- Kirchhoff's laws (KCL and KVL) solve any circuit network
Important Formulae
| Coulomb's force | F = kq₁q₂/r²; k = 9×10⁹ |
| Field & potential | E = kq/r²; V = kq/r |
| Parallel-plate capacitor | C = ε₀A/d |
| Energy in capacitor | U = ½CV² = Q²/2C |
| Power | P = VI = I²R = V²/R |
| Combinations | Series: R add, 1/C add · Parallel: 1/R add, C add |
Quick Tips
- A dielectric increases capacitance by a factor κ (dielectric constant).
- Maximum power is transferred to the load when R_external = r (internal).
- Drift velocity is tiny (~mm/s) even though the signal travels near c.
Sample Practice Questions
-
The value of Coulomb's constant k is approximately:
Show answer
Answer: 9×10⁹ N·m²/C²
-
Using Ohm's law, the voltage across a 5 Ω resistor carrying 2 A is:
Show answer
Answer: 10 V
-
Two 2 Ω resistors connected in series give a total resistance of:
Show answer
Answer: 4 Ω
-
Two 2 Ω resistors connected in parallel give a total resistance of:
Show answer
Answer: 1 Ω
Practice Questions
Practise randomly selected JEE questions on Electrostatics & Current Electricity. Answers are revealed after each question.
Start Practice →Frequently Asked Questions
What are the key concepts in Electrostatics & Current Electricity for JEE?
Coulomb's law gives the force between point charges; Gauss's law: total flux through a closed surface = Q_enc/ε₀; Electric field inside a conductor is zero; surface is equipotential; Ohm's law (V = IR) holds for ohmic conductors at constant temperature; Kirchhoff's laws (KCL and KVL) solve any circuit network.
Which formulae are important for Electrostatics & Current Electricity in JEE?
Coulomb's force: F = kq₁q₂/r²; k = 9×10⁹; Field & potential: E = kq/r²; V = kq/r; Parallel-plate capacitor: C = ε₀A/d; Energy in capacitor: U = ½CV² = Q²/2C; Power: P = VI = I²R = V²/R; Combinations: Series: R add, 1/C add · Parallel: 1/R add, C add.
How can I practise JEE Electrostatics & Current Electricity questions for free?
Practise free JEE Electrostatics & Current Electricity questions with instant answers and explanations on Safe Learning for Kids — no sign-up required.