The Static Electricity Formula describes the electrostatic force between two charged objects, expressed mathematically as Coulomb’s Law: \( F = k rac{q_1 q_2}{r^2} \). This fundamental formula is a core topic in NCERT Class 12 Physics, Chapter 1 (Electric Charges and Fields). It is equally critical for JEE Main, JEE Advanced, and NEET, where electrostatics questions appear every year. This article covers every key static electricity formula, derivations, solved examples at progressive difficulty levels, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET applications.
Key Static Electricity Formulas at a Glance
Quick reference for the most important static electricity formulas used in CBSE and competitive exams.
- Coulomb’s Law: \( F = k \frac{q_1 q_2}{r^2} \)
- Electric Field: \( E = \frac{F}{q} = k \frac{Q}{r^2} \)
- Electric Potential: \( V = k \frac{Q}{r} \)
- Potential Energy: \( U = k \frac{q_1 q_2}{r} \)
- Gauss’s Law: \( \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0} \)
- Capacitance: \( C = \frac{Q}{V} \)
- Energy stored in capacitor: \( U = \frac{1}{2}CV^2 \)
What is Static Electricity Formula?
The Static Electricity Formula refers to the set of mathematical expressions that quantify the behaviour of electric charges at rest. The most fundamental of these is Coulomb’s Law, which calculates the electrostatic force between two point charges. Static electricity arises when electric charges accumulate on the surface of an object without flowing as a current.
In NCERT Class 12 Physics, Chapter 1 (Electric Charges and Fields) and Chapter 2 (Electrostatic Potential and Capacitance) cover these formulas in detail. The concept also appears briefly in NCERT Class 8 Science, Chapter 15, as an introduction to static charge phenomena.
The static electricity formula set includes Coulomb’s Law, electric field intensity, electric potential, electrostatic potential energy, Gauss’s Law, and capacitor-related formulas. Together, these expressions form the backbone of electrostatics. Understanding each formula helps students solve CBSE board problems and competitive exam questions with confidence. The SI unit of charge is the Coulomb (C), and the SI unit of force is the Newton (N).
Static Electricity Formula — Expression and Variables
The primary static electricity formula is Coulomb’s Law:
\[ F = k \frac{q_1 q_2}{r^2} \]
where \( k = \frac{1}{4\pi\varepsilon_0} \approx 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \) is Coulomb’s constant.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | Electrostatic Force | Newton (N) |
| k | Coulomb’s Constant | N m² C² |
| q₁ | Charge on first object | Coulomb (C) |
| q₂ | Charge on second object | Coulomb (C) |
| r | Distance between charges | Metre (m) |
| ϵ₀ | Permittivity of free space | C² N¹ m² |
Derivation of Coulomb’s Law
Coulomb derived this law experimentally using a torsion balance in 1785. He observed that the force between two point charges is directly proportional to the product of the magnitudes of the charges: \( F \propto q_1 q_2 \). He also found that the force is inversely proportional to the square of the distance between them: \( F \propto \frac{1}{r^2} \). Combining these two observations gives \( F \propto \frac{q_1 q_2}{r^2} \). Introducing the proportionality constant \( k \), we get the final expression \( F = k \frac{q_1 q_2}{r^2} \). In a medium with relative permittivity \( \varepsilon_r \), the constant becomes \( k’ = \frac{k}{\varepsilon_r} \), reducing the force in that medium.
Complete Electrostatics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Coulomb’s Law | \( F = k \frac{q_1 q_2}{r^2} \) | F=Force, q=charges, r=distance | N | Class 12, Ch 1 |
| Electric Field Intensity | \( E = \frac{F}{q_0} = k \frac{Q}{r^2} \) | E=field, F=force, q₀=test charge | N C¹ or V m¹ | Class 12, Ch 1 |
| Electric Field due to a dipole (axial) | \( E = \frac{2kp}{r^3} \) | p=dipole moment, r=distance | N C¹ | Class 12, Ch 1 |
| Electric Field due to a dipole (equatorial) | \( E = \frac{kp}{r^3} \) | p=dipole moment, r=distance | N C¹ | Class 12, Ch 1 |
| Electric Potential | \( V = k \frac{Q}{r} \) | V=potential, Q=charge, r=distance | Volt (V) | Class 12, Ch 2 |
| Electrostatic Potential Energy | \( U = k \frac{q_1 q_2}{r} \) | U=potential energy, q=charges | Joule (J) | Class 12, Ch 2 |
| Gauss’s Law | \( \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0} \) | Qₛₙₘ=enclosed charge | N m² C¹ | Class 12, Ch 1 |
| Capacitance | \( C = \frac{Q}{V} \) | C=capacitance, Q=charge, V=potential | Farad (F) | Class 12, Ch 2 |
| Parallel Plate Capacitor | \( C = \frac{\varepsilon_0 A}{d} \) | A=area, d=separation | Farad (F) | Class 12, Ch 2 |
| Energy Stored in Capacitor | \( U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} \) | C=capacitance, V=voltage | Joule (J) | Class 12, Ch 2 |
| Electric Dipole Moment | \( p = q \times 2l \) | p=dipole moment, q=charge, 2l=separation | C m | Class 12, Ch 1 |
| Torque on Dipole in Electric Field | \( \tau = pE\sin\theta \) | τ=torque, p=dipole moment, θ=angle | N m | Class 12, Ch 1 |
Static Electricity Formula — Solved Examples
Example 1 (Class 9-10 Level): Force Between Two Charges
Problem: Two point charges of \( +4 \, \mu C \) and \( +6 \, \mu C \) are placed 30 cm apart in air. Calculate the electrostatic force between them.
Given:
- \( q_1 = 4 \times 10^{-6} \, C \)
- \( q_2 = 6 \times 10^{-6} \, C \)
- \( r = 0.30 \, m \)
- \( k = 9 \times 10^9 \, N\,m^2\,C^{-2} \)
Step 1: Write Coulomb’s Law: \( F = k \frac{q_1 q_2}{r^2} \)
Step 2: Substitute the values:
\[ F = 9 \times 10^9 \times \frac{(4 \times 10^{-6})(6 \times 10^{-6})}{(0.30)^2} \]
Step 3: Calculate numerator: \( 4 \times 10^{-6} \times 6 \times 10^{-6} = 24 \times 10^{-12} \)
Step 4: Calculate denominator: \( (0.30)^2 = 0.09 \)
Step 5: Final calculation: \( F = 9 \times 10^9 \times \frac{24 \times 10^{-12}}{0.09} = 9 \times 10^9 \times 2.667 \times 10^{-10} = 2.4 \, N \)
Answer
The electrostatic force between the two charges is 2.4 N (repulsive, since both charges are positive).
Example 2 (Class 11-12 Level): Electric Field and Potential at a Point
Problem: A point charge of \( Q = +5 \, \mu C \) is placed in free space. Find the electric field intensity and electric potential at a point 0.5 m away from the charge.
Given:
- \( Q = 5 \times 10^{-6} \, C \)
- \( r = 0.5 \, m \)
- \( k = 9 \times 10^9 \, N\,m^2\,C^{-2} \)
Step 1: Electric field formula: \( E = k \frac{Q}{r^2} \)
Step 2: Substitute values:
\[ E = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{(0.5)^2} = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{0.25} \]
Step 3: \( E = 9 \times 10^9 \times 2 \times 10^{-5} = 1.8 \times 10^5 \, N\,C^{-1} \)
Step 4: Electric potential formula: \( V = k \frac{Q}{r} \)
Step 5: Substitute values:
\[ V = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{0.5} = 9 \times 10^4 \, V \]
Answer
Electric field \( E = 1.8 \times 10^5 \, N\,C^{-1} \) and electric potential \( V = 9 \times 10^4 \, V \) at the given point.
Example 3 (JEE/NEET Level): Energy Stored in a Capacitor with Dielectric
Problem: A parallel plate capacitor has plate area \( A = 200 \, cm^2 \), plate separation \( d = 2 \, mm \), and is connected to a 100 V battery. A dielectric slab with relative permittivity \( \varepsilon_r = 5 \) is inserted between the plates while the battery remains connected. Find (a) capacitance before and after insertion, and (b) energy stored in each case.
Given:
- \( A = 200 \times 10^{-4} = 2 \times 10^{-2} \, m^2 \)
- \( d = 2 \times 10^{-3} \, m \)
- \( V = 100 \, V \)
- \( \varepsilon_0 = 8.85 \times 10^{-12} \, C^2\,N^{-1}\,m^{-2} \)
- \( \varepsilon_r = 5 \)
Step 1: Capacitance without dielectric:
\[ C_0 = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 2 \times 10^{-2}}{2 \times 10^{-3}} = 8.85 \times 10^{-11} \, F \approx 88.5 \, pF \]
Step 2: Energy stored without dielectric:
\[ U_0 = \frac{1}{2} C_0 V^2 = \frac{1}{2} \times 8.85 \times 10^{-11} \times (100)^2 = 4.425 \times 10^{-7} \, J \]
Step 3: Capacitance with dielectric (battery connected, so V stays at 100 V):
\[ C = \varepsilon_r C_0 = 5 \times 8.85 \times 10^{-11} = 4.425 \times 10^{-10} \, F \approx 442.5 \, pF \]
Step 4: Energy stored with dielectric:
\[ U = \frac{1}{2} C V^2 = \frac{1}{2} \times 4.425 \times 10^{-10} \times (100)^2 = 2.2125 \times 10^{-6} \, J \]
Step 5: Note that energy increases by a factor of \( \varepsilon_r = 5 \) because the battery maintains constant voltage and supplies extra charge.
Answer
(a) \( C_0 \approx 88.5 \, pF \) and \( C \approx 442.5 \, pF \). (b) \( U_0 \approx 4.43 \times 10^{-7} \, J \) and \( U \approx 2.21 \times 10^{-6} \, J \). Energy increases fivefold when the dielectric is inserted with the battery connected.
CBSE Exam Tips 2025-26
- Memorise the constant: Always write \( k = 9 \times 10^9 \, N\,m^2\,C^{-2} \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \, C^2\,N^{-1}\,m^{-2} \) at the start of your answer. CBSE often awards one mark for correctly stating the constant.
- State the law before applying it: In 2025-26 board exams, examiners expect you to write the formula first, then substitute values. Skipping this step costs marks even if the answer is correct.
- Direction of force matters: Clearly state whether the force is attractive or repulsive. Like charges repel; unlike charges attract. We recommend adding a one-line statement after your numerical answer.
- Unit conversions: Always convert microcoulombs (\( \mu C \)) to Coulombs and centimetres to metres before substituting. This is the most common source of errors in CBSE answers.
- Gauss’s Law applications: Practice the standard cases — infinite line charge, infinite plane sheet, and spherical shell. These are high-frequency 3-mark questions in CBSE 2025-26.
- Capacitor problems: Know both series and parallel combinations. For CBSE 2025-26, energy stored in a capacitor and the effect of a dielectric are very likely to appear as 5-mark questions.
Common Mistakes to Avoid
- Forgetting to square the distance: The most frequent error is writing \( F = k \frac{q_1 q_2}{r} \) instead of \( F = k \frac{q_1 q_2}{r^2} \). Always double-check the exponent on \( r \).
- Ignoring the sign of charges: The magnitude of force uses \( |q_1||q_2| \), but the sign determines attraction or repulsion. Do not substitute negative signs into the formula blindly; determine direction separately.
- Confusing electric field and electric force: Electric field \( E = \frac{F}{q_0} \) is force per unit positive test charge. Many students write \( E = F \) or \( E = kQ \) without the \( r^2 \) denominator.
- Mixing up potential and potential energy: Electric potential \( V = k \frac{Q}{r} \) is a scalar. Potential energy \( U = k \frac{q_1 q_2}{r} \) involves two charges. Substituting only one charge into the potential energy formula is a common error.
- Wrong formula for capacitor energy: Students often write \( U = CV^2 \) instead of \( U = \frac{1}{2}CV^2 \). The factor of \( \frac{1}{2} \) is essential and must not be omitted.
JEE/NEET Application of Static Electricity Formula
In our experience, JEE aspirants encounter static electricity formulas in at least 3–5 questions per paper. NEET typically includes 2–3 questions from electrostatics every year. Understanding the deeper applications of the static electricity formula is essential for scoring in these exams.
Pattern 1: Superposition of Forces
JEE problems frequently place three or more charges and ask for the net force on one charge. You must apply Coulomb’s Law for each pair and use vector addition. The formula remains \( F = k \frac{q_1 q_2}{r^2} \), but you resolve each force into x and y components before adding. Our experts suggest practising at least 10 such problems to build speed.
Pattern 2: Equipotential Surfaces and Work Done
NEET and JEE both test the concept that work done in moving a charge along an equipotential surface is zero. This uses the relation \( W = q(V_A – V_B) \). If \( V_A = V_B \), then \( W = 0 \). Questions often present a charge moved between two points and ask for work done, combining the electric potential formula \( V = k \frac{Q}{r} \) with the work-energy theorem.
Pattern 3: Capacitor with Dielectric — Battery Connected vs. Disconnected
This is a classic JEE Advanced topic. When a battery remains connected, voltage \( V \) stays constant and capacitance increases by \( \varepsilon_r \), so charge and energy both increase. When the battery is disconnected, charge \( Q \) stays constant and capacitance increases by \( \varepsilon_r \), so voltage and energy both decrease. Mixing up these two cases is the single most common error in JEE electrostatics. In our experience, JEE aspirants who master this distinction gain an easy 4 marks per paper.
For further reading on related phenomena, visit the official NCERT Class 12 Physics Textbook (Chapter 1) on the NCERT official website.
FAQs on Static Electricity Formula
Explore More Physics Formulas
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