JEE Physics: Electricity Complete Guide

Electricity is a cornerstone topic in JEE Physics, with questions spanning from fundamental concepts like electrostatics to advanced circuit analysis. This guide provides an in-depth overview, formulas, and explanations to help you master the Electricity section for JEE Mains and Advanced exams.

1. Electrostatics

1.1 Electric Charge and Coulomb’s Law

Electric charge is a fundamental property of matter responsible for electric forces. Coulomb’s law quantifies the force between two point charges:

$$\vec{F} = k \frac{q_1 q_2}{r^2} \hat{r}$$

Where:

$q_1, q_2$ are the magnitudes of the charges,
$r$ is the distance between the charges,
$k = \frac{1}{4\pi \epsilon_0} \approx 9 \times 10^{9} \, \mathrm{Nm^2/C^2}$,
$\hat{r}$ is the unit vector from charge 1 to charge 2.

1.2 Electric Field

The electric field $\vec{E}$ at a point in space is the force experienced by a unit positive charge placed there:

$$\vec{E} = \frac{\vec{F}}{q} = k \frac{Q}{r^2} \hat{r}$$

It is a vector quantity and points away from positive charges and toward negative charges.

1.3 Electric Potential

Electric potential $V$ at a point is the work done per unit charge in bringing a positive test charge from infinity to that point:

$$V = k \frac{Q}{r}$$

Potential difference between two points $A$ and $B$ is $V_{AB} = V_A - V_B$.

1.4 Potential Energy of a System of Charges

For a system of point charges, the total electrostatic potential energy is:

$$U = k \sum_{i

Where $r_{ij}$ is the distance between charges $q_i$ and $q_j$.

1.5 Electric Dipole

An electric dipole consists of two equal and opposite charges separated by a distance $2a$. Its dipole moment is:

$$\vec{p} = q \times 2a$$

The electric field at an axial point (along the line of the dipole) at distance $r$ ($r \gg a$) is:

$$E_{\text{axial}} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$$

At an equatorial point (perpendicular bisector):

$$E_{\text{equatorial}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$$

2. Gauss's Law

Gauss's law relates the net electric flux $\Phi_E$ through a closed surface to the charge enclosed $Q_{\text{enc}}$:

$$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$$

Useful for calculating electric fields with symmetrical charge distributions such as spheres, planes, and cylinders.

3. Capacitors and Dielectrics

3.1 Capacitance

Capacitance $C$ is defined as the charge stored per unit potential difference:

$$C = \frac{Q}{V}$$

For a parallel-plate capacitor:

$$C = \frac{\epsilon_0 A}{d}$$

Where $A$ is the plate area and $d$ is the separation between plates.

3.2 Energy Stored in a Capacitor

$$U = \frac{1}{2} C V^2 = \frac{Q^2}{2C}$$

3.3 Capacitors in Series and Parallel

Series: $ \frac{1}{C_{\text{eq}}} = \sum \frac{1}{C_i} $
Parallel: $ C_{\text{eq}} = \sum C_i $

3.4 Dielectrics

Introducing a dielectric material increases capacitance by dielectric constant $K$:

$$C' = K C$$

4. Current Electricity

4.1 Electric Current

Electric current $I$ is the rate of flow of charge through a conductor:

$$I = \frac{dQ}{dt}$$

4.2 Drift Velocity

Current relates to drift velocity $v_d$ of charge carriers:

$$I = nq A v_d$$

Where $n$ is the number density of electrons, $q$ is charge, and $A$ is cross-sectional area.

4.3 Ohm’s Law

The relationship between voltage, current, and resistance is given by:

$$V = IR$$

$R$ depends on material and geometry:

$$R = \rho \frac{L}{A}$$

Where $\rho$ is resistivity, $L$ length, and $A$ cross-sectional area.

4.4 Temperature Dependence of Resistance

$$R_T = R_0 [1 + \alpha (T - T_0)]$$

Where $\alpha$ is temperature coefficient.

4.5 Resistivity and Conductivity

$$\sigma = \frac{1}{\rho}$$

$\sigma$ is conductivity.

5. Kirchhoff's Laws

5.1 Kirchhoff's Current Law (KCL)

Sum of currents entering a junction equals sum leaving it:

$$\sum I_{\text{in}} = \sum I_{\text{out}}$$

5.2 Kirchhoff's Voltage Law (KVL)

Sum of emfs and voltage drops around any closed loop is zero:

$$\sum V = 0$$

6. Wheatstone Bridge

Used to measure unknown resistance by balancing two legs of a bridge circuit:

$$\frac{R_1}{R_2} = \frac{R_3}{R_x}$$

When balanced, no current flows through the galvanometer.

7. Meter Bridge

Practical Wheatstone bridge with uniform wire; used to find unknown resistance:

$$\frac{R_1}{R_2} = \frac{l}{100 - l}$$

$l$ is length in cm on wire.

8. Potentiometer

Measures emf and potential difference accurately without drawing current.

Principle: potential drop per unit length is constant on uniform wire:

$$E = k l$$

Where $k$ is potential gradient, $l$ balancing length.

9. Electrical Power and Energy

9.1 Electrical Power

$$P = VI = I^2 R = \frac{V^2}{R}$$

9.2 Electrical Energy

$$E = P t = V I t$$

10. Series and Parallel Circuits

10.1 Series Circuits

Current $I$ is same through all resistors.
Equivalent resistance: $R_{\text{eq}} = R_1 + R_2 + \cdots$
Voltage divides: $V = V_1 + V_2 + \cdots$

10.2 Parallel Circuits

Voltage across each resistor is same.
Equivalent resistance: $\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$
Current divides: $I = I_1 + I_2 + \cdots$

11. Drift and Mobility

11.1 Drift Velocity

$$v_d = \frac{I}{n q A}$$

11.2 Mobility

$$\mu = \frac{v_d}{E}$$

Where $E$ is electric field intensity.

12. Thermoelectricity

Phenomenon where temperature difference produces an emf.

Seebeck Effect: emf is generated due to temperature difference between two dissimilar conductors.
Thermocouple: a circuit of two different metals producing emf when junctions are at different temperatures.

13. Superconductivity

Some materials show zero resistance below a critical temperature $T_c$, enabling lossless power transmission and powerful magnets.

14. Important Formulas Summary

Concept	Formula	Remarks
Coulomb's Law	$$F = k \frac{q_1 q_2}{r^2}$$	Force between charges
Electric Field	$$E = k \frac{Q}{r^2}$$	Force per unit charge
Electric Potential	$$V = k \frac{Q}{r}$$	Potential energy per unit charge
Capacitance (Parallel Plates)	$$C = \frac{\epsilon_0 A}{d}$$	Geometry dependent
Ohm's Law	$$V = IR$$	Voltage-Current relation
Resistance	$$R = \rho \frac{L}{A}$$	Depends on material
Electrical Power	$$P = VI = I^2 R = \frac{V^2}{R}$$	Power dissipated

15. Sample Problems and Solutions

Problem 1: Force Between Two Charges

Calculate the force between two charges of $2\,\mu C$ and $3\,\mu C$ placed 0.5 m apart in air.

Solution:

Given $q_1 = 2 \times 10^{-6} C$, $q_2 = 3 \times 10^{-6} C$, $r = 0.5\, m$, $k = 9 \times 10^9$.

$$F = k \frac{q_1 q_2}{r^2} = 9 \times 10^9 \times \frac{2 \times 10^{-6} \times 3 \times 10^{-6}}{(0.5)^2} = 0.216\, N$$

Problem 2: Equivalent Capacitance

Three capacitors of $2\,\mu F$, $3\,\mu F$, and $6\,\mu F$ are connected in series. Find equivalent capacitance.

Solution:

$$\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{1}{1} \implies C_{eq} = 1\, \mu F$$

Problem 3: Current in Series Circuit

Three resistors $5\, \Omega$, $10\, \Omega$, and $15\, \Omega$ are connected in series to a 12 V battery. Find current.

Solution:

$R_{eq} = 5 + 10 + 15 = 30\, \Omega$

$$I = \frac{V}{R_{eq}} = \frac{12}{30} = 0.4\, A$$

Problem 4: Using Kirchhoff’s Laws

In a circuit with two loops and resistors $R_1=4\,\Omega$, $R_2=6\,\Omega$, $R_3=3\,\Omega$, and batteries $E_1=12\,V$, $E_2=6\,V$, find the currents.

Solution: Set up loop equations using KVL and solve simultaneously (details depend on circuit). Practice is essential.

Conclusion

Mastery of Electricity in JEE Physics requires clear understanding of electrostatics, current electricity, circuit laws, and related formulas. Regular practice of numerical problems and conceptual questions is key. Use this guide for reference and strengthen your problem-solving skills to excel in your JEE exam.