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.

KCET 2024Set D-21 markMCQ
$+10\,\mu\text{C}$
›Reveal solutionSolution
Newton’s Third Law applies to electrostatic forces: the force on A due to B is equal in magnitude and opposite in direction to the force on B due to A. Therefore $\vec{F}_1 = -\vec{F}_2$ , making option (B) correct.

The key idea here is that electrostatic forces obey Newton’s Third Law of motion. No matter how large or small the charges are, the force one charge exerts on another is always matched by an equal and opposite force from the second charge back on the first. This is a fundamental symmetry built into Coulomb’s law.

Let’s walk through it step by step.

Recall Coulomb’s law. The magnitude of the electrostatic force between two point charges $q_1$ and $q_2$ separated by a distance $r$ is

$F = \frac{1}{4\pi\varepsilon_0} \frac{|q_1 q_2|}{r^2}.$
This force acts along the line joining the charges. The direction is attractive for opposite signs and repulsive for like signs.

Apply it to the given charges. Here $q_A = +10\,\mu\text{C}$ and $q_B = +20\,\mu\text{C}$ , both positive, so the force is repulsive. The magnitude of the force on A due to B is

$|\vec{F}_1| = \frac{1}{4\pi\varepsilon_0} \frac{(10\times10^{-6})(20\times10^{-6})}{(1)^2}.$
The magnitude of the force on B due to A is exactly the same:
$|\vec{F}_2| = \frac{1}{4\pi\varepsilon_0} \frac{(20\times10^{-6})(10\times10^{-6})}{(1)^2} = |\vec{F}_1|.$
So the magnitudes are equal.

Now consider direction. Since both charges are positive, they repel each other. The force $\vec{F}_1$ on A due to B points away from B (from A to the direction opposite B). The force $\vec{F}_2$ on B due to A points away from A (from B to the direction opposite A). These two vectors point in exactly opposite directions along the line joining the charges. …
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KCET 2023Set A-31 markMCQ
$1.6 \times 10^{-19}$
›Reveal solutionSolution
Electrostatic induction temporarily separates charges in the sphere, but when the rod is removed without grounding, the charges recombine — the sphere returns to zero net charge.

The key here is understanding electrostatic induction — the redistribution of electric charge in a conductor when a charged object is brought near, without direct contact.

When the positively charged glass rod approaches the uncharged metal sphere, it repels positive charges in the sphere and attracts negative charges. Since the sphere is a conductor, electrons (negative charges) move freely toward the side closest to the rod, leaving the far side with a net positive charge. This separation happens instantly, but the sphere as a whole still has zero net charge — it’s just that the charges have rearranged themselves.

Now, the critical question: does the sphere gain or lose any charge during this process? No — because there is no path for charge to flow to or from the sphere. It sits on an insulated stand, so it’s isolated. The rod never touches it. The only thing that happens is a temporary polarization.

When you remove the glass rod, the electric field that was holding the charges apart disappears. The electrons that had gathered on the near side are no longer attracted to the rod, so they spread back evenly throughout the sphere. The positive charges on the far side also redistribute. The sphere returns to its original, uniform, uncharged state.

Before the rod approaches: sphere has zero net charge, charges uniformly distributed.

Rod brought near: induction occurs — negative charges shift toward the rod, positive charges shift away. The sphere remains neutral overall. …
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KCET 2022Set B-31 markMCQ

\gamma

›Reveal solution

Deflection in an electric field requires an electric charge ( $\vec{F} = q\vec{E}$ ); of the four, only the $\alpha$ -particle is charged.

Step 1 — The governing law.

The force exerted by an electric field $\vec{E}$ on a particle of charge $q$ is

\vec{F} = q\,\vec{E}

So the deflection is directly proportional to the charge. If $q = 0$ , then $\vec{F} = 0$ and the particle travels straight through, no matter how strong the field.

Step 2 — Check the charge of each radiation.

Option	Identity	Charge	Deflected by $\vec E$ ?
(A) $\gamma$ -rays	high-energy photons (EM radiation)	$0$	No
(B) $\alpha$ -particles	helium nucleus $^4_2\text{He}^{2+}$	$+2e$	Yes
(C) X-rays	photons (EM radiation)	$0$	No
(D) Neutrons	neutral nucleon	$0$	No

Step 3 — Conclude. …

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COMEDK 2021Set 2021-B1 markMCQ
$Q. Which of the following statements is FALSE in case of electrostatic force between 2 particles? (A) It is always attractive (B) It is a central force (C) It is a conservative force (D) It obeys inverse square law$
›Reveal solutionSolution
The Coulomb force between two charges is repulsive for like charges and attractive for unlike charges — so the claim that it is "always attractive" is the false statement.

The electrostatic force $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}$ :

is central (acts along the line joining the charges) — true;

is conservative (work is path-independent) — true; …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. The force between 2 point charge separated by a distance l is 18 N. If the separation is 3 times, then the force between them becomes (A) 1 N (B) 2 N (C) 6 N (D) 54 N$
›Reveal solutionSolution

$F\propto1/r^2$ ; $r\to3r \Rightarrow F\to F/9 = 18/9 = 2$ N. …

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KCET 2019Set A-11 markMCQ

F_n

›Reveal solution

At 10 nm the protons are millions of nuclear-force ranges apart, so the nuclear force has vanished while the long-range Coulomb force survives.

Step 1 — Compare the two forces' ranges.

Force	Range	Behaviour with distance
Strong nuclear force $F_n$	$\sim$ a few fm ( $10^{-15}$ m); essentially zero beyond $\sim$ 10 fm	falls off exponentially — not a power law
Electromagnetic (Coulomb) $F_e$	infinite	$F_e=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$ — inverse square, never truly zero

Step 2 — Put the given separation on that scale.

r=10\ \text{nm}=10\times10^{-9}\ \text{m}=10^{-8}\ \text{m}=10^{7}\ \text{fm}.

So the protons are ten million femtometres apart — about a million times further than the nuclear force can reach. $F_n\approx0$ .

Step 3 — Is the Coulomb force still there?

Yes, and it is easily computed: …

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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

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