The Inverse Square Law Formula states that a physical quantity varies inversely with the square of the distance from its source, expressed as \ ( I \propto \frac{1}{r^2} \). This fundamental principle appears across multiple chapters in NCERT Physics for Class 11 and Class 12, covering gravitation, electrostatics, and radiation. It is equally critical for JEE Main, JEE Advanced, and NEET, where questions on intensity, force, and field strength are regularly set. This article covers the formula expression, derivation, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Inverse Square Law Formulas at a Glance
Quick reference for the most important Inverse Square Law expressions used in CBSE and competitive exams.
- General form: \( I \propto \dfrac{1}{r^2} \)
- Gravitational force: \( F = \dfrac{Gm_1 m_2}{r^2} \)
- Coulomb's electrostatic force: \( F = \dfrac{kq_1 q_2}{r^2} \)
- Light intensity: \( I = \dfrac{P}{4\pi r^2} \)
- Electric field due to point charge: \( E = \dfrac{kq}{r^2} \)
- Ratio form: \( \dfrac{I_1}{I_2} = \dfrac{r_2^2}{r_1^2} \)
- Gravitational field: \( g = \dfrac{GM}{r^2} \)
What is the Inverse Square Law Formula?
The Inverse Square Law Formula describes a fundamental relationship in physics: when a physical quantity spreads outward from a point source, its intensity decreases in proportion to the square of the distance from that source. In simple terms, doubling the distance reduces the intensity to one-quarter of its original value.
This law applies to many natural phenomena. It governs gravitational attraction between masses, electrostatic force between charges, the intensity of light from a bulb, sound intensity from a speaker, and radiation from a radioactive source. The law arises because energy or force spreads over the surface area of an expanding sphere. That surface area grows as \( 4\pi r^2 \), so the intensity per unit area falls as \( 1/r^2 \).
In the NCERT curriculum, the Inverse Square Law Formula appears in Class 11 Chapter 8 (Gravitation) and Class 12 Chapter 1 (Electric Charges and Fields). It is one of the most tested concepts in both CBSE board examinations and competitive entrance tests. Understanding this formula deeply helps students connect gravitation, electrostatics, optics, and nuclear physics under one unifying principle.
Inverse Square Law Formula — Expression and Variables
The general form of the Inverse Square Law Formula is written as:
\[ I = \frac{k}{r^2} \]
where \( k \) is a constant that depends on the source strength and the nature of the quantity being measured. For light intensity from a point source of power \( P \), the formula becomes:
\[ I = \frac{P}{4\pi r^2} \]
For gravitational force between two masses, the Inverse Square Law takes the form of Newton's Law of Gravitation:
\[ F = \frac{G m_1 m_2}{r^2} \]
For electrostatic force between two point charges, Coulomb's Law gives:
\[ F = \frac{k q_1 q_2}{r^2} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( I \) | Intensity of the physical quantity | W/m² (for light/radiation) |
| \( r \) | Distance from the point source | metre (m) |
| \( P \) | Power of the source | Watt (W) |
| \( k \) | Proportionality constant (context-dependent) | Varies |
| \( G \) | Universal gravitational constant | N m² kg−² |
| \( m_1, m_2 \) | Masses of two objects | kilogram (kg) |
| \( q_1, q_2 \) | Electric charges | Coulomb (C) |
| \( F \) | Force (gravitational or electrostatic) | Newton (N) |
| \( E \) | Electric field intensity | N/C or V/m |
Derivation of the Inverse Square Law Formula
Consider a point source that emits energy uniformly in all directions. At a distance \( r \) from the source, the energy spreads over the surface of a sphere of radius \( r \). The surface area of a sphere is \( A = 4\pi r^2 \). If the total power emitted by the source is \( P \), then the intensity at distance \( r \) is:
\[ I = \frac{P}{A} = \frac{P}{4\pi r^2} \]
Since \( P \) and \( 4\pi \) are constants for a given source, we get \( I \propto \frac{1}{r^2} \). This is the mathematical basis of the Inverse Square Law. The same geometric argument applies to gravitational and electric fields, which spread outward from a point mass or point charge respectively.
Complete Physics Formula Sheet — Inverse Square Law and Related Formulas
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| General Inverse Square Law | \( I = k/r^2 \) | I = intensity, k = constant, r = distance | W/m² | Class 11, Ch 8 / Class 12, Ch 1 |
| Light Intensity from Point Source | \( I = P / (4\pi r^2) \) | P = power, r = distance | W/m² | Class 12, Ch 10 |
| Newton's Law of Gravitation | \( F = Gm_1 m_2 / r^2 \) | G = 6.674 × 10−¹¹ N m² kg−² | Newton (N) | Class 11, Ch 8 |
| Gravitational Field Intensity | \( g = GM / r^2 \) | M = source mass, r = distance | m/s² | Class 11, Ch 8 |
| Coulomb's Law (Electrostatic Force) | \( F = kq_1 q_2 / r^2 \) | k = 9 × 10&sup9; N m² C−² | Newton (N) | Class 12, Ch 1 |
| Electric Field due to Point Charge | \( E = kq / r^2 \) | q = charge, r = distance | N/C | Class 12, Ch 1 |
| Intensity Ratio Form | \( I_1/I_2 = r_2^2/r_1^2 \) | I = intensity at distances r&sub1; and r&sub2; | Dimensionless ratio | Class 12, Ch 10 |
| Sound Intensity | \( I = P / (4\pi r^2) \) | P = acoustic power, r = distance | W/m² | Class 11, Ch 15 |
| Radiation Intensity (Nuclear) | \( I \propto 1/r^2 \) | r = distance from radioactive source | W/m² | Class 12, Ch 13 |
| Electric Flux through Sphere | \( \Phi = q / \varepsilon_0 \) | q = enclosed charge, ε&sub0; = permittivity | N m² C−¹ | Class 12, Ch 1 |
Inverse Square Law Formula — Solved Examples
Example 1 (Class 9-10 Level): Light Intensity Comparison
Problem: A lamp has a power output of 100 W. Calculate the light intensity at a distance of 2 m from the lamp. Then find the intensity at 4 m. Assume the lamp radiates uniformly in all directions.
Given: P = 100 W, r&sub1; = 2 m, r&sub2; = 4 m
Step 1: Write the formula for light intensity: \( I = \dfrac{P}{4\pi r^2} \)
Step 2: Calculate intensity at r&sub1; = 2 m:
\( I_1 = \dfrac{100}{4\pi (2)^2} = \dfrac{100}{4\pi \times 4} = \dfrac{100}{16\pi} \approx \dfrac{100}{50.27} \approx 1.99 \) W/m²
Step 3: Calculate intensity at r&sub2; = 4 m:
\( I_2 = \dfrac{100}{4\pi (4)^2} = \dfrac{100}{4\pi \times 16} = \dfrac{100}{64\pi} \approx \dfrac{100}{201.06} \approx 0.497 \) W/m²
Step 4: Verify the inverse square relationship: \( I_1 / I_2 = (r_2/r_1)^2 = (4/2)^2 = 4 \). Indeed, \( 1.99 / 0.497 \approx 4 \). ✓
Answer
Intensity at 2 m ≈ 1.99 W/m². Intensity at 4 m ≈ 0.50 W/m². Doubling the distance reduces intensity to one-quarter, confirming the Inverse Square Law.
Example 2 (Class 11-12 Level): Gravitational Force Between Two Masses
Problem: The gravitational force between two spheres of masses 5 kg and 10 kg separated by 0.5 m is F&sub1;. If the separation is increased to 1.5 m, find the new gravitational force F&sub2;. Also calculate the ratio F&sub1;/F&sub2;.
Given: m&sub1; = 5 kg, m&sub2; = 10 kg, r&sub1; = 0.5 m, r&sub2; = 1.5 m, G = 6.674 × 10−¹¹ N m² kg−²
Step 1: Write Newton's Law of Gravitation: \( F = \dfrac{G m_1 m_2}{r^2} \)
Step 2: Calculate F&sub1; at r&sub1; = 0.5 m:
\( F_1 = \dfrac{6.674 \times 10^{-11} \times 5 \times 10}{(0.5)^2} = \dfrac{6.674 \times 10^{-11} \times 50}{0.25} = \dfrac{3.337 \times 10^{-9}}{0.25} = 1.335 \times 10^{-8} \) N
Step 3: Calculate F&sub2; at r&sub2; = 1.5 m:
\( F_2 = \dfrac{6.674 \times 10^{-11} \times 50}{(1.5)^2} = \dfrac{3.337 \times 10^{-9}}{2.25} = 1.483 \times 10^{-9} \) N
Step 4: Find the ratio using the inverse square relationship:
\( \dfrac{F_1}{F_2} = \left(\dfrac{r_2}{r_1}\right)^2 = \left(\dfrac{1.5}{0.5}\right)^2 = 3^2 = 9 \)
Answer
F&sub1; = 1.335 × 10−&sup8; N, F&sub2; = 1.483 × 10−&sup9; N. The ratio F&sub1;/F&sub2; = 9. Tripling the distance reduces the gravitational force to one-ninth of its original value.
Example 3 (JEE/NEET Level): Electrostatic Force and Intensity Combined
Problem: A point charge q = 4 μC is placed at the origin. (a) Find the electric field intensity at r = 0.3 m. (b) At what distance from the charge will the electric field be reduced to one-ninth of the value found in (a)?
Given: q = 4 × 10−&sup6; C, r&sub1; = 0.3 m, k = 9 × 10&sup9; N m² C−²
Step 1: Write the electric field formula: \( E = \dfrac{kq}{r^2} \)
Step 2: Calculate E at r&sub1; = 0.3 m:
\( E_1 = \dfrac{9 \times 10^9 \times 4 \times 10^{-6}}{(0.3)^2} = \dfrac{3.6 \times 10^4}{0.09} = 4 \times 10^5 \) N/C
Step 3: Use the Inverse Square Law ratio to find r&sub2; where \( E_2 = E_1 / 9 \):
\( \dfrac{E_1}{E_2} = \dfrac{r_2^2}{r_1^2} \Rightarrow 9 = \dfrac{r_2^2}{(0.3)^2} \Rightarrow r_2^2 = 9 \times 0.09 = 0.81 \Rightarrow r_2 = 0.9 \) m
Step 4: Verify: \( E_2 = \dfrac{9 \times 10^9 \times 4 \times 10^{-6}}{(0.9)^2} = \dfrac{3.6 \times 10^4}{0.81} \approx 4.44 \times 10^4 \) N/C \( = E_1/9 \). ✓
Answer
(a) Electric field at 0.3 m = 4 × 10&sup5; N/C. (b) The field reduces to one-ninth at r&sub2; = 0.9 m, which is three times the original distance. This confirms the Inverse Square Law: tripling the distance reduces field intensity to 1/9.
CBSE Exam Tips 2025-26
- Memorise all three forms: Know the gravitational, electrostatic, and radiation forms of the Inverse Square Law. CBSE 2025-26 papers test all three in different chapters.
- Use the ratio method: When only the ratio of distances is given, use \( I_1/I_2 = r_2^2/r_1^2 \) directly. This saves time and avoids calculation errors in board exams.
- State the law in words first: In 3-mark and 5-mark answers, always write the statement of the law before the formula. CBSE awards one mark for the statement alone.
- Check units carefully: Intensity is in W/m², force is in Newtons, and electric field is in N/C or V/m. Mixing units is a common reason for losing marks.
- Link to Gauss's Law: In Class 12, the Inverse Square Law for electric fields is directly derived from Gauss's Law. Mentioning this connection in answers demonstrates deeper understanding and can earn bonus marks.
- We recommend practising at least 10 numerical problems on the ratio form of the law. This builds speed for the 2025-26 board paper.
Common Mistakes to Avoid
- Squaring only one variable: A frequent error is writing \( I \propto 1/r \) instead of \( I \propto 1/r^2 \). Always remember the exponent is 2, not 1. The law is called “inverse square” for this reason.
- Forgetting the \( 4\pi \) factor: For light and radiation intensity, the formula is \( I = P/(4\pi r^2) \), not \( I = P/r^2 \). The \( 4\pi \) comes from the surface area of a sphere and must not be omitted.
- Confusing force with field: Gravitational force \( F = Gm_1 m_2/r^2 \) requires two masses, but gravitational field \( g = GM/r^2 \) uses only the source mass. Mixing these two formulas leads to wrong answers.
- Incorrect ratio setup: When using the ratio method, students sometimes write \( I_1/I_2 = r_1^2/r_2^2 \) (inverted). The correct form is \( I_1/I_2 = r_2^2/r_1^2 \). Since intensity decreases with distance, the larger distance must appear in the numerator of the ratio.
- Using diameter instead of radius: Always use the distance from the source (radius), not the diameter of any object, when substituting into the formula.
JEE/NEET Application of the Inverse Square Law Formula
In our experience, JEE aspirants encounter the Inverse Square Law Formula in at least 3 to 5 questions per paper, spread across gravitation, electrostatics, and optics. Understanding the unifying principle behind all these applications is key to scoring full marks.
Application Pattern 1: Comparing Intensities at Different Distances
JEE Main frequently asks questions of the type: “If the intensity at distance d is I, what is the intensity at distance 3d?” Use the ratio form directly:
\[ \frac{I_1}{I_2} = \frac{r_2^2}{r_1^2} \Rightarrow I_2 = I_1 \times \frac{r_1^2}{r_2^2} = I \times \frac{d^2}{9d^2} = \frac{I}{9} \]
This pattern appears in optics (photometry), gravitation, and electrostatics questions in both JEE and NEET.
Application Pattern 2: Finding Distance for a Given Field or Force
NEET questions often give a target intensity or force value and ask for the corresponding distance. Rearrange the formula to solve for \( r \):
\[ r = \sqrt{\frac{kq}{E}} \quad \text{or} \quad r = \sqrt{\frac{Gm_1 m_2}{F}} \]
This type of question tests algebraic manipulation skills alongside conceptual understanding.
Application Pattern 3: Multi-Source Superposition
JEE Advanced problems combine the Inverse Square Law with vector superposition. Two charges or two sources are placed at known positions. The student must find the net field or intensity at a third point by calculating individual contributions using \( E = kq/r^2 \) and adding them vectorially. This tests both the Inverse Square Law and vector addition simultaneously.
Our experts suggest practising these three patterns using previous years' JEE and NEET papers from 2018 to 2024. The Inverse Square Law Formula appears consistently across all years, making it a high-priority topic for revision.
FAQs on the Inverse Square Law Formula
Explore More Physics Formulas
To strengthen your understanding of related concepts, explore our detailed articles on the Electric Flux Formula, which directly connects to the Inverse Square Law through Gauss's Law, and the Angular Speed Formula for rotational mechanics. For optics and wave topics, our guide on the Critical Angle Formula provides equally comprehensive coverage. You can also browse the complete Physics Formulas hub for a full list of NCERT-aligned formula guides for Class 11 and Class 12. For official NCERT textbook content, refer to the NCERT official website.