Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force betwe

Worked Examples · Example 1.3

Q. Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively.

(a) Compare the strength of these forces by determining the ratio of their magnitudes

(i) for an electron and a proton and

(ii) for two protons.

(b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are

1\,\text{Å}\ (=10^{-10}\,\text{m})

apart? (

m_p = 1.67 \times 10^{-27}\,\text{kg}

m_e = 9.11 \times 10^{-31}\,\text{kg}

)

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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.

Where This Law Applies (and Where It Doesn't)

Applies to:

Light from a point source (a star, a bulb)
Sound from a small source (a speaker in open air)
Gravitational force between two masses
Electric force between two charges
Radiation from a radioactive point source

Does NOT apply to:

Light from a laser beam (it stays collimated)
Sound inside a pipe (it's guided)
Gravity inside a planet (the mass distribution changes the law)
Very large distances in cosmology (space itself is expanding)

The One-Line Takeaway

Important

Inverse square law comparison: When distance multiplies by $n$ , intensity divides by $n^2$ . Always compare using the square of the distance ratio, not the distance itself.

That's the whole idea. The rest is just practice applying it to different problems — but the geometry never changes.

Searches such as "inverse square law examples physics" and "intensity distance relationship formula" recur across gravitation, electrostatics, and optics topics in the NCERT/CBSE Class 11-12 Physics curriculum, since this same $1/r^2$ pattern underlies all of them. Comparing intensities at two distances is a very common numerical-question format in JEE Main and NEET.

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:

\text{Intensity} \propto \frac{1}{r^2}

because the area over which the quantity spreads is proportional to $r^2$ .

6. Key Exam Points

Quantity	Formula	Reason
Gravitational force	$F = \frac{G M m}{r^2}$	Flux of field lines over sphere area
Electric force	$F = \frac{k q_1 q_2}{r^2}$	Gauss’s law + spherical symmetry
Light intensity	$I = \frac{P}{4\pi r^2}$	Power spread over sphere surface
Sound intensity	$I = \frac{P}{4\pi r^2}$	Same geometric spreading

Important: The inverse square law holds only for point sources in 3D space with no absorption or reflection. For extended sources or non-isotropic emission, the law is modified.

7. Common Misconception

Not because "the force gets weaker with distance" — that’s vague.
The exact $1/r^2$ comes from the geometry of a sphere, not from any arbitrary assumption.
If space had 2 dimensions, the law would be $1/r$ ; if 4 dimensions, $1/r^3$ .

Final takeaway: The inverse square law is a geometric necessity for any conserved quantity spreading uniformly from a point in three-dimensional space. Memorize the formula, but understand the sphere-area argument — it’s the core reasoning for every exam question.

The key idea here involves comparing the magnitudes of electrostatic and gravitational forces, both of which follow an inverse-square law, and then applying Newton's second law to determine accelerations.

To compare the strength of these forces, we determine the ratio of their magnitudes. The electrostatic force is $F_e = k \frac{|q_1 q_2|}{r^2}$ $F_{e} = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}}$ , and the gravitational force is $F_g = G \frac{m_1 m_2}{r^2}$ $F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}$ . The ratio is $\frac{F_e}{F_g} = \frac{k |q_1 q_2|}{G m_1 m_2}$ $\frac{F _{e}}{F _{g}} = \frac{k ∣ q _{1} q _{2} ∣}{G m _{1} m _{2}}$ . Using constants: $k = 9 \times 10^9\,\text{N m}^2/\text{C}^2$ $k = 9 \times 1 0^{9} N m^{2} / C^{2}$ , $G = 6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2$ $G = 6.67 \times 1 0^{- 11} N m^{2} / kg^{2}$ , $e = 1.602 \times 10^{-19}\,\text{C}$ $e = 1.602 \times 1 0^{- 19} C$ .
1. For an electron and a proton ( $q_1=e, q_2=e, m_1=m_e, m_2=m_p$ ): $\frac{F_e}{F_g} = \frac{(9 \times 10^9)(1.602 \times 10^{-19})^2}{(6.67 \times 10^{-11})(9.11 \times 10^{-31})(1.67 \times 10^{-27})} = \frac{2.30976 \times 10^{-28}}{1.0140 \times 10^{-67}} \approx 2.28 \times 10^{39}$
2. For two protons ( $q_1=e, q_2=e, m_1=m_p, m_2=m_p$ ): $\frac{F_e}{F_g} = \frac{(9 \times 10^9)(1.602 \times 10^{-19})^2}{(6.67 \times 10^{-11})(1.67 \times 10^{-27})^2} = \frac{2.30976 \times 10^{-28}}{1.8592 \times 10^{-64}} \approx 1.24 \times 10^{36}$
  ✓Final answer
  $\boxed{2.28 \times 10^{39}}$
  
  ✓Final answer
  $\boxed{1.24 \times 10^{36}}$
To estimate the accelerations, we first calculate the electrostatic force between an electron and a proton $1\,\text{Å}$ apart ( $r = 10^{-10}\,\text{m}$ ). $F_e = k \frac{e^2}{r^2} = (9 \times 10^9) \frac{(1.602 \times 10^{-19})^2}{(10^{-10})^2} = 2.30976 \times 10^{-8}\,\text{N}$ . Using Newton's second law, $a = F/m$ : For the electron ( $m_e = 9.11 \times 10^{-31}\,\text{kg}$ ): $a_e = \frac{F_e}{m_e} = \frac{2.30976 \times 10^{-8}\,\text{N}}{9.11 \times 10^{-31}\,\text{kg}} \approx 2.54 \times 10^{22}\,\text{m/s}^2$ . For the proton ( $m_p = 1.67 \times 10^{-27}\,\text{kg}$ ): $a_p = \frac{F_e}{m_p} = \frac{2.30976 \times 10^{-8}\,\text{N}}{1.67 \times 10^{-27}\,\text{kg}} \approx 1.38 \times 10^{19}\,\text{m/s}^2$ .
✓Final answer
$\boxed{2.54 \times 10^{22}\,\text{m/s}^2}$

The electrostatic force is immensely stronger than the gravitational force, by factors of about $2.3 \times 10^{39}$ for an electron-proton pair and $1.2 \times 10^{36}$ for two protons. At $1\,\text{Å}$ separation, the electron experiences an acceleration of approximately $2.54 \times 10^{22}\,\text{m/s}^2$ , while the proton experiences an acceleration of about $1.38 \times 10^{19}\,\text{m/s}^2$ due to their mutual electrical attraction.

Both Coulomb's law for electrostatic force and Newton's law for gravitational force describe interactions that follow an inverse-square dependence on the distance between the interacting particles. This means the force strength decreases rapidly as the separation increases. However, the fundamental nature and magnitudes of these forces are vastly different. Electrostatic force arises from charge and can be attractive or repulsive, while gravitational force arises from mass and is always attractive.

To truly grasp the relative importance of these forces, especially at the microscopic scale where atoms and subatomic particles interact, we compare their magnitudes directly. By forming a ratio of the electrostatic force to the gravitational force, we can quantify which force dominates. A key insight here is that since both forces depend on $1/r^2$ , the distance $r$ will cancel out in the ratio. This means the relative strength of these fundamental forces is a constant, independent of how far apart the particles are.

(a) Comparing the strength of electrostatic and gravitational forces

Formulate the force equations: The magnitude of the electrostatic force ( $F_e$ ) between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb's law:

F_e = k \frac{|q_1 q_2|}{r^2}

where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

The magnitude of the gravitational force ($F_g$) between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by Newton's law of gravitation:

F_g = G \frac{m_1 m_2}{r^2}

where $G$ is the universal gravitational constant.

2. Determine the ratio of forces:

To compare their strengths, we take the ratio $\frac{F_e}{F_g}$ :

\frac{F_e}{F_g} = \frac{k \frac{|q_1 q_2|}{r^2}}{G \frac{m_1 m_2}{r^2}}

As discussed, the $r^2$ terms cancel out, simplifying the ratio to:

\frac{F_e}{F_g} = \frac{k |q_1 q_2|}{G m_1 m_2}

List fundamental constants:

We use the following standard values for our calculations:
- Elementary charge, $e = 1.602 \times 10^{-19}\,\text{C}$
- Coulomb's constant, $k = 9 \times 10^9\,\text{N m}^2/\text{C}^2$
- Universal gravitational constant, $G = 6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2$
- Mass of electron, $m_e = 9.11 \times 10^{-31}\,\text{kg}$
- Mass of proton, $m_p = 1.67 \times 10^{-27}\,\text{kg}$
Watch out

Precision in calculations involving powers of ten is crucial. Ensure correct substitution and arithmetic to avoid significant errors in the final magnitude.

(i) For an electron and a proton:

For an electron and a proton, the magnitude of their charges is $e$ for both, so $|q_1 q_2| = e^2$ . Their masses are $m_e$ and $m_p$ .

Substituting these into the ratio formula:

\frac{F_e}{F_g} = \frac{k e^2}{G m_e m_p}

\frac{F_e}{F_g} = \frac{(9 \times 10^9\,\text{N m}^2/\text{C}^2) (1.602 \times 10^{-19}\,\text{C})^2}{(6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2) (9.11 \times 10^{-31}\,\text{kg}) (1.67 \times 10^{-27}\,\text{kg})}

Calculating the numerator:

(9 \times 10^9) \times (1.602)^2 \times 10^{-38} = 9 \times 2.566404 \times 10^{-29} \approx 2.310 \times 10^{-28}

Calculating the denominator:

(6.67 \times 10^{-11}) \times (9.11 \times 10^{-31}) \times (1.67 \times 10^{-27}) = (6.67 \times 9.11 \times 1.67) \times 10^{-11-31-27}

= 101.399 \times 10^{-69} \approx 1.014 \times 10^{-67}

Now, divide the numerator by the denominator:

\frac{F_e}{F_g} = \frac{2.310 \times 10^{-28}}{1.014 \times 10^{-67}} \approx 2.278 \times 10^{39}

Rounding to two significant figures, consistent with the precision of $k$ and $G$:

\frac{F_e}{F_g} \approx \mathbf{2.3 \times 10^{39}}

This result highlights that the electrostatic force is astronomically stronger than the gravitational force at the atomic scale.

**(ii) For two protons:**
For two protons, the magnitude of their charges is $e$ for both, so $|q_1 q_2| = e^2$. Their masses are both $m_p$.
Substituting these into the ratio formula:

\frac{F_e}{F_g} = \frac{k e^2}{G m_p^2}

\frac{F_e}{F_g} = \frac{(9 \times 10^9\,\text{N m}^2/\text{C}^2) (1.602 \times 10^{-19}\,\text{C})^2}{(6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2) (1.67 \times 10^{-27}\,\text{kg})^2}

The numerator is the same as before: $2.310 \times 10^{-28}$.
Calculating the denominator:

(6.67 \times 10^{-11}) \times (1.67 \times 10^{-27})^2 = (6.67 \times (1.67)^2) \times 10^{-11-54}

= (6.67 \times 2.7889) \times 10^{-65} = 18.598 \times 10^{-65} \approx 1.860 \times 10^{-64}

Now, divide the numerator by the denominator:

\frac{F_e}{F_g} = \frac{2.310 \times 10^{-28}}{1.860 \times 10^{-64}} \approx 1.242 \times 10^{36}

Rounding to two significant figures:

\frac{F_e}{F_g} \approx \mathbf{1.2 \times 10^{36}}

Even for two protons, the electrostatic repulsion is vastly stronger than their gravitational attraction. The ratio is smaller than for the electron-proton pair because protons are much more massive than electrons, leading to a larger gravitational force in the denominator.

(b) Estimating accelerations

Understand acceleration from force: According to Newton's second law, the acceleration ( $a$ ) of an object is directly proportional to the net force ( $F$ ) acting on it and inversely proportional to its mass ( $m$ ):

a = \frac{F}{m}

We need to calculate the acceleration of both the electron and the proton due to their mutual electrical attraction.

2. Calculate the electrostatic force:

The electron and proton are separated by a distance $r = 1\,\text{Å} = 10^{-10}\,\text{m}$ .

The magnitude of the electrostatic force between them is:

F_e = k \frac{e^2}{r^2}

F_e = (9 \times 10^9\,\text{N m}^2/\text{C}^2) \frac{(1.602 \times 10^{-19}\,\text{C})^2}{(10^{-10}\,\text{m})^2}

F_e = (9 \times 10^9) \frac{2.566404 \times 10^{-38}}{10^{-20}}

F_e = 9 \times 2.566404 \times 10^{9 - 38 + 20}

F_e = 23.097636 \times 10^{-9}\,\text{N}

F_e \approx \mathbf{2.31 \times 10^{-8}\,\text{N}}

This is the magnitude of the attractive force acting on both the electron and the proton.

3. Calculate the acceleration of the electron:

The force on the electron is $F_e$ , and its mass is $m_e = 9.11 \times 10^{-31}\,\text{kg}$ .

a_e = \frac{F_e}{m_e}

a_e = \frac{2.30976 \times 10^{-8}\,\text{N}}{9.11 \times 10^{-31}\,\text{kg}}

a_e \approx 0.2535 \times 10^{23}\,\text{m/s}^2

a_e \approx \mathbf{2.54 \times 10^{22}\,\text{m/s}^2}

Calculate the acceleration of the proton: The force on the proton is also $F_e$ (by Newton's third law), and its mass is $m_p = 1.67 \times 10^{-27}\,\text{kg}$ .

a_p = \frac{F_e}{m_p}

a_p = \frac{2.30976 \times 10^{-8}\,\text{N}}{1.67 \times 10^{-27}\,\text{kg}}

a_p \approx 1.383 \times 10^{19}\,\text{m/s}^2

a_p \approx \mathbf{1.38 \times 10^{19}\,\text{m/s}^2}

> [!TIP]
> The electron, being significantly less massive than the proton ($m_p \approx 1836 m_e$), experiences a proportionally larger acceleration for the same magnitude of force. This is a direct consequence of Newton's second law, $a = F/m$.

✓Final answer

\boxed{2.3 \times 10^{39}}

Method: Ratio of Forces & Newton’s Second Law

This problem uses the Inverse Square Law Comparison method — comparing two forces that both follow a $\frac{1}{r^2}$ dependence, then applying Newton’s Second Law ( $F = ma$ ) to find accelerations.

Part (a) — Ratio of Electrostatic to Gravitational Force

Step 1: Write the two force laws

Coulomb’s law (electrostatic):

F_e = \frac{1}{4\pi\varepsilon_0} \cdot \frac{|q_1 q_2|}{r^2}

Newton’s law (gravitational):

F_g = G \cdot \frac{m_1 m_2}{r^2}

Step 2: Form the ratio

Since both have $r^2$ in the denominator, the distance cancels out:

\frac{F_e}{F_g} = \frac{\frac{1}{4\pi\varepsilon_0} |q_1 q_2|}{G \, m_1 m_2}

Step 3: Plug constants

$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$
$G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$
$e = 1.6 \times 10^{-19} \, \text{C}$

(i) For an electron and a proton

$|q_1 q_2| = e^2$
$m_1 m_2 = m_e m_p$

\frac{F_e}{F_g} = \frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{(6.67 \times 10^{-11})(9.11 \times 10^{-31})(1.67 \times 10^{-27})}

Step 4: Calculate

Numerator: $9 \times 10^9 \times 2.56 \times 10^{-38} = 2.304 \times 10^{-28}$
Denominator: $6.67 \times 10^{-11} \times 1.52 \times 10^{-57} = 1.014 \times 10^{-67}$

\frac{F_e}{F_g} = \frac{2.304 \times 10^{-28}}{1.014 \times 10^{-67}} \approx 2.27 \times 10^{39}

Result (i): $\boxed{2.27 \times 10^{39}}$ — electrostatic force is enormously stronger than gravitational force for an electron-proton pair.

(ii) For two protons

$|q_1 q_2| = e^2$ (same as above)
$m_1 m_2 = m_p^2$

\frac{F_e}{F_g} = \frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{(6.67 \times 10^{-11})(1.67 \times 10^{-27})^2}

Denominator: $6.67 \times 10^{-11} \times 2.79 \times 10^{-54} = 1.86 \times 10^{-64}$

\frac{F_e}{F_g} = \frac{2.304 \times 10^{-28}}{1.86 \times 10^{-64}} \approx 1.24 \times 10^{36}

Result (ii): $\boxed{1.24 \times 10^{36}}$ — still huge, but smaller than (i) because protons are much heavier than electrons.

Part (b) — Accelerations due to Electrical Force at $r = 1\,\text{Å}$

Step 1: Calculate the electrostatic force

F_e = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r^2}

F_e = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(10^{-10})^2}

F_e = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{10^{-20}}

F_e = 2.304 \times 10^{-8} \, \text{N}

Step 2: Apply Newton’s Second Law ( $F = ma$ )

For the electron:

a_e = \frac{F_e}{m_e} = \frac{2.304 \times 10^{-8}}{9.11 \times 10^{-31}}

a_e \approx 2.53 \times 10^{22} \, \text{m/s}^2

For the proton:

a_p = \frac{F_e}{m_p} = \frac{2.304 \times 10^{-8}}{1.67 \times 10^{-27}}

a_p \approx 1.38 \times 10^{19} \, \text{m/s}^2

Final results:

Electron acceleration: $\boxed{2.53 \times 10^{22} \, \text{m/s}^2}$

Proton acceleration: $\boxed{1.38 \times 10^{19} \, \text{m/s}^2}$

The electron accelerates ~1836 times more than the proton (inverse ratio of their masses), as expected from Newton’s third law — equal force, but vastly different masses.

Here are the most common mistakes students make on this inverse-square law comparison problem, and how to avoid each.

1. Forgetting the Direction of the Force (Vector vs. Scalar)

The Mistake: Students often write the ratio of forces as a vector, or they forget that Coulomb and Newton forces act along the same line (the line joining the two particles). They might also incorrectly add or subtract the forces.
Why it’s wrong: The question asks for the ratio of magnitudes. Force is a vector, but a ratio of magnitudes is a scalar. You must take the absolute values.
How to avoid: Always write the magnitude formulas first:
- $F_{\text{electrostatic}} = k \frac{|q_1 q_2|}{r^2}$
- $F_{\text{gravitational}} = G \frac{m_1 m_2}{r^2}$
- Then, divide one by the other. The $r^2$ cancels out immediately.

2. Using the Wrong Constants or Units

The Mistake: Using $k = 9 \times 10^9$ but forgetting it’s in SI units (N m²/C²), or using $G = 6.67 \times 10^{-11}$ but in wrong units (N m²/kg²). Also, mixing up $m_p$ and $m_e$ in the ratio.
Why it’s wrong: The ratio is dimensionless, but only if you use consistent SI units. If you plug in $r = 1\, \text{Å} = 10^{-10}\,\text{m}$ , you must use meters.
How to avoid: Write down the constants clearly:
- $k = 8.99 \times 10^9 \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2$
- $G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$
- $e = 1.6 \times 10^{-19} \, \text{C}$
- $m_p = 1.67 \times 10^{-27} \, \text{kg}$
- $m_e = 9.11 \times 10^{-31} \, \text{kg}$
Pro tip: For the ratio, you don’t even need $r$ — it cancels. So you only need $k$ , $G$ , charges, and masses.

3. Incorrectly Handling the Charges for (a)(i) and (a)(ii)

The Mistake: For (a)(i) (electron & proton), students use $q_1 = -e$ and $q_2 = +e$ , then get confused about the sign. For (a)(ii) (two protons), they forget that both charges are $+e$ .
Why it’s wrong: The magnitude of force uses $|q_1 q_2|$ . For electron-proton, $|(-e)(+e)| = e^2$ . For two protons, $|(+e)(+e)| = e^2$ . Both are $e^2$ ! The ratio is the same for both cases.
How to avoid: Always take the absolute value of the product of charges. Write:
- $|q_1 q_2| = e^2$ for both (i) and (ii).
- Then the ratio $\frac{F_e}{F_g} = \frac{k e^2}{G m_1 m_2}$ .
- For (i): $m_1 m_2 = m_e m_p$ .
- For (ii): $m_1 m_2 = m_p m_p = m_p^2$ .

4. Forgetting to Square the Masses in (a)(ii)

The Mistake: In the two-proton case, students write $m_1 m_2 = m_p$ instead of $m_p^2$ .
Why it’s wrong: Newton’s law uses the product of both masses. For two identical protons, that’s $m_p \times m_p = m_p^2$ .
How to avoid: Always write the product explicitly: $m_1 m_2$ . If both are protons, it’s $m_p^2$ . If one is electron and one is proton, it’s $m_e m_p$ .

5. Misinterpreting “Acceleration” in Part (b)

The Mistake: Students calculate the force correctly but then use $F = ma$ with the wrong mass — they use the total mass or the mass of the other particle.
Why it’s wrong: Newton’s second law says $F = m a$ for each particle individually. The same magnitude of force acts on both, but their accelerations are different because their masses are different.
How to avoid:
- First, calculate the electrostatic force $F_e = k \frac{e^2}{r^2}$ (with $r = 10^{-10}\,\text{m}$ ).
- Then, separately compute:
  - $a_e = \frac{F_e}{m_e}$ (acceleration of electron)
  - $a_p = \frac{F_e}{m_p}$ (acceleration of proton)
- Notice: $a_e$ will be much larger than $a_p$ because $m_e \ll m_p$ .

6. Not Checking the Magnitude of the Ratio

The Mistake: Students get a number like $10^{39}$ or $10^{36}$ but don’t realize how huge it is, so they don’t double-check their exponent arithmetic.
Why it’s wrong: A small error in exponent (e.g., using $10^{-31}$ vs $10^{-27}$ ) changes the ratio by a factor of $10^4$ .
How to avoid: Do a quick order-of-magnitude check:
- $k \approx 10^{10}$ , $G \approx 10^{-10}$ , so $k/G \approx 10^{20}$ .
- $e^2 \approx (10^{-19})^2 = 10^{-38}$ .
- For electron-proton: $m_e m_p \approx (10^{-30})(10^{-27}) = 10^{-57}$ .
- Ratio $\approx \frac{10^{20} \times 10^{-38}}{10^{-57}} = 10^{20-38+57} = 10^{39}$ . That’s correct.
- For two protons: $m_p^2 \approx 10^{-54}$ , ratio $\approx \frac{10^{20} \times 10^{-38}}{10^{-54}} = 10^{20-38+54} = 10^{36}$ . Also correct.

7. Forgetting to Convert Å to Meters in Part (b)

The Mistake: Using $r = 1$ instead of $r = 10^{-10}\,\text{m}$ .
Why it’s wrong: The force formula requires SI units. $1\,\text{Å} = 10^{-10}\,\text{m}$ .
How to avoid: Write the conversion explicitly: $r = 1\,\text{Å} = 1 \times 10^{-10}\,\text{m}$ . Then plug into $F = k e^2 / r^2$ .

Quick Summary Table of Mistakes & Fixes

Mistake	Why It’s Wrong	How to Avoid
Using vector instead of magnitude	Ratio is scalar	Take absolute values of charges
Wrong constants/units	Ratio becomes dimensionally wrong	Write $k$ , $G$ , $e$ , $m_p$ , $m_e$ in SI
Sign confusion in charges	Magnitude is always $e^2$	Use $
Forgetting $m_p^2$ in (a)(ii)	Product of masses is $m_p^2$	Write $m_1 m_2$ explicitly
Using wrong mass for acceleration	Each particle has its own $a$	$a = F/m$ for each particle separately
Exponent errors in ratio	Off by factors of $10^4$	Do order-of-magnitude check
Not converting Å to m	Force value is wrong	$1\,\text{Å} = 10^{-10}\,\text{m}$

Final tip: Always write the formulas in full, cancel common terms ( $r^2$ ), and then plug numbers. This minimizes errors and shows the examiner you understand the concept.

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.

Combine magnitude and direction. Equal magnitudes and opposite directions mean

$\vec{F}_1 = -\vec{F}_2.$
Watch out

A common mistake is to think the force depends on which charge is “source” and which is “test,” leading to a ratio like $2:1$ because one charge is twice the other. But Coulomb’s law is symmetric: the product $q_1 q_2$ is the same either way, so the force magnitude is identical. Newton’s Third Law guarantees the vector relation.

Tip

You never need to compute the actual numerical force here. The symmetry of Coulomb’s law and Newton’s Third Law give the answer directly. The charges could be $1\,\text{C}$ and $1000\,\text{C}$ — the forces would still be equal and opposite.

✓Final answer
$\vec{F}_1 = -\vec{F}_2$
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.

Rod removed: the inducing field vanishes. The separated charges recombine uniformly because there’s no external force to keep them apart.

Final state: sphere is once again uncharged — net charge is zero.

Watch out

A common mistake is to think the sphere acquires a permanent charge just because charges moved. Induction alone does not transfer charge — it only redistributes it. Without grounding (connecting to Earth), no net charge enters or leaves the sphere.

Tip

Think of it like this: induction is like pushing water in a bucket to one side — the water level tilts, but the total amount of water hasn’t changed. Let go, and it sloshes back flat.

✓Final answer
$The net charge on the metal sphere is zero, so the correct option is (A) .$

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.

Only the $\alpha$ -particle carries charge ( $+2e$ ), so it alone experiences $\vec F = 2e\vec E$ and is deflected — towards the negative plate, since its charge is positive.

(For context: $\beta$ -particles, being electrons with charge $-e$ , are also deflected, and much more strongly than $\alpha$ 's because their mass is ~7300 times smaller — but $\beta$ is not among the options.)

✓Final answer

\alpha

ANSWER: B

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;

obeys the inverse-square law — true;

is attractive OR repulsive depending on charge signs — so "always attractive" is false.

✓Final answer
$The correct option is (A) — It is always attractive$
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.

By Coulomb's law $F = \dfrac{kq_1q_2}{r^2}$ , so force is inversely proportional to the square of separation. Tripling $r$ multiplies the force by $\frac{1}{3^2}=\frac19$ :
$F' = \frac{18}{9} = 2\ \text{N}.$
✓Final answer
$The correct option is (B) — 2 N$

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:

F_e=\frac{(9\times10^{9})(1.6\times10^{-19})^{2}}{(10^{-8})^{2}}=\frac{9\times10^{9}\times2.56\times10^{-38}}{10^{-16}}\approx 2.3\times10^{-12}\ \text{N}.

Small in absolute terms, but finite — whereas $F_n$ at this range is utterly negligible.

Step 4 — Conclude.

F_e\ \gg\ F_n .

Note the contrast: inside a nucleus ( $r\sim1$ fm) the situation reverses — there $F_n\gg F_e$ , which is why nuclei hold together at all despite proton–proton repulsion. The question deliberately places the protons far outside nuclear dimensions.

✓Final answer

F_e \gg F_n

ANSWER: B

←Previous section · 1.5Coulomb's Law →Next question · 1.5Example 1.4 — A charged metallic sphere A is suspended by a nylon thread. Another ch…

Q.Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively.

(a) Comparing the strength of electrostatic and gravitational forces

(b) Estimating accelerations

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