What are the differences between coulomb force and gravitational force ?

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Q. What are the differences between coulomb force and gravitational force ?

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2025Subjective· 3mImportance★★★★★

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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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The Coulomb force (between charges) can be attractive or repulsive, depends on the medium, and is enormously stronger than the always-attractive, medium-independent gravitational force (between masses).

Coulomb (electrostatic) force	Gravitational force
Acts between electric charges	Acts between any two masses
Can be attractive or repulsive, depending on the sign of the charges	Always attractive only
Depends on the intervening medium (via permittivity/dielectric constant)	Independent of the intervening medium
Governed by Coulomb's constant $k=9\times10^9\ \text{Nm}^2\text{C}^{-2}$	Governed by gravitational constant $G=6.67\times10^{-11}\ \text{Nm}^2\text{kg}^{-2}$

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

CBSE 2026Set ANNUAL1 markMCQ
$Q. The electric field E at a distance r due to a point charge is:$
(a) E ∝ r
(b) E ∝ 1/r²
(c) E ∝ 1/r
(d) E ∝ 1/r³
›Reveal solutionSolution

By Coulomb's law the field of a point charge is $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ , so $E\propto 1/r^2$ .

The force on a small test charge $q_0$ placed at distance $r$ from a point charge $q$ is $F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{qq_0}{r^2}$ (Coulomb's law). The electric field is defined as force per unit test charge, $E=F/q_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ . Since $q$ and the constant $\dfrac{1}{4\pi\varepsilon_0}$ do not depend on $r$ , the field varies only through the $1/r^2$ te …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The force between two point charges placed at distance r apart, is F. If the distance increased to 2r between the charges then force will be$
(a) F
(b) F/2
(c) F/4
(d) F/8
›Reveal solutionSolution

Coulomb's force follows an inverse-square law, so doubling the separation cuts the force to one-quarter.

Coulomb's law: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2} = \dfrac{k q_1 q_2}{r^2}$

At separation $r$ : $F = \dfrac{k q_1 q_2}{r^2}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. When the distance between the charges is halved, the force between the charges become :$
(a) half
(b) twice
(c) four times
(d) none of these
›Reveal solutionSolution

By Coulomb's law $F \propto 1/r^2$ ; halving the separation $r$ quadruples the force.

Coulomb's law states

$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$

If the distance is halved, $r' = r/2$ , then …

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CBSE 2025Set ANNUAL1 mark
$R_1$
›Reveal solutionSolution

Surface field $E=kQ/R^2$ ; with equal charge $Q$ on each sphere, $E \propto 1/R^2$ .

The electric field at the surface of a charged conducting sphere of radius $R$ carrying charge $Q$ is $E = \dfrac{kQ}{R^2}$ . Given that after separation both spheres carry the same charge $Q$ (as stated), the field at each surface is inversely proportiona …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If both the charges and distance between them is doubled, then the new electrostatic force will be$
(a) F
(b) 2F
(c) 3F
(d) 4F
›Reveal solutionSolution

Doubling both charges and the distance leaves the electrostatic force unchanged because the charge-product factor of 4 exactly cancels the distance-squared factor of 4.

Coulomb's law: $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2} = k\frac{q_1q_2}{r^2}$

Let the new charges be $q_1' = 2q_1$ , $q_2' = 2q_2$ , and the new separation $r' = 2r$ .

…

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CBSE 2023Set MODEL1 markMCQ
$q$
(a) $E \propto \dfrac{1}{r}$
(b) $E \propto \dfrac{1}{r^2}$
(c) $E \propto r$
(d) $E \propto \dfrac{1}{r^3}$
›Reveal solutionSolution

The electric field of a point charge falls off as the inverse square of distance.

By Coulomb's law, the electric field due to a point charge $q$ at a distance $r$ is

$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ …

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CBSE 2023Set F1 markMCQ
$Q. The ratio of electric force and gravitational force acting between two charges is in the order of (A) 10^42 (B) 10^39 (C) 10^36 (D) 1$
›Reveal solutionSolution

For two electrons, $F_e/F_g \approx 4\times10^{42}$ — order 10⁴².

Both forces between two identical charges (electrons) obey an inverse-square law, so the distance cancels in the ratio:

$\frac{F_e}{F_g} = \frac{\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2}}{G\dfrac{m_e^2}{r^2}} = \frac{e^2}{4\pi\varepsilon_0 G m_e^2}.$

Putting $e = 1.6\times10^{-19}$ C and $m_e = 9.1\times10^{-31}$ kg gives $\approx 4\times10^{42}$ — i.e. of order $10^{42}$ .

…

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CBSE 2023Set ANNUAL1 mark
$Q. Between an electron and a proton, which is stronger, the electrostatic force or the gravitational force?$
›Reveal solutionSolution

Coulomb's law force between an electron and proton vastly exceeds Newton's gravitational force between them, because e²/(4πε₀) is enormously larger than Gm_em_p.

Electrostatic force: $F_e=\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2}$ . Gravitational force: $F_g=\dfrac{Gm_em_p}{r^2}$ .

Taking the ratio (the $r^2$ cancels): $\dfrac{F_e}{F_g}=\dfrac{e^2/4\pi\varepsilon_0}{Gm_em_p}\approx 2.3\times10^{39}$

…

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CBSE 2022Set ANNUAL1 markMCQ
$+1\mu C$
(a) 1 : 8
(b) 8 : 1
(c) 1 : 1
(d) 1 : 16
›Reveal solutionSolution

The forces are an action–reaction pair, so they are equal in magnitude: ratio 1 : 1, option (C).

The Coulomb force each charge exerts on the other is

$F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}.$

Both charges share the same product $q_1 q_2$ and the same separation $r$ , so the magnitude is identical for each. This is also required by Newton's third law: the force on charge 1 due to charge 2 is equal and …

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CBSE 2020Set ANNUAL1 mark
$Q. The force of attraction between a positively charged particle and a negatively charged particle is F. When distance between them is made one fourth, what will be the value of this force?$
›Reveal solutionSolution

By Coulomb's law $F\propto 1/r^2$ ; reducing distance to one-fourth increases force 16 times.

Coulomb's law: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ , so $F \propto \dfrac{1}{r^2}$ .

If new distance $r' = r/4$ : …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. When the distance between two charged particles is doubled, the force between them becomes: (A) one-fourth (B) half (C) double (D) four times$
›Reveal solutionSolution

By Coulomb's law, force varies as $1/r^2$ , so doubling $r$ divides the force by 4.

Coulomb's law gives the force between two point charges $q_1, q_2$ separated by distance $r$ as

$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$

If the separation is doubled, $r \to 2r$ , then …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. The intensity of electric field E due to charge Q at distance r :$
(a) E ∝ r
(b) E ∝ 1/r
(c) E ∝ 1/r²
(d) E ∝ 1/r³
›Reveal solutionSolution

By Coulomb's law the field of a point charge is E = kQ/r², so E ∝ 1/r².

The electric field intensity at a point is the force per unit positive test charge. For a point charge Q, the magnitude of the field at distance r is

E = (1/4πε₀) × (Q / r²) = kQ/r²,

…

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Q.What are the differences between coulomb force and gravitational force ?

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