JEE Electrostatics Formula 2026
Electrostatics is one of the most important and high-scoring chapters in JEE Physics. Every year, around 2–4 questions are asked from this topic in both JEE Main and JEE Advanced, making it a must-prepare section for aspirants. The chapter introduces fundamental concepts such as electric charge, Coulomb’s law, electric field, electric potential, Gauss’s law, electric dipoles, and capacitors. Since many problems in electromagnetism are directly based on these concepts, mastering the formulas and their applications is essential for solving numerical questions quickly and accurately in the exam.
To strengthen your preparation, students should regularly revise the key equations and practice applying them to different types of problems. A quick revision resource like a JEE Physics Formula PDF can be extremely helpful for remembering important electrostatics formulas during last-minute preparation. Referring to a well-organized JEE Physics Formula PDF also allows students to connect electrostatics concepts with other related topics in electromagnetism, helping them build a strong conceptual foundation and improve problem-solving speed in competitive exams.
Electric Charge — Properties and Quantization
Everything around us is made of atoms, and atoms contain tiny particles called protons (positive charge) and electrons (negative charge). Normally, an object has equal numbers of protons and electrons, so it appears electrically neutral. When electrons are transferred from one object to another, one becomes positively charged (lost electrons) and the other becomes negatively charged (gained electrons). This is how objects get "charged."
Electric Charge: A fundamental property of matter that causes it to experience a force when placed near other charged matter. It is measured in coulombs (C).
Coulomb: The SI unit of charge. One coulomb is the charge carried by approximately $$6.25 \times 10^{18}$$ electrons (or protons).
Properties of Electric Charge
- Two types: Positive ($$+$$) and Negative ($$-$$). Like charges repel; unlike charges attract.
- Quantization: Charge always comes in integer multiples of the elementary charge $$e$$. $$$q = ne \quad \text{where } n = 0, \pm 1, \pm 2, \ldots \text{ and } e = 1.6 \times 10^{-19} \text{ C}$$$
- Conservation: Charge can neither be created nor destroyed. The total charge of an isolated system remains constant.
- Additivity: Total charge is the algebraic sum of all individual charges: $$q_{\text{total}} = q_1 + q_2 + \cdots$$
- Invariance: Charge does not depend on the speed of the charged particle.
Worked Example
A body has a charge of $$-3.2 \times 10^{-18}$$ C. How many excess electrons does it have?
Using $$q = ne$$:
$$n = \dfrac{q}{e} = \dfrac{3.2 \times 10^{-18}}{1.6 \times 10^{-19}} = 20$$ excess electrons.
Tip: Remember: protons are tightly bound inside the nucleus and do not move. Charging happens only by transfer of electrons. A positively charged body has lost electrons, not gained protons.
Coulomb's Law
Two charged objects exert a force on each other. In 1785, Charles-Augustin de Coulomb measured this force and found a simple law: the force depends on how much charge each object has and how far apart they are.
Coulomb's Law
The electrostatic force between two point charges $$q_1$$ and $$q_2$$ separated by a distance $$r$$ is:
$$$F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{|q_1 \, q_2|}{r^2}$$$
where:
- $$\varepsilon_0 = 8.85 \times 10^{-12} \; \text{C}^2 \text{N}^{-1}\text{m}^{-2}$$ is the permittivity of free space
- $$\dfrac{1}{4\pi\varepsilon_0} = k = 9 \times 10^9 \; \text{N m}^2 \text{C}^{-2}$$ is Coulomb's constant
- The force is attractive if charges are opposite, repulsive if same sign
- In a medium with dielectric constant $$\kappa$$: $$F_{\text{medium}} = \dfrac{F_{\text{vacuum}}}{\kappa}$$
In vector form, the force on charge $$q_2$$ due to charge $$q_1$$ is:
$$$\vec{F}_{12} = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q_1 \, q_2}{r^2} \, \hat{r}_{12}$$$
where $$\hat{r}_{12}$$ is the unit vector from $$q_1$$ to $$q_2$$. The sign of charges automatically gives the correct direction.
Note: Coulomb's law applies only to point charges (or spherical charge distributions where $$r$$ is measured from the centre). For extended charge distributions, we must integrate.
Worked Example
Two charges $$q_1 = +3\;\mu\text{C}$$ and $$q_2 = -4\;\mu\text{C}$$ are separated by 0.2 m. Find the force between them.
$$F = k \cdot \dfrac{|q_1 q_2|}{r^2} = 9 \times 10^9 \times \dfrac{3 \times 10^{-6} \times 4 \times 10^{-6}}{(0.2)^2}$$
$$= 9 \times 10^9 \times \dfrac{12 \times 10^{-12}}{0.04} = 9 \times 10^9 \times 3 \times 10^{-10} = 2.7 \; \text{N}$$
Since the charges are opposite, the force is attractive.
Superposition Principle
When there are more than two charges, the net force on any one charge is the vector sum of the forces due to all other charges, calculated independently using Coulomb's law.
$$$\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \cdots$$$
Electric Field Formulas
Instead of thinking about "action at a distance," we say that every charge creates an electric field in the space around it. When another charge is placed in this field, it experiences a force. The electric field is the "messenger" that carries the force.
Electric Field: The force experienced per unit positive test charge placed at a point: $$\vec{E} = \vec{F}/q_0$$. Its SI unit is N/C or equivalently V/m.
Electric Field Due to a Point Charge
$$$\vec{E} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r^2} \, \hat{r}$$$
- $$\vec{E}$$ points away from a positive charge (radially outward)
- $$\vec{E}$$ points toward a negative charge (radially inward)
- Magnitude: $$E = \dfrac{kq}{r^2}$$
Electric Field Due to Special Charge Distributions
For JEE, you need to know the electric field produced by several standard configurations. These are derived by integrating Coulomb's law over the charge distribution.
Important Electric Field Formulas
- On the axis of a dipole (at distance $$r \gg l$$ from centre): $$$E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r^3}$$$
- On the equatorial plane of a dipole (at distance $$r \gg l$$): $$$E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{r^3}$$$
- Infinite line charge (charge per unit length $$\lambda$$, at distance $$r$$): $$$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$$
- On the axis of a uniformly charged ring (radius $$R$$, total charge $$Q$$, at distance $$x$$): $$$E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qx}{(R^2 + x^2)^{3/2}}$$$
- On the axis of a uniformly charged disc (surface charge $$\sigma$$, radius $$R$$, at distance $$x$$): $$$E = \frac{\sigma}{2\varepsilon_0}\left[1 - \frac{x}{\sqrt{R^2 + x^2}}\right]$$$
- Infinite plane sheet (surface charge $$\sigma$$): $$$E = \frac{\sigma}{2\varepsilon_0} \quad \text{(uniform, independent of distance)}$$$
Tip: For the charged ring, $$E = 0$$ at the centre ($$x = 0$$) and maximum at $$x = R/\sqrt{2}$$. This is a frequently asked JEE question.
Worked Example
Find the electric field at the centre of a uniformly charged ring of radius 0.1 m carrying a total charge of $$5\;\mu\text{C}$$.
At the centre of the ring, $$x = 0$$:
$$E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q \cdot 0}{(R^2 + 0)^{3/2}} = 0$$
The electric field at the centre of a uniformly charged ring is always zero by symmetry.
Electric Dipole
An electric dipole consists of two equal and opposite charges ($$+q$$ and $$-q$$) separated by a small distance $$2l$$. Many molecules (like water) behave as dipoles, making this an important concept.
Electric Dipole Moment: A vector quantity $$\vec{p} = q \cdot 2l \, \hat{d}$$, directed from the negative charge to the positive charge. SI unit: C·m.
Dipole in a Uniform Electric Field
- Torque: $$\vec{\tau} = \vec{p} \times \vec{E}$$, with magnitude $$\tau = pE\sin\theta$$
- Potential energy: $$U = -\vec{p} \cdot \vec{E} = -pE\cos\theta$$
- Net force: Zero in a uniform field (both charges feel equal and opposite forces)
- In a non-uniform field, there is a net force: $$F = p \dfrac{dE}{dx}$$
Electric Field Lines — Properties and Rules
Electric field lines are imaginary curves drawn to visualize the electric field. They are not the actual path a charge follows; they show the direction of force on a positive test charge at each point.
Properties of Electric Field Lines
- Start from positive charges and end on negative charges
- Never cross each other (field has unique direction at every point)
- Closer lines = stronger field; farther apart = weaker field
- Perpendicular to the surface of a conductor at every point
- Do not form closed loops in electrostatics
- Number of lines proportional to the magnitude of the charge
Electric Potential and Potential Energy
Moving a charge against an electric field requires work (like lifting a ball against gravity). This leads to the concept of electric potential — the electric field version of "height" in gravity.
Electric Potential: The work done per unit positive charge in bringing a test charge from infinity to a point: $$V = W/q_0$$. SI unit: volt (V) = J/C.
Potential Difference: The work done per unit charge in moving a charge between two points: $$V_A - V_B = W_{B \to A}/q_0$$.
Electric Potential Formulas
- Due to a point charge: $$V = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q}{r}$$
- Due to a dipole:
- Axial point: $$V = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{p}{r^2}$$
- Equatorial point: $$V = 0$$
- General point: $$V = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{p\cos\theta}{r^2}$$
- Due to a uniformly charged sphere (radius $$R$$, total charge $$Q$$):
- Outside ($$r > R$$): $$V = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r}$$ (behaves like a point charge)
- On the surface ($$r = R$$): $$V = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{R}$$
- Inside ($$r < R$$): $$V = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{R}$$ (constant, same as surface)
- Due to multiple charges: $$V = V_1 + V_2 + V_3 + \cdots$$ (scalar addition)
The relationship between electric field and potential is:
$$$\vec{E} = -\dfrac{dV}{dr}\,\hat{r}$$$ (field points from higher to lower potential)
Electrostatic Potential Energy
- Two point charges: $$U = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q_1 q_2}{r}$$
- System of charges: $$U = \dfrac{1}{4\pi\varepsilon_0} \sum_{i
- For $$n$$ charges, there are $$\dfrac{n(n-1)}{2}$$ pairs
Worked Example
Three charges $$q_1 = 1\;\mu\text{C}$$, $$q_2 = 2\;\mu\text{C}$$, $$q_3 = -3\;\mu\text{C}$$ are at the vertices of an equilateral triangle of side 1 m. Find the total potential energy.
$$U = k\left(\dfrac{q_1 q_2}{r_{12}} + \dfrac{q_1 q_3}{r_{13}} + \dfrac{q_2 q_3}{r_{23}}\right)$$
$$= 9 \times 10^9 \left(\dfrac{(1)(2) \times 10^{-12}}{1} + \dfrac{(1)(-3) \times 10^{-12}}{1} + \dfrac{(2)(-3) \times 10^{-12}}{1}\right)$$
$$= 9 \times 10^9 \times (2 - 3 - 6) \times 10^{-12} = 9 \times 10^9 \times (-7) \times 10^{-12}$$
$$= -6.3 \times 10^{-2} \; \text{J} = -0.063 \; \text{J}$$
Equipotential Surfaces
An equipotential surface is a surface where every point has the same electric potential. No work is done when a charge moves along an equipotential surface.
Properties of Equipotential Surfaces
- Electric field lines are always perpendicular to equipotential surfaces
- No work is done in moving a charge along an equipotential surface ($$W = q\Delta V = 0$$)
- Closer equipotential surfaces = stronger electric field
- For a point charge: concentric spheres
- For a uniform field: parallel planes perpendicular to the field
- The surface of a conductor is always an equipotential surface
Gauss's Law
Gauss's law provides a powerful shortcut for calculating electric fields when the charge distribution has symmetry — spherical, cylindrical, or planar. Instead of integrating Coulomb's law, we use electric flux.
Electric Flux: The number of electric field lines passing through a surface. For a uniform field $$\vec{E}$$ through a flat surface of area $$\vec{A}$$: $$\Phi = \vec{E} \cdot \vec{A} = EA\cos\theta$$. SI unit: N·m²/C (or V·m).
Gauss's Law
The total electric flux through any closed surface is equal to the net charge enclosed divided by $$\varepsilon_0$$:
$$$\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$$
- The closed surface is called a Gaussian surface (an imaginary surface we choose)
- Only the charge inside the surface contributes to the flux
- Charges outside contribute zero net flux through the surface
Applications of Gauss's Law
Electric Field Using Gauss's Law
- Uniformly charged sphere (total charge $$Q$$, radius $$R$$):
- Outside ($$r > R$$): $$E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r^2}$$
- On surface ($$r = R$$): $$E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{R^2}$$
- Inside (conducting): $$E = 0$$
- Inside (non-conducting, uniform $$\rho$$): $$E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Qr}{R^3}$$
- Infinite line charge ($$\lambda$$ = charge per unit length): $$$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$$
- Infinite plane sheet ($$\sigma$$ = surface charge density): $$$E = \frac{\sigma}{2\varepsilon_0}$$$
- Two parallel plates (with $$+\sigma$$ and $$-\sigma$$): $$$E_{\text{between}} = \frac{\sigma}{\varepsilon_0}, \quad E_{\text{outside}} = 0$$$
Worked Example
A solid non-conducting sphere of radius 10 cm has a uniform charge density. The total charge is $$Q = 4\;\mu\text{C}$$. Find $$E$$ at $$r = 5$$ cm (inside) and $$r = 20$$ cm (outside).
Inside ($$r = 0.05$$ m $$< R = 0.10$$ m):
$$E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Qr}{R^3} = 9 \times 10^9 \times \dfrac{4 \times 10^{-6} \times 0.05}{(0.10)^3} = 9 \times 10^9 \times \dfrac{2 \times 10^{-7}}{10^{-3}} = 1.8 \times 10^6$$ N/C
Outside ($$r = 0.20$$ m $$> R$$):
$$E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r^2} = 9 \times 10^9 \times \dfrac{4 \times 10^{-6}}{(0.20)^2} = 9 \times 10^9 \times 10^{-4} = 9 \times 10^5$$ N/C
Tip: When choosing a Gaussian surface, pick one that exploits symmetry: sphere for point/spherical charges, cylinder for line charges, pillbox (short cylinder) for plane sheets.
Capacitor Formulas — Series, Parallel, and Energy
A capacitor is a device that stores electric charge (and therefore energy) by maintaining a potential difference between two conductors. The simplest example is two parallel metal plates separated by a small gap.
Capacitance: The ratio of the charge stored on a capacitor to the potential difference across it: $$C = Q/V$$. SI unit: farad (F). Typical values are in $$\mu$$F, nF, or pF.
Parallel Plate Capacitor
For two parallel plates, each of area $$A$$, separated by distance $$d$$:
$$$C = \frac{\varepsilon_0 A}{d}$$$
- $$E$$ between plates: $$E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{V}{d}$$
- $$V$$ across plates: $$V = Ed$$
- Charge on each plate: $$Q = CV$$
Capacitors with Dielectrics
When an insulating material (called a dielectric) is placed between the plates of a capacitor, it increases the capacitance. The dielectric gets polarized — positive and negative charges within it shift slightly, partially cancelling the internal field.
Dielectric Constant ($$\kappa$$): The factor by which the capacitance increases when a dielectric fills the space between the plates. Also called relative permittivity $$\varepsilon_r$$. Always $$\kappa \geq 1$$.
Effect of Dielectric
- Capacitance: $$C = \dfrac{\kappa \varepsilon_0 A}{d}$$ (increases by factor $$\kappa$$)
- Electric field: $$E = \dfrac{E_0}{\kappa}$$ (decreases by factor $$\kappa$$)
- If battery stays connected: $$V$$ stays same, $$Q$$ increases by $$\kappa$$
- If battery is disconnected first: $$Q$$ stays same, $$V$$ decreases by $$\kappa$$
Combination of Capacitors
Series and Parallel Combination
Parallel (same voltage across each):
$$$C_{\text{eq}} = C_1 + C_2 + C_3 + \cdots$$$
Series (same charge on each):
$$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots$$$
For two capacitors in series: $$C_{\text{eq}} = \dfrac{C_1 C_2}{C_1 + C_2}$$
Tip: Capacitors in parallel add directly (like resistors in series), and capacitors in series add reciprocally (like resistors in parallel). They are opposite to resistors!
Energy Stored in a Capacitor
A charged capacitor stores electrical energy in the electric field between its plates. This energy can be released when the capacitor is discharged.
Energy Stored in a Capacitor
$$$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$$$
- $$C$$ = capacitance, $$V$$ = potential difference, $$Q$$ = charge stored
- Energy density (energy per unit volume) in an electric field: $$$u = \frac{1}{2}\varepsilon_0 E^2$$$
Worked Example
A parallel plate capacitor has plate area $$200\;\text{cm}^2$$ and plate separation 1 mm. Find: (a) capacitance, (b) charge when connected to a 100 V battery, (c) energy stored.
(a) $$C = \dfrac{\varepsilon_0 A}{d} = \dfrac{8.85 \times 10^{-12} \times 200 \times 10^{-4}}{1 \times 10^{-3}} = \dfrac{8.85 \times 10^{-12} \times 0.02}{0.001} = 1.77 \times 10^{-10}$$ F $$= 177$$ pF
(b) $$Q = CV = 1.77 \times 10^{-10} \times 100 = 1.77 \times 10^{-8}$$ C $$= 17.7$$ nC
(c) $$U = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 1.77 \times 10^{-10} \times (100)^2 = 8.85 \times 10^{-7}$$ J $$= 0.885\;\mu$$J
Worked Example
Two capacitors $$C_1 = 3\;\mu\text{F}$$ and $$C_2 = 6\;\mu\text{F}$$ are connected in series across a 12 V battery. Find the equivalent capacitance, charge on each, and voltage across each.
$$C_{\text{eq}} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{3 \times 6}{3 + 6} = \dfrac{18}{9} = 2\;\mu\text{F}$$
In series, charge is same on both: $$Q = C_{\text{eq}} \times V = 2 \times 10^{-6} \times 12 = 24\;\mu\text{C}$$
$$V_1 = \dfrac{Q}{C_1} = \dfrac{24}{3} = 8$$ V, $$V_2 = \dfrac{Q}{C_2} = \dfrac{24}{6} = 4$$ V
Check: $$V_1 + V_2 = 8 + 4 = 12$$ V ✓
Note: When a dielectric slab of thickness $$t$$ and dielectric constant $$\kappa$$ is inserted in a capacitor of plate separation $$d$$: $$$C = \frac{\varepsilon_0 A}{d - t + t/\kappa}$$$ If the slab completely fills the gap ($$t = d$$), this reduces to $$C = \kappa\varepsilon_0 A/d$$.
JEE Electrostatics Formula 2026: Conclusion
The JEE Electrostatics Formula 2026 chapter forms the foundation of many important topics in electromagnetism. Concepts such as electric charge, Coulomb’s law, electric field, electric potential, Gauss’s law, and capacitors play a crucial role in solving numerical problems in JEE Physics. Since this topic frequently appears in both JEE Main and JEE Advanced, mastering the formulas and understanding their applications can significantly improve a student’s problem-solving speed and accuracy.
For effective revision, students should maintain short notes or refer to a well-organized JEE Physics Formula PDF that includes all the important electrostatics equations in one place. Regular practice of numerical problems, along with quick formula revision, helps build strong conceptual clarity and confidence. With consistent practice and a strong command of the JEE Electrostatics formulas, aspirants can score well in the physics section and strengthen their overall JEE performance.
