A paisa coin is made up of Al-Mg alloy and weighs $0.75\text{ g}$. It has a square shape and its diagonal measures $17\text{ mm}$. It is electrically neutral and contains equal amounts of positive and negative charges. Treating the paisa coin as made up of only Al, find the magnitude of equal number of positive and negative charges. What conclusion do you draw from this magnitude?

A paisa coin is made up of Al-Mg alloy and weighs 0.75 g. It has a square shape and its diagonal measures 17 m

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0.75\text{ g}

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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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We calculate the total number of atoms in the coin, then multiply by the atomic number of Aluminum to find the total number of protons (or electrons), and finally multiply by the elementary charge to get the total positive/negative charge. The magnitude of this charge is $\boxed{3.49 \times 10^4 \text{ C}}$ .

The problem asks us to determine the total magnitude of positive and negative charges within a paisa coin, assuming it's made entirely of Aluminum (Al). The key to solving this lies in understanding that matter is composed of atoms, and each atom contains a specific number of protons (positive charge) and electrons (negative charge). Since the coin is electrically neutral, the total positive charge must exactly balance the total negative charge. Our task is to find this magnitude.

We can approach this by first determining how many Aluminum atoms are present in the given mass of the coin. Once we know the number of atoms, we can use Aluminum's atomic number to find the total number of protons (or electrons), and then multiply by the elementary charge to get the total charge.

Identify the given parameters and necessary constants.
- Mass of the coin, $m = 0.75\text{ g}$ .
- The coin is treated as made up of only Aluminum (Al).
- From the periodic table, the atomic number of Aluminum is $Z = 13$ . This means each Al atom has 13 protons and 13 electrons.
- The molar mass of Aluminum is $M = 26.98\text{ g/mol}$ .
- Avogadro's number, $N_A = 6.022 \times 10^{23}\text{ mol}^{-1}$ . This tells us the number of atoms in one mole of any substance.
- The elementary charge, $e = 1.602 \times 10^{-19}\text{ C}$ . This is the magnitude of charge on a single proton or electron.
- The diagonal measurement ( $17\text{ mm}$ ) is extraneous information for this problem, as we are given the mass directly.
Calculate the number of moles of Aluminum in the coin.

The number of moles ( $n$ ) is found by dividing the given mass by the molar mass of Aluminum.

n = \frac{m}{M}

n = \frac{0.75\text{ g}}{26.98\text{ g/mol}}

n \approx 0.027798\text{ mol}

Calculate the total number of Aluminum atoms in the coin. Multiply the number of moles by Avogadro's number to find the total number of atoms ( $N_{atoms}$ ).

N_{atoms} = n \times N_A

N_{atoms} = 0.027798\text{ mol} \times (6.022 \times 10^{23}\text{ atoms/mol})

N_{atoms} \approx 1.6749 \times 10^{22}\text{ atoms}

Calculate the total number of protons (or electrons) in the coin. Since each Aluminum atom has $Z=13$ protons and 13 electrons, the total number of positive charges (protons) or negative charges (electrons) is the total number of atoms multiplied by the atomic number.

N_{charge\_carriers} = N_{atoms} \times Z

N_{charge\_carriers} = (1.6749 \times 10^{22}\text{ atoms}) \times 13

$$N_{charge\_carriers} \approx 2.1774 \times 10^{23}$$ …

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Here’s a breakdown of the common mistakes students make on this problem and how to avoid each.

Mistake 1: Confusing mass of the coin with mass of Aluminium

Students often take the given mass ( $0.75\text{ g}$ ) as the mass of pure Aluminium directly, forgetting the coin is an alloy (Al-Mg). The problem explicitly says: “Treating the paisa coin as made up of only Al” — this is a simplifying assumption for calculation.

How to avoid:

Read the problem statement carefully.
When it says “treat as made up of only Al,” you must use the atomic mass of Al ( $27\text{ u}$ ) and the given mass ( $0.75\text{ g}$ ) to find the number of Al atoms.
Do not look up or use the density of Al — the coin’s shape and diagonal are distractors (they are irrelevant for charge calculation).

Mistake 2: Using the coin’s dimensions to find mass or number of atoms

The diagonal ( $17\text{ mm}$ ) and square shape are given, but they are not needed to find the number of atoms or charges. Students sometimes try to compute volume, then mass via density — this is unnecessary and leads to errors.

How to avoid:

Identify the only data you need: mass of coin ( $0.75\text{ g}$ ) and atomic mass of Al ( $27\text{ g/mol}$ ).
The diagonal is a red herring — ignore it for the charge calculation.
Use the formula:

\text{Number of atoms} = \frac{\text{mass of coin}}{\text{atomic mass of Al}} \times N_A

Mistake 3: Forgetting Avogadro’s number or using it incorrectly

Students sometimes forget to multiply by Avogadro’s number ( $N_A = 6.022 \times 10^{23}$ ) or use it in the wrong place.

How to avoid:

Remember:

\text{Number of atoms} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \times N_A

Write the units clearly:

\frac{0.75}{27} \times 6.022 \times 10^{23}

Double-check that the result is a huge number (~ $1.67 \times 10^{22}$ atoms) — if you get a small number, you likely missed $N_A$ .

Mistake 4: Confusing number of atoms with number of charges

Each Al atom has 13 protons and 13 electrons (atomic number $Z=13$ ). Students often stop after finding the number of atoms, forgetting to multiply by 13.

How to avoid:

Clearly separate the steps:
1. Find number of Al atoms ( $n$ ).
2. Multiply by 13 to get number of protons (or electrons).
Write:

\text{Number of positive charges} = n \times 13

\text{Number of negative charges} = n \times 13

Mistake 5: Using charge of an electron incorrectly or mixing units

Students sometimes use $1.6 \times 10^{-19}$ C correctly but forget to multiply by the total number of charges, or they use the charge of a proton instead.

How to avoid:

The magnitude of charge on one proton = $1.6 \times 10^{-19}$ C (same as electron).
Total charge magnitude:

Q = (\text{number of charges}) \times (1.6 \times 10^{-19})

Keep units consistent — all in SI (Coulombs).

Mistake 6: Not drawing the conclusion from the magnitude …

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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

Concept: Quantization of Charge & Avogadro's Number

Method: Avogadro-Based Charge Calculation

Steps

Mistake 1: Confusing mass of the coin with mass of Aluminium

Mistake 2: Using the coin’s dimensions to find mass or number of atoms

Mistake 3: Forgetting Avogadro’s number or using it incorrectly

Mistake 4: Confusing number of atoms with number of charges

Mistake 5: Using charge of an electron incorrectly or mixing units

Mistake 6: Not drawing the conclusion from the magnitude …

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