NEET Physics — Electrostatics and Current Electricity…

Chapter Overview & Weightage

Electrostatics and Current Electricity together form the second-largest scoring unit in NEET Physics. Coulomb’s law, electric field/potential, capacitors, Ohm’s law, Kirchhoff’s laws, and Wheatstone bridge — this unit covers a massive range with consistent question patterns.

This combined unit carries 10-12% weightage in NEET Physics — 5-6 questions per paper. Coulomb’s law numericals, capacitor combinations, and Kirchhoff’s law circuit problems are near-guaranteed.

Key Concepts You Must Know

Tier 1 (Core)

Coulomb’s law: $F = kq_1q_2/r^2$ , $k = 9 \times 10^9$ Nm $^2$ /C $^2$
Electric field: $E = kq/r^2$ (point charge), $E = \sigma/(2\epsilon_0)$ (infinite plane)
Electric potential: $V = kq/r$ , relation $E = -dV/dr$
Capacitance: $C = Q/V$ , parallel plate: $C = \epsilon_0 A/d$
Capacitors: series $1/C = 1/C_1 + 1/C_2$ , parallel $C = C_1 + C_2$
Ohm’s law: $V = IR$ , resistivity $\rho = RA/l$
Kirchhoff’s laws: junction rule ( $\sum I = 0$ ), loop rule ( $\sum V = 0$ )
Wheatstone bridge: balance condition $P/Q = R/S$

Tier 2 (Frequently tested)

Electric dipole: moment $\vec{p} = q\vec{d}$ , field on axial/equatorial line
Energy stored in capacitor: $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$
Effect of dielectric: $C' = KC$ (capacitance increases)
Resistors: series $R = R_1 + R_2$ , parallel $1/R = 1/R_1 + 1/R_2$
Meter bridge (Wheatstone bridge application)
Internal resistance and EMF: $V = E - Ir$

Important Formulas

F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} = k\frac{q_1 q_2}{r^2}

E = \frac{kq}{r^2}, \quad V = \frac{kq}{r}

C = \frac{\epsilon_0 A}{d} \text{ (parallel plate)}

With dielectric: $C' = \frac{K\epsilon_0 A}{d} = KC$

Energy: $U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}$

V = IR, \quad R = \frac{\rho l}{A}

Series: $R_{eq} = R_1 + R_2 + R_3 + ...$

Parallel: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$

Wheatstone bridge balance: $\frac{P}{Q} = \frac{R}{S}$ → galvanometer current = 0

Power: $P = VI = I^2R = V^2/R$

EMF and internal resistance: $V = E - Ir$ (terminal voltage)

For NEET circuit problems, always check for Wheatstone bridge first. If 4 resistors form a bridge and the balance condition is satisfied ( $P/Q = R/S$ ), the bridge resistance is zero — ignore it. This simplifies many circuits dramatically.

Solved Previous Year Questions

PYQ 1 — NEET 2024

Problem: Two capacitors of 4 μF and 6 μF are connected in series. Find the equivalent capacitance.

Solution:

\frac{1}{C_{eq}} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}

$C_{eq} = \frac{12}{5} = \mathbf{2.4 \text{ }\mu\text{F}}$

PYQ 2 — NEET 2023

Problem: In a Wheatstone bridge, $P = 100$ ohm, $Q = 200$ ohm, $R = 40$ ohm. For balance, $S$ should be:

Solution:

Balance condition: $P/Q = R/S$

$100/200 = 40/S$

$S = 40 \times 200/100 = \mathbf{80 \text{ ohm}}$

PYQ 3 — NEET 2022

Problem: A cell of EMF 2V and internal resistance 1 ohm is connected to an external resistance of 4 ohm. Find the current.

Solution:

$I = E/(R + r) = 2/(4 + 1) = \mathbf{0.4 \text{ A}}$

Terminal voltage: $V = E - Ir = 2 - 0.4 \times 1 = 1.6$ V

Expert Strategy

Week 1: Electrostatics — Coulomb’s law, electric field, potential. These are formula-driven problems. Practice numerical substitution with proper units.

Week 2: Capacitors — series/parallel combinations, effect of dielectric, energy stored. Also cover the parallel plate capacitor formula and how inserting a dielectric changes capacitance.

Week 3: Current electricity — Ohm’s law, Kirchhoff’s laws, Wheatstone bridge. Solve at least 15-20 circuit problems. Meter bridge is a direct application of Wheatstone bridge.

Common Traps

Trap 1 — Capacitors in series: charges are equal, voltages add up. In parallel: voltages are equal, charges add up. This is opposite to resistors. Capacitor series gives LESS capacitance (like resistors in parallel give less resistance).

Trap 2 — Terminal voltage is LESS than EMF when the cell delivers current. $V = E - Ir$ . Only when no current flows ( $I = 0$ ) does $V = E$ . NEET tests this with “a cell of EMF E and internal resistance r” — always account for the voltage drop across internal resistance.

Trap 3 — Inserting a dielectric INCREASES capacitance. $C' = KC$ where $K > 1$ . If the capacitor is isolated (charge constant), voltage decreases. If connected to a battery (voltage constant), charge increases.

Trap 4 — In Wheatstone balance, current through the galvanometer is ZERO. The bridge arm can be removed for analysis. If not balanced, you need Kirchhoff’s laws for the full circuit.