There is another useful system of units, besides the SI/mks A system, called the cgs (centimeter-gram-second) system. In this system Coulomb's law is given by $\vec{F} = \dfrac{Qq}{r^2}\hat{r}$ where the distance $r$ is measured in cm ($= 10^{-2}\text{ m}$), $F$ in dynes ($= 10^{-5}\text{ N}$) and the charges in electrostatic units (es units), where $1\text{ es unit of charge} = \dfrac{1}{[3]}\times 10^{-9}\text{ C}$. The number $[3]$ actually arises from the speed of light in vacuum which is now taken to be exactly given by $c = 2.99792458 \times 10^8\text{ m/s}$. An approximate value of $c$ then is $c = [3] \times 10^8\text{ m/s}$. (i) Show that the Coulomb law in cgs units yields $1\text{ esu of charge} = 1\ (\text{dyne})^{1/2}\text{ cm}$. Obtain the dimensions of units of charge in terms of mass $M$, length $L$ and time $T$. Show that it is given in terms of fractional powers of $M$ and $L$. (ii) Write $1\text{ esu of charge} = x\text{ C}$, where $x$ is a dimensionless number. Show that this gives $\dfrac{1}{4\pi\varepsilon_0} = \dfrac{10^{-9}}{x^2}\dfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}$. With $x = \dfrac{1}{[3]}\times 10^{-9}$, we have $\dfrac{1}{4\pi\varepsilon_0} = [3]^2 \times 10^9\ \dfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}$, or $\dfrac{1}{4\pi\varepsilon_0} = (2.99792458)^2 \times 10^9\ \dfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}$ (exactly).

There is another useful system of units, besides the SI/mks A system, called the cgs (centimeter-gram-second)

NCERT Exemplar · Q29

\vec{F} = \dfrac{Qq}{r^2}\hat{r}

(i) Show that the Coulomb law in cgs units yields

1\text{ esu of charge} = 1\ (\text{dyne})^{1/2}\text{ cm}

. Obtain the dimensions of units of charge in terms of mass

M

, length

L

and time

T

. Show that it is given in terms of fractional powers of

M

and

L

(ii) Write

1\text{ esu of charge} = x\text{ C}

, where

x

is a dimensionless number. Show that this gives

\dfrac{1}{4\pi\varepsilon_0} = \dfrac{10^{-9}}{x^2}\dfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}

. With

x = \dfrac{1}{[3]}\times 10^{-9}

, we have

\dfrac{1}{4\pi\varepsilon_0} = [3]^2 \times 10^9\ \dfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}

, or

\dfrac{1}{4\pi\varepsilon_0} = (2.99792458)^2 \times 10^9\ \dfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}

(exactly).

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Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

Imagine you're measuring the length of your desk. You pull out a ruler marked in centimetres and find it's 120 cm long. Your friend, using a metre stick, says it's 1.2 m. You're both right — you've just used different units to describe the same physical length.

That's the core idea: unit conversion is the process of changing how you express a quantity without changing the quantity itself.

The Intuition: Same Quantity, Different Labels

Think of a pizza. Whether you call it "one pizza" or "8 slices," the amount of pizza hasn't changed. You've just used a different unit (pizza vs. slice) to describe it.

Similarly, 1 metre and 100 centimetres are the same length — just like 1 pizza and 8 slices are the same amount. The number changes (1 becomes 100, or 1 becomes 8), but the actual thing being measured stays identical.

Note

This is the most important idea to hold onto: conversion changes the number, not the quantity. If you ever feel like the quantity has changed, you've made a mistake.

The Precise Statement

Unit conversion is the multiplication of a quantity by a conversion factor — a fraction equal to 1 — that cancels the old unit and introduces the new one.

A conversion factor looks like this:

\frac{\text{new unit}}{\text{old unit}} = 1

For length: $\frac{1 \text{ m}}{100 \text{ cm}} = 1$ and $\frac{100 \text{ cm}}{1 \text{ m}} = 1$ .

Why are these fractions equal to 1? Because 1 metre is 100 centimetres. The numerator and denominator describe the same physical length, so their ratio is exactly 1.

How to Convert: The Only Rule You Need

Multiply by a conversion factor that cancels the unit you have and leaves the unit you want.

Let's convert 120 cm to metres:

Start with what you have: $120 \text{ cm}$
Choose the conversion factor that has "cm" in the denominator (to cancel it) and "m" in the numerator: $\frac{1 \text{ m}}{100 \text{ cm}}$
Multiply:

120 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = \frac{120}{100} \text{ m} = 1.2 \text{ m}

The "cm" units cancel just like numbers do: $\frac{\text{cm}}{\text{cm}} = 1$ .

Tip

Always write the units explicitly. If the units don't cancel correctly, you've used the wrong conversion factor. This catches 90% of conversion mistakes.

The Reverse: Metres to Centimetres

Now convert 1.2 m to cm. This time, you want "cm" to remain and "m" to cancel. Use $\frac{100 \text{ cm}}{1 \text{ m}}$ :

1.2 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 1.2 \times 100 \text{ cm} = 120 \text{ cm}

Notice: when going from a larger unit (m) to a smaller unit (cm), the number gets larger (1.2 → 120). When going from smaller to larger, the number gets smaller (120 → 1.2). This is a useful sanity check.

Common Conversion Factors You'll Use

Quantity	Relationship	Conversion Factors
Length	1 m = 100 cm	$\frac{1 \text{ m}}{100 \text{ cm}}$ , $\frac{100 \text{ cm}}{1 \text{ m}}$
Mass	1 kg = 1000 g	$\frac{1 \text{ kg}}{1000 \text{ g}}$ , $\frac{1000 \text{ g}}{1 \text{ kg}}$
Time	1 h = 60 min	$\frac{1 \text{ h}}{60 \text{ min}}$ , $\frac{60 \text{ min}}{1 \text{ h}}$
Speed	1 km/h = $\frac{1000}{3600}$ m/s	$\frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

Dimensional Analysis is a powerful tool in physics and engineering that lets us check the consistency of equations, derive relationships, and convert units. But why does it work? Let's build the reasoning from the ground up.

1. The Core Idea: Physical Quantities Have Dimensions

Every physical quantity (like length, time, mass) can be expressed in terms of fundamental dimensions. The most common set in mechanics is:

L = Length
M = Mass
T = Time

For example:

Speed has dimensions $[L T^{-1}]$
Force has dimensions $[M L T^{-2}]$
Energy has dimensions $[M L^2 T^{-2}]$

Why this matters: Two quantities can only be meaningfully compared or equated if they have the same dimensions. You cannot add apples to oranges — and you cannot add length to time.

2. The Principle of Dimensional Homogeneity

The key formula that underpins everything is:

Every valid physical equation must be dimensionally homogeneous.

This means: the dimensions on the left-hand side must equal the dimensions on the right-hand side.

Why must this hold?

Consider an equation like:

v = u + at

Left side: $[v] = L T^{-1}$
Right side: $[u] = L T^{-1}$ , $[at] = (L T^{-2})(T) = L T^{-1}$

Both sides have dimensions $L T^{-1}$ . If they didn't match, the equation would be physically meaningless — you'd be comparing quantities that cannot be equal in any real experiment.

Reasoning: Physical laws describe relationships between measurable quantities. If the dimensions don't match, the equation cannot represent a real physical relationship, because the numerical value would depend on the arbitrary choice of units.

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

This is the deeper mathematical reason. The Buckingham Pi Theorem states:

If a physical problem involves $n$ variables and $k$ fundamental dimensions, then it can be reduced to $n-k$ independent dimensionless groups (called $\pi$ groups).

Why does this work?

Imagine you have a relationship:

f(Q_1, Q_2, \dots, Q_n) = 0

where each $Q_i$ has dimensions. Because the equation must be dimensionally homogeneous, we can rearrange it into a function of dimensionless products only:

F(\pi_1, \pi_2, \dots, \pi_{n-k}) = 0

The reasoning: Dimensions act as constraints. Each fundamental dimension (M, L, T) gives one constraint. So if you have $n$ variables and $k$ constraints, you only have $n-k$ independent dimensionless combinations.

Example: For a simple pendulum, the period $T$ depends on length $L$ , mass $m$ , and gravity $g$ . That's 4 variables with 3 dimensions (M, L, T). So $4-3 = 1$ dimensionless group: $\pi = \frac{T^2 g}{L}$ . This tells us $T \propto \sqrt{L/g}$ without solving any differential equation.

4. Why We Can Convert Units Using Dimensional Analysis

The conversion factor formula:

\text{Value in new unit} = \text{Value in old unit} \times \left(\frac{\text{old unit}}{\text{new unit}}\right)^{\text{dimension exponent}}

Why this works: …

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Concept: Unit Conversion and Dimensional Analysis

Derivation of esu unit and dimensions of charge: Coulomb's law in cgs units is given by $F = \dfrac{Qq}{r^2}$ . From this, the unit of charge can be expressed as $[\text{charge}] = \sqrt{[\text{force}][\text{distance}]^2}$ . Substituting the cgs units: $[\text{esu}] = \sqrt{[\text{dyne}][\text{cm}]^2} = (\text{dyne})^{1/2}\text{ cm}$ . To find the dimensions of charge in terms of $M, L, T$ , we use the dimensions of force: $1\text{ dyne}$ has dimensions of $M L T^{-2}$ (since $1\text{ N} = 1\text{ kg}\cdot\text{m/s}^2$ ). Therefore, the dimensions of charge are $[Q] = (M L T^{-2})^{1/2} L = M^{1/2} L^{3/2} T^{-1}$ . This expression clearly shows fractional powers for $M$ and $L$ .
Relation between SI and cgs Coulomb's law constant: Coulomb's law in SI units is $F_{\text{SI}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_{\text{SI}}q_{\text{SI}}}{r_{\text{SI}}^2}$ . Coulomb's law in cgs units is $F_{\text{cgs}} = \dfrac{Q_{\text{cgs}}q_{\text{cgs}}}{r_{\text{cgs}}^2}$ . We use the conversion factors: $1\text{ N} = 10^5\text{ dynes}$ , $1\text{ m} = 10^2\text{ cm}$ , and $1\text{ C} = \dfrac{1}{x}\text{ esu}$ (since $1\text{ esu} = x\text{ C}$ ). Substituting these into the cgs equation: $F_{\text{SI}} \times 10^5 = \dfrac{(Q_{\text{SI}}/x)(q_{\text{SI}}/x)}{(r_{\text{SI}} \times 10^2)^2}$ $F_{\text{SI}} \times 10^5 = \dfrac{Q_{\text{SI}}q_{\text{SI}}}{x^2 r_{\text{SI}}^2 \cdot 10^4}$ $F_{\text{SI}} = \dfrac{1}{10^9 x^2} \dfrac{Q_{\text{SI}}q_{\text{SI}}}{r_{\text{SI}}^2}$ . …

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In the cgs system, $1\text{ esu of charge} = 1\ (\text{dyne})^{1/2}\text{ cm}$ , and its dimensions are $[M^{1/2}L^{3/2}T^{-1}]$ . By converting units between cgs and SI, we show that $\dfrac{1}{4\pi\varepsilon_0} = \dfrac{10^{-9}}{x^2}\dfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}$ , which, with $x = \dfrac{1}{[3]}\times 10^{-9}$ , yields $\dfrac{1}{4\pi\varepsilon_0} = [3]^2 \times 10^9\ \dfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}$ .

The problem asks us to explore the cgs (centimeter-gram-second) system of units, specifically in the context of Coulomb's law, and relate it to the SI system. This involves understanding how physical laws are expressed in different unit systems and performing careful unit conversions. The core concept is dimensional analysis, which ensures that equations remain consistent regardless of the units chosen, and unit conversion, which allows us to translate quantities from one system to another.

Part (i): Unit and Dimensions of Charge in cgs

Understanding Coulomb's Law in cgs: In the cgs system, Coulomb's law for the force $\vec{F}$ between two point charges $Q$ and $q$ separated by a distance $r$ is given by:

\vec{F} = \dfrac{Qq}{r^2}\hat{r}

Unlike the SI system, there is no explicit constant like $\dfrac{1}{4\pi\varepsilon_0}$ in this cgs formulation. This is because the unit of charge in the cgs electrostatic system (esu) is defined such that the constant of proportionality in Coulomb's law becomes unity.

2. Showing $1\text{ esu of charge} = 1\ (\text{dyne})^{1/2}\text{ cm}$ :

To find the unit of charge (esu), we can rearrange the magnitude of the force equation: $F = \dfrac{Qq}{r^2}$ .

If we consider two unit charges ( $Q=q=1\text{ esu}$ ) separated by a unit distance ( $r=1\text{ cm}$ ), the force between them is $1\text{ dyne}$ .

Substituting these unit values into the equation:

1\text{ dyne} = \dfrac{(1\text{ esu})(1\text{ esu})}{(1\text{ cm})^2}

1\text{ dyne} = \dfrac{(1\text{ esu})^2}{1\text{ cm}^2}

Rearranging to solve for $(1\text{ esu})^2$:

(1\text{ esu})^2 = 1\text{ dyne} \cdot (1\text{ cm})^2

Taking the square root of both sides gives the unit of charge:

1\text{ esu} = \sqrt{1\text{ dyne} \cdot \text{cm}^2} = 1\ (\text{dyne})^{1/2}\text{ cm}

This confirms the first part of the statement.

> [!IMPORTANT]
> The electrostatic unit (esu) of charge is defined such that two point charges of 1 esu each, separated by 1 cm in vacuum, exert a force of 1 dyne on each other.

3. Obtaining the dimensions of charge in terms of $M, L, T$ :

To find the dimensions of charge, we need to express the dimensions of dyne and cm in terms of fundamental dimensions: mass ( $M$ ), length ( $L$ ), and time ( $T$ ).

* The unit of length is cm, so its dimension is $[L]$ .

* The unit of force is dyne. Force is defined as mass times acceleration ( $F=ma$ ). In cgs, $1\text{ dyne} = 1\text{ g}\cdot\text{cm/s}^2$ .

Therefore, the dimensions of force are $[M][L][T]^{-2}$ .

Now, substitute these dimensions into the expression for $1\text{ esu}$:

[1\text{ esu}] = [(\text{dyne})^{1/2}\text{ cm}]

[1\text{ esu}] = ([M][L][T]^{-2})^{1/2} [L]^1

Distributing the power $1/2$ to each dimension within the parenthesis:

[1\text{ esu}] = [M]^{1/2}[L]^{1/2}[T]^{-1} [L]^1

Combining the powers of $[L]$:

[1\text{ esu}] = [M]^{1/2}[L]^{(1/2)+1}[T]^{-1}

[1\text{ esu}] = [M]^{1/2}[L]^{3/2}[T]^{-1}

This shows that the dimensions of charge in the cgs system are indeed given in terms of fractional powers of $M$ and $L$.

Part (ii): Relating esu to Coulombs and Deriving $1/(4\pi\varepsilon_0)$

Setting up the conversion factors:

We are given the following conversion factors:
- Charge: $1\text{ esu of charge} = x\text{ C}$
- Force: $1\text{ dyne} = 10^{-5}\text{ N}$
- Distance: $1\text{ cm} = 10^{-2}\text{ m}$
We need to relate Coulomb's law in cgs to Coulomb's law in SI.
- Coulomb's law in cgs: $F_{\text{cgs}} = \dfrac{Q_{\text{cgs}}q_{\text{cgs}}}{r_{\text{cgs}}^2}$
- Coulomb's law in SI: $F_{\text{SI}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_{\text{SI}}q_{\text{SI}}}{r_{\text{SI}}^2}$
Our strategy is to take the cgs form of Coulomb's law and convert all its quantities to SI units. Then, we will compare the resulting expression with the standard SI form to find the relationship for $\dfrac{1}{4\pi\varepsilon_0}$ .
Converting cgs Coulomb's law to SI units:

Let's express each cgs quantity in terms of its SI equivalent using the given conversion factors.
- Force: If $F_{\text{cgs}}$ is measured in dynes, then $F_{\text{SI}}$ (in Newtons) is $F_{\text{cgs}} \times 10^{-5}$ . So, $F_{\text{cgs}} (\text{dyne}) = \dfrac{F_{\text{SI}} (\text{N})}{10^{-5}}$ .
- Charge: If $Q_{\text{cgs}}$ is measured in esu, then $Q_{\text{SI}}$ (in Coulombs) is $Q_{\text{cgs}} \times x$ . So, $Q_{\text{cgs}} (\text{esu}) = \dfrac{Q_{\text{SI}} (\text{C})}{x}$ . Similarly for $q_{\text{cgs}}$ .
- Distance: If $r_{\text{cgs}}$ is measured in cm, then $r_{\text{SI}}$ (in meters) is $r_{\text{cgs}} \times 10^{-2}$ . So, $r_{\text{cgs}} (\text{cm}) = \dfrac{r_{\text{SI}} (\text{m})}{10^{-2}}$ .
Now, substitute these expressions into the cgs Coulomb's law:

\dfrac{F_{\text{SI}}}{10^{-5}} = \dfrac{\left(\dfrac{Q_{\text{SI}}}{x}\right)\left(\dfrac{q_{\text{SI}}}{x}\right)}{\left(\dfrac{r_{\text{SI}}}{10^{-2}}\right)^2}

Simplify the right-hand side:

\dfrac{F_{\text{SI}}}{10^{-5}} = \dfrac{\dfrac{Q_{\text{SI}}q_{\text{SI}}}{x^2}}{\dfrac{r_{\text{SI}}^2}{10^{-4}}}

\dfrac{F_{\text{SI}}}{10^{-5}} = \dfrac{Q_{\text{SI}}q_{\text{SI}}}{x^2} \cdot \dfrac{10^{-4}}{r_{\text{SI}}^2}

Now, isolate $F_{\text{SI}}$:

F_{\text{SI}} = 10^{-5} \cdot \dfrac{10^{-4}}{x^2} \dfrac{Q_{\text{SI}}q_{\text{SI}}}{r_{\text{SI}}^2}

F_{\text{SI}} = \dfrac{10^{-9}}{x^2} \dfrac{Q_{\text{SI}}q_{\text{SI}}}{r_{\text{SI}}^2}

> [!WARNING]
> A common mistake is to invert the conversion factors. Remember: if $1\text{ unit}_A = k\text{ unit}_B$, then a quantity $X$ expressed in $\text{unit}_A$ is $X_A$, and in $\text{unit}_B$ is $X_B$. So $X_A \cdot k = X_B$. When substituting into an equation, if you have $X_A$ and want to replace it with $X_B$, you use $X_A = X_B/k$.

3. Comparing with SI Coulomb's law:

We have derived the cgs law in SI units as:

F_{\text{SI}} = \dfrac{10^{-9}}{x^2} \dfrac{Q_{\text{SI}}q_{\text{SI}}}{r_{\text{SI}}^2}

The standard SI form of Coulomb's law is: …

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Method: Dimensional Analysis & Unit Conversion

This method uses the principle of dimensional homogeneity — physical laws must have the same form in any consistent system of units. We convert between SI and cgs by equating the Coulomb force law in both systems.

(i) Show $1\ \text{esu} = 1\ (\text{dyne})^{1/2}\ \text{cm}$ and find dimensions

Step 1: Write Coulomb’s law in cgs units

In cgs:

\vec{F} = \frac{Qq}{r^2} \hat{r}

where $F$ is in dynes, $r$ in cm, and $Q,q$ in esu.

Step 2: Solve for the unit of charge

Take $Q = q = 1\ \text{esu}$ , $r = 1\ \text{cm}$ , then:

1\ \text{dyne} = \frac{(1\ \text{esu})^2}{(1\ \text{cm})^2}

So:

1\ \text{esu} = \sqrt{1\ \text{dyne} \times 1\ \text{cm}^2} = 1\ (\text{dyne})^{1/2}\ \text{cm}

Step 3: Find dimensions of charge in cgs

We know:

$1\ \text{dyne} = 1\ \text{g} \cdot \text{cm/s}^2$ → dimensions $[M L T^{-2}]$
$1\ \text{cm}$ → dimensions $[L]$

Therefore:

[\text{esu}] = [\text{dyne}]^{1/2} \cdot [\text{cm}] = (M L T^{-2})^{1/2} \cdot L = M^{1/2} L^{3/2} T^{-1}

Result: Charge in cgs has dimensions $\boxed{M^{1/2} L^{3/2} T^{-1}}$ — fractional powers of $M$ and $L$ .

(ii) Relate esu to Coulomb and find $\frac{1}{4\pi\varepsilon_0}$

Step 1: Write the conversion

Let $1\ \text{esu} = x\ \text{C}$ , where $x$ is dimensionless.

Step 2: Equate forces for the same physical situation

Take two charges $Q = q = 1\ \text{esu}$ at $r = 1\ \text{cm} = 10^{-2}\ \text{m}$ .

In cgs:

F_{\text{cgs}} = \frac{(1\ \text{esu})^2}{(1\ \text{cm})^2} = 1\ \text{dyne} = 10^{-5}\ \text{N}

In SI:

F_{\text{SI}} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(x\ \text{C})^2}{(10^{-2}\ \text{m})^2}

Step 3: Equate the forces

Since it’s the same physical situation: …

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Below are the common mistakes and how to avoid each.

1. Confusing the form of Coulomb’s law in cgs vs SI

Mistake:

Students write Coulomb’s law in cgs as

$\vec{F} = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2} \hat{r}$

— which is wrong for cgs. In cgs, the constant is 1 (dimensionless), not $1/(4\pi\varepsilon_0)$ .

How to avoid:

Memorise:
- SI: $\vec{F} = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2} \hat{r}$
- cgs (esu): $\vec{F} = \frac{Qq}{r^2} \hat{r}$
The constant is hidden inside the definition of the charge unit (esu).

2. Forgetting to convert units properly in part (i)

Mistake:

Plugging $F$ in dynes and $r$ in cm directly into $F = Qq/r^2$ without realising that this defines the esu.

How to avoid:

Start from the given: $1 \text{ esu}$ is that charge which, when placed $1 \text{ cm}$ from an equal charge, experiences a force of $1 \text{ dyne}$ .
So set $F = 1 \text{ dyne}$ , $r = 1 \text{ cm}$ , $Q = q = 1 \text{ esu}$ :

1 = \frac{(1)(1)}{1^2} \quad \Rightarrow \quad 1 \text{ esu} = 1 \ (\text{dyne})^{1/2} \cdot \text{cm}

Then replace dyne with $\text{g} \cdot \text{cm/s}^2$ :

1 \text{ esu} = (1 \ \text{g} \cdot \text{cm/s}^2)^{1/2} \cdot \text{cm} = \text{g}^{1/2} \ \text{cm}^{3/2} \ \text{s}^{-1}

3. Getting the dimensions wrong in part (i)

Mistake:

Writing $[Q] = M^{1/2} L^{3/2} T^{-1}$ but forgetting the fractional powers or writing $T^{-2}$ .

How to avoid:

Use dimensional analysis step-by-step:
- $[F] = M L T^{-2}$
- $[r^2] = L^2$
- From $F = Qq/r^2$ , we have $[Q^2] = [F][r^2] = (M L T^{-2})(L^2) = M L^3 T^{-2}$
- So $[Q] = M^{1/2} L^{3/2} T^{-1}$
Key point: Fractional powers are allowed in cgs — this is unusual for students used to integer exponents in SI.

4. Mishandling the conversion factor $[3]$ in part (ii)

Mistake:

Treating $[3]$ as exactly 3, then later using $c = 2.99792458 \times 10^8$ without realising $[3]$ is an approximation.

How to avoid:

The problem says: $1 \text{ esu} = \frac{1}{[3]} \times 10^{-9} \text{ C}$ $1 esu = \frac{1}{[ 3 ]} \times 1 0^{- 9} C$
- Here $[3]$ is a placeholder for the approximate value $3$ (from $c \approx 3 \times 10^8$ ).
When doing exact work, replace $[3]$ with $2.99792458$ .
Never cancel $[3]$ with $3$ unless the problem explicitly says "approximate".

5. Algebraic slip in deriving $\frac{1}{4\pi\varepsilon_0}$

Mistake:

Writing $1 \text{ esu} = x \text{ C}$ and then incorrectly substituting into $F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2}$ .

How to avoid:

Write the same force in both systems:
- In cgs: $F = \frac{Qq}{r^2}$ with $F$ in dynes, $r$ in cm, $Q$ in esu.
- In SI: $F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2}$ with $F$ in N, $r$ in m, $Q$ in C.
Convert:
- $1 \text{ dyne} = 10^{-5} \text{ N}$
- $1 \text{ cm} = 10^{-2} \text{ m}$
- $1 \text{ esu} = x \text{ C}$
Equate the two expressions for force and solve for $\frac{1}{4\pi\varepsilon_0}$ : …

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Part (i): Unit and Dimensions of Charge in cgs

Part (ii): Relating esu to Coulombs and Deriving 1/(4πε0)1/(4\pi\varepsilon_0)1/(4πε0​)

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Part (ii): Relating esu to Coulombs and Deriving $1/(4\pi\varepsilon_0)$

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