Consider a coin of Question 1.20. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8\text{ kC}$. Suppose that these equal charges were concentrated in two point charges separated by (i) $1\text{ cm}$ $\left(\sim \tfrac{1}{2} \times \text{diagonal of the one paisa coin}\right)$, (ii) $100\text{ m}$ ($\sim$ length of a long building), and (iii) $10^6\text{ m}$ (radius of the earth). Find the force on each such point charge in each of the three cases. What do you conclude from these results?

Consider a coin of Question 1.20. It is electrically neutral and contains equal amounts of positive and negati

NCERT Exemplar · Q21

34.8\text{ kC}

(i)

1\text{ cm}

\left(\sim \tfrac{1}{2} \times \text{diagonal of the one paisa coin}\right)

(ii)

100\text{ m}

(

\sim

length of a long building), and

(iii)

10^6\text{ m}

(radius of the earth). Find the force on each such point charge in each of the three cases. What do you conclude from these results?

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Concept understanding — Inverse Square Law Comparison

The Intuition: Why Does Light Get Dimmer So Fast?

Imagine you're standing near a campfire. You feel its warmth on your face. Now take ten steps back. Does the warmth feel half as strong? No — it feels much weaker, maybe a quarter as strong. That's not an accident. It's a pattern that shows up everywhere in physics: gravity, light, sound, electric fields, even radiation.

The reason is simple: as you move away from a source, the same amount of energy (or force) has to spread out over a larger area. And that area grows with the square of the distance.

The Core Idea in One Picture

Think of a light bulb at the centre of a balloon. As you inflate the balloon, the light hitting the inner surface spreads thinner and thinner. If you double the radius of the balloon, the surface area becomes four times larger. So each patch of the balloon gets only one-fourth the light.

That's the inverse square law in a nutshell: double the distance → one-fourth the intensity.

The Precise Statement

I \propto \frac{1}{r^2} \quad \text{or} \quad I = \frac{k}{r^2}

where:

$I$ = intensity (brightness, force per unit area, etc.)
$r$ = distance from the source
$k$ = a constant that depends on the source's strength

If you compare two distances $r_1$ and $r_2$ , the ratio of intensities is:

\frac{I_2}{I_1} = \left( \frac{r_1}{r_2} \right)^2

This is the inverse square law comparison — you compare how strong a quantity is at two different distances by taking the inverse ratio of the squares of those distances.

Why "Inverse Square" and Not Just "Inverse"?

Because the geometry of space is three-dimensional. The surface of a sphere is $4\pi r^2$ . As $r$ grows, the sphere's surface grows as $r^2$ . Whatever is radiating outward (light, sound, gravity) must pass through that entire surface. So the amount per unit area drops as $1/r^2$ .

If we lived in a flat, two-dimensional world, the law would be $1/r$ (like ripples on a pond). In one dimension, it would be constant. The inverse square law is a direct consequence of living in three dimensions.

The Comparison: What It Really Means

When you compare two situations, you're not calculating absolute intensity — you're finding the ratio. For example:

A star is 3 times farther away than another identical star. How much dimmer does it appear?

\frac{I_{\text{far}}}{I_{\text{near}}} = \left( \frac{1}{3} \right)^2 = \frac{1}{9}

The farther star is 9 times dimmer. Not 3 times — 9 times. That's the punch of the square.

Watch out

A common mistake: thinking "twice the distance means half the intensity." It's actually one-fourth. The square makes the drop much steeper than linear intuition suggests. …

Why this formula?

Inverse Square Law Comparison — Why the Formula Holds

The Inverse Square Law appears in physics wherever a quantity spreads out uniformly from a point source in three-dimensional space. The core idea is that the intensity (or field strength) decreases as the square of the distance from the source.

1. The Intuition: Spreading Over a Sphere

Imagine a point source emitting energy, light, sound, or gravitational force equally in all directions.

At a distance $r$ , the energy is spread uniformly over the surface area of a sphere of radius $r$ .
The surface area of a sphere is:

A = 4\pi r^2

If the total power (or flux) emitted by the source is $P$ , then the intensity $I$ (power per unit area) at distance $r$ is:

I = \frac{P}{4\pi r^2}

Key insight: The same total power is spread over a larger and larger area as $r$ increases. Hence, intensity is inversely proportional to $r^2$ .

2. Derivation for Gravitational Force (Newton's Law)

Newton’s law of gravitation states:

F = \frac{G M m}{r^2}

Why $1/r^2$ ?

The gravitational field lines from a point mass $M$ radiate outward uniformly.
The number of field lines crossing a sphere of radius $r$ is constant (conservation of flux).
The density of field lines (force per unit mass) at distance $r$ is:

g = \frac{GM}{r^2}

This is because the total flux $\Phi = 4\pi G M$ is spread over $4\pi r^2$ , giving:

g = \frac{\Phi}{4\pi r^2} = \frac{GM}{r^2}

Thus, the force on a test mass $m$ is $F = m g = \frac{G M m}{r^2}$ .

3. Derivation for Coulomb's Law (Electrostatics)

Coulomb’s law for electric force between two point charges $q_1$ and $q_2$ :

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

Why $1/r^2$ ?

Electric field lines from a point charge $q$ radiate radially outward (or inward for negative charge).
Gauss’s law states that the total electric flux through a closed surface is proportional to the enclosed charge:

\oint \mathbf{E} \cdot d\mathbf{A} = \frac{q}{\varepsilon_0}

For a sphere of radius $r$ centered on the charge, the field is radial and constant in magnitude:

E \cdot 4\pi r^2 = \frac{q}{\varepsilon_0}

Therefore:

E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}

The force on a test charge $q_2$ is $F = q_2 E = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$ .

4. Derivation for Light/Radiation Intensity

For a point source of light emitting power $P$ :

At distance $r$ , the power is spread over a sphere of area $4\pi r^2$ .
Illuminance (intensity) is:

I = \frac{P}{4\pi r^2}

Why not $1/r$ ?

In 2D (e.g., a line source), intensity falls as $1/r$ because the circumference of a circle is $2\pi r$ .
In 3D, the surface area grows as $r^2$ , so intensity falls as $1/r^2$ .

5. The Common Mathematical Reason

All inverse square laws arise from conservation of flux in three-dimensional space with isotropic emission. The geometry forces: …

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By Coulomb's law $F=\dfrac{kq^2}{r^2}$ with $q=34.8$ kC, the forces are (i) $1.09\times10^{23}\ \text{N}$ ,

(ii) $1.09\times10^{15}\ \text{N}$ ,

(iii) $1.09\times10^{7}\ \text{N}$ . Even at Earth-radius separation the force is colossal, so a coin's positive and negative charges must be intimately mixed — matter is stable only because it is electrically neutral at every macroscopic scale.

Set up the constant part

The magnitude of the force between the two point charges is

F=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2}=\frac{kq^2}{r^2},\qquad k=8.99\times10^{9}\ \text{N m}^2\,\text{C}^{-2}

With $q=34.8\ \text{kC}=3.48\times10^{4}\ \text{C}$ :

q^2=(3.48\times10^{4})^2=1.211\times10^{9}\ \text{C}^2,\qquad kq^2=8.99\times10^{9}\times1.211\times10^{9}=1.09\times10^{19}\ \text{N m}^2

This numerator is the same in all three cases; only $r$ changes.

Case (i): $r=1\ \text{cm}=10^{-2}\ \text{m}$

F=\frac{1.09\times10^{19}}{(10^{-2})^2}=\frac{1.09\times10^{19}}{10^{-4}}=1.09\times10^{23}\ \text{N}

Case (ii): $r=100\ \text{m}=10^{2}\ \text{m}$

F=\frac{1.09\times10^{19}}{(10^{2})^2}=\frac{1.09\times10^{19}}{10^{4}}=1.09\times10^{15}\ \text{N}

Case (iii): $r=10^{6}\ \text{m}$

F=\frac{1.09\times10^{19}}{(10^{6})^2}=\frac{1.09\times10^{19}}{10^{12}}=1.09\times10^{7}\ \text{N}

Conclusion …

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Common Mistakes on Inverse Square Law Comparison Problems

Students often make predictable errors when comparing electrostatic forces across vastly different distances. Here are the most frequent ones — and how to avoid each.

1. Forgetting the Square in the Denominator

The Mistake

Students treat the force as inversely proportional to distance ( $F \propto 1/r$ ) instead of distance squared ( $F \propto 1/r^2$ ). This leads to underestimating how rapidly force drops.

Example of Error

If $r$ increases by $100\times$ , a student might think $F$ becomes $1/100$ of its original value — but the correct factor is $1/100^2 = 1/10,000$ .

How to Avoid

Write Coulomb’s law every time before substituting:

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

Circle the $r^2$ term. Remind yourself: double the distance → force drops to one-fourth.

2. Unit Conversion Errors

The Mistake

Plugging in distances in cm or km without converting to metres. Since $k = 9 \times 10^9\ \text{N m}^2/\text{C}^2$ , the SI unit for $r$ is metres.

Example of Error

Using $r = 1\ \text{cm}$ as $1$ instead of $0.01\ \text{m}$ gives a force $10^4$ times too large.

How to Avoid

Convert all distances to metres before calculation:
- $1\ \text{cm} = 1 \times 10^{-2}\ \text{m}$
- $100\ \text{m}$ stays as is
- $10^6\ \text{m}$ stays as is
Write the conversion step explicitly:

r = 1\ \text{cm} = 0.01\ \text{m}

3. Misinterpreting "Force on Each Point Charge"

The Mistake

Students calculate the total force between the two charges but forget that each charge experiences the same magnitude of force (Newton’s Third Law). Some then halve the result incorrectly.

How to Avoid

Remember: $F_{12} = F_{21}$ in magnitude.
The question asks for the force on each — the answer is the same number for both charges.
No need to divide by 2.

4. Not Recognising the Scale of the Numbers

The Mistake

After computing, students don’t check if the answer is physically plausible. For $q = 34.8\ \text{kC}$ (that’s $3.48 \times 10^4\ \text{C}$ ), forces are enormous — even at large distances.

Example of Error

Getting a force like $10^{-5}\ \text{N}$ for $r = 1\ \text{cm}$ and not realising it’s absurdly small for such huge charges.

Quick Sanity Check

For $r = 1\ \text{cm}$ :

F = \frac{(9 \times 10^9)(3.48 \times 10^4)^2}{(0.01)^2} \approx 1.09 \times 10^{20}\ \text{N}

That’s huge — comparable to the weight of a mountain. If your answer is tiny, you’ve made a unit or exponent error.

How to Avoid

Estimate orders of magnitude before calculating:
- $q^2 \approx 10^9$
- $k \approx 10^{10}$
- $r^2$ for 1 cm $\approx 10^{-4}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2025Set eng-2025-04264 marksMCQ
$r$
›Reveal solutionSolution

By Coulomb's law the force is proportional to the product of the charges. Since all pairs sum to $1\ \text{C}$ , the product (and hence force) is largest when the charges are equal, $0.5\ \text{C}$ and $0.5\ \text{C}$ .

Coulomb's law at fixed separation $r$ : $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2} \propto q_1 q_2$ .

Every pair has $q_1 + q_2 = 1\ \text{C}$ . For a fixed sum, the product $q_1 q_2$ is maximised when the two are equal: …
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KEAM 2025Set eng-2025-04284 marksMCQ
$r$
›Reveal solutionSolution

Gravity $\propto 1/d^2$ . Centres go from $2r$ to $4r$ , so force scales by $(2r/4r)^2=1/4$ , giving $F/4$ .

Two identical spheres of radius $r$ in contact have their centres separated by
$d_1=2r,$
and the gravitational force between them is $F=\dfrac{Gm^2}{(2r)^2}$ .

When the centre-to-centre distance is made $d_2=4r$ : …
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KEAM 2025Set pha-2025-0424A4 marksMCQ
$d$
›Reveal solutionSolution

New separation $= \tfrac{3d}{2}$ , and Coulomb force scales as $1/r^2$ , so $F' = \tfrac{4}{9}F$ .

Coulomb's law: $F = \dfrac{k q_1 q_2}{r^2}$ , so at separation $d$ , $F = \dfrac{k q_1 q_2}{d^2}$ .

When one charge is moved away by an additional $d/2$ , the new separation is
$r' = d + \frac{d}{2} = \frac{3d}{2}.$
The new force is …
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KEAM 2024Set eng-2024-06094 marksMCQ
$Q. The ratio of the magnitudes of electrostatic force between an electron and a proton separated by a distance r to that between a proton and an alpha particle separated by the same distance r is (A) 1:1 (B) 1:4 (C) 4:1 (D) 2:1 (E) 1:2$
›Reveal solutionSolution
Coulomb force $\propto q_1 q_2$ at fixed $r$ . Electron–proton uses $e\cdot e$ ; proton–alpha uses $e\cdot2e$ . Ratio $=1:2$ .

At the same separation $r$ , $F\propto q_1 q_2$ .

Electron–proton: $|q_1 q_2|=e\cdot e=e^2$ . …
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KEAM 2024Set pha-2024-06104 marksMCQ
$Q.When two spheres of radii $r$ and $\frac{r}{2}$ are brought in contact, the gravitational force of attraction between them is proportional to (A) $r^6$ (B) $r^4$ (C) $r^{-6}$ (D) $r^{-4}$ (E) $r^{-2}$$
›Reveal solutionSolution
[!TLDR]

Masses go as $r^3$ and the separation of centres goes as $r$ , so $F \propto (r^3\cdot r^3)/r^2 = r^4$ .

Concept

The gravitational attraction between two masses is $F = G m_1 m_2 / d^2$ , where $d$ is the distance between their centres. For solid spheres of the same material, mass is proportional to volume, i.e. to the cube of the radius.

Solution

Let the spheres have radii $r$ and $\tfrac{r}{2}$ and equal density $\rho$ .

Masses:
$m_1 = \rho\cdot\tfrac{4}{3}\pi r^3 \propto r^3,\qquad m_2 = \rho\cdot\tfrac{4}{3}\pi\left(\tfrac{r}{2}\right)^3 \propto \tfrac{r^3}{8}.$
When the spheres are in contact, the distance between centres is the sum of the radii:
$d = r + \tfrac{r}{2} = \tfrac{3r}{2} \propto r.$
Hence …
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KEAM 2022Set eng-2022-P1-A14 marksMCQ
$F_1$
›Reveal solutionSolution
Coulomb force depends only on the charge magnitudes, and all four particles carry charge of magnitude $e$ , so $F_1 : F_2 = 1 : 1$ .

Concept and Intuition

The electrostatic (Coulomb) force between two charges is $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$ . It depends on the magnitudes of the charges and the separation, not on the masses of the particles. A proton, electron and positron all carry a charge of the same magnitude $e$ (only the sign differs).

Step-by-Step Solution

Proton–electron: $|q_1 q_2| = e \cdot e = e^2$ , so $F_1 = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}$ .

Electron–positron: $|q_1 q_2| = e \cdot e = e^2$ , so $F_2 = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}$ . …
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KEAM 2021Set eng-2021-P1-A14 marksMCQ
$r$
›Reveal solutionSolution
[!TLDR]

Mass $\propto r^3$ and centre-to-centre distance $=2r$ , so Newtonian gravitation gives $F \propto r^4$ .

Concept

Newton's law of gravitation $F = \dfrac{G m_1 m_2}{d^2}$ (NCERT/CBSE class 11) applies between the centres of uniform spheres. For identical spheres in contact the separation of centres equals $2r$ .

Solution

Each sphere has mass
$m = \rho \cdot \tfrac{4}{3}\pi r^3 \ \propto\ r^3.$
Centre-to-centre distance for spheres in contact:
$d = 2r.$
Hence …
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KEAM 2021Set eng-2021-P1-A14 marksMCQ
$r$
›Reveal solutionSolution
The two forces are equal: ratio $1:1$ .

Concept and Intuition

Coulomb's law uses charge magnitudes, not masses. Protons and electrons both have magnitude $e$ , so proton–proton and electron–electron repulsions are equal at equal distance.

Step-by-Step Solution

$F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ .

For protons and for electrons, $|q_1|=|q_2|=e$ and $r$ is the same. …
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Q.Consider a coin of Question 1.20. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8\text{ kC}$ . Suppose that these equal charges were concentrated in two point charges separated by

Concept understanding — Inverse Square Law Comparison

The Intuition: Why Does Light Get Dimmer So Fast?

The Core Idea in One Picture

The Precise Statement

Why "Inverse Square" and Not Just "Inverse"?

The Comparison: What It Really Means

Why this formula?

Inverse Square Law Comparison — Why the Formula Holds

1. The Intuition: Spreading Over a Sphere

2. Derivation for Gravitational Force (Newton's Law)

3. Derivation for Coulomb's Law (Electrostatics)

4. Derivation for Light/Radiation Intensity

5. The Common Mathematical Reason

Method: Coulomb’s Law (Inverse Square Law Comparison)

Steps

Common Mistakes on Inverse Square Law Comparison Problems

1. Forgetting the Square in the Denominator

2. Unit Conversion Errors

3. Misinterpreting "Force on Each Point Charge"

4. Not Recognising the Scale of the Numbers

Concept

Solution

Concept and Intuition

Step-by-Step Solution

Concept

Solution

Concept and Intuition

Step-by-Step Solution

Q.Consider a coin of Question 1.20. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude 34.8 kC34.8\text{ kC}34.8 kC. Suppose that these equal charges were concentrated in two point charges separated by

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Q.Consider a coin of Question 1.20. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8\text{ kC}$ . Suppose that these equal charges were concentrated in two point charges separated by

Unlock everything free for 14 days