Q.The electrostatic force on a small sphere of charge $0.4\,\mu\text{C}$ due to another small sphere of charge $-0.8\,\mu\text{C}$ in air is $0.2\,\text{N}$.

Concept: Inverse Square Law Comparison — Coulomb’s law gives the force between two point charges as $F = k \frac{|q_1 q_2|}{r^2}$, where $k = 9 \times 10^9\ \text{N m}^2/\text{C}^2$ in air.

(a)

$F = 0.2\ \text{N}$, $q_1 = 0.4 \times 10^{-6}\ \text{C}$, $q_2 = -0.8 \times 10^{-6}\ \text{C}$.

Using $r^2 = k \frac{|q_1 q_2|}{F}$:

$r^2 = 9 \times 10^9 \times \frac{(0.4 \times 10^{-6})(0.8 \times 10^{-6})}{0.2} = 9 \times 10^9 \times \frac{3.2 \times 10^{-13}}{0.2} = 9 \times 10^9 \times 1.6 \times 10^{-12} = 1.44 \times 10^{-2}$.

So $r = \sqrt{1.44 \times 10^{-2}} = 0.12\ \text{m}$.

(b)

By Newton’s third law, the force on the second sphere due to the first is equal in magnitude and opposite in direction: $0.2\ \text{N}$, attractive.

The problem is a direct application of Coulomb’s law for the electrostatic force between two point charges. The key idea is that the force magnitude depends only on the product of the charges and the square of the distance between them — the sign of the charges tells us the direction (attractive or repulsive), but the magnitude is given by the absolute values.

Because the two charges are opposite in sign, the force is attractive. The problem gives the force on the first sphere, and part (b) simply asks for the force on the second sphere — which, by Newton’s third law, must be equal in magnitude and opposite in direction.

Let’s work through it step by step.

1. Write down Coulomb’s law in magnitude form The electrostatic force between two point charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum (or air, which has nearly the same permittivity) is:

where $k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \,\text{N m}^2/\text{C}^2$.

2. Identify the given quantities

   • $q_1 = 0.4\,\mu\text{C} = 0.4 \times 10^{-6}\,\text{C} = 4 \times 10^{-7}\,\text{C}$
   • $q_2 = -0.8\,\mu\text{C} = -0.8 \times 10^{-6}\,\text{C} = -8 \times 10^{-7}\,\text{C}$
   • $F = 0.2\,\text{N}$ (magnitude of force on $q_1$ due to $q_2$)

   The product $|q_1 q_2| = (4 \times 10^{-7})(8 \times 10^{-7}) = 32 \times 10^{-14} = 3.2 \times 10^{-13}\,\text{C}^2$.

3. Solve for the distance $r$

   Rearranging Coulomb’s law:

Substitute the values:

First compute the fraction:

Then:

Taking the square root:

So the distance between the spheres is $0.12$ metres (or $12$ cm).

4. Answer part (b) using Newton’s third law The force on the second sphere due to the first is equal in magnitude and opposite in direction to the force on the first sphere due to the second. Magnitude: $0.2\,\text{N}$ Direction: Since the charges are opposite, the force is attractive — so the second sphere is pulled toward the first. Thus the force on the second sphere is $0.2\,\text{N}$ (attractive).

Method: Coulomb’s Law (Inverse Square Law)

We use Coulomb’s Law for electrostatic force between two point charges:

where:

• $F$ = magnitude of electrostatic force (N)
• $k = 9 \times 10^9\ \text{N m}^2/\text{C}^2$ (Coulomb’s constant for air)
• $q_1, q_2$ = charges (C)
• $r$ = distance between charges (m)

Step 1: Identify given values

• $q_1 = 0.4\ \mu\text{C} = 0.4 \times 10^{-6}\ \text{C}$
• $q_2 = -0.8\ \mu\text{C} = -0.8 \times 10^{-6}\ \text{C}$
• $F = 0.2\ \text{N}$

Note: Force magnitude uses absolute values of charges. The negative sign on $q_2$ only tells us the force is attractive.

Step 2: Solve for distance $r$ (part a)

From Coulomb’s Law:

Substitute values:

Simplify numerator:

So $|q_1 q_2| = 0.32 \times 10^{-12}\ \text{C}^2$

Now:

Take square root:

Answer (a): $\boxed{0.12\ \text{m}}$ (or $12\ \text{cm}$)

Step 3: Force on second sphere (part b)

By Newton’s Third Law, the force on the second sphere due to the first is equal in magnitude and opposite in direction to the force on the first sphere due to the second.

• Magnitude: $0.2\ \text{N}$
• Direction: attractive (toward the first sphere)

Answer (b): $\boxed{0.2\ \text{N}}$ (attractive, toward the first sphere)

Key Concept Check

• The inverse square law means: if distance doubles, force becomes $\frac{1}{4}$.
• Force magnitude depends only on product of charges and distance — not on which charge we consider.
• The negative sign on $q_2$ indicates opposite charges → attraction.

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting to convert microcoulombs ($\mu\text{C}$) to coulombs (C)

The error: Students plug $0.4$ and $0.8$ directly into Coulomb's law, forgetting the $\mu$ (micro) means $10^{-6}$.

Why it happens: The problem gives charges in $\mu\text{C}$, but Coulomb's law requires SI units (C). The conversion factor is easy to overlook under time pressure.

How to avoid: Always write the conversion explicitly before substituting:

Pro tip: Circle or underline the unit in the question. If it's not in base SI, convert first — every time.

Mistake 2: Using the wrong value of $k$ or forgetting it entirely

The error: Some students use $k = 9 \times 10^9$ but forget the units, or mistakenly use $k = 1$ (thinking "in air" means vacuum permittivity is irrelevant).

Why it happens: The constant $k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,\text{N m}^2/\text{C}^2$ is often memorised without understanding its role.

How to avoid: Write Coulomb's law fully:

Then substitute $k = 9 \times 10^9$ with its units. This helps you check that your final distance comes out in metres.

Mistake 3: Ignoring the sign of the charges when calculating force magnitude

The error: Students include the negative sign of $q_2 = -0.8\,\mu\text{C}$ in the product $q_1 q_2$, getting a negative value, then panic or get confused.

Why it happens: Coulomb's law for magnitude uses absolute values. The sign only tells you attraction (opposite signs) or repulsion (same sign).

How to avoid: For part (a), use only magnitudes:

So $|q_1| = 4 \times 10^{-7}\,\text{C}$ and $|q_2| = 8 \times 10^{-7}\,\text{C}$. The force $0.2\,\text{N}$ is already positive — it's the magnitude.

Key insight: The sign of the force (attractive/repulsive) is a direction concept, not a magnitude concept. Part (a) only asks for distance, so signs are irrelevant.

Mistake 4: Solving for $r$ incorrectly (algebra errors)

The error: After substituting, students make mistakes like:

• Forgetting to take the square root
• Inverting the fraction
• Misplacing powers of 10

How to avoid: Solve step-by-step:

1. Write: $r^2 = k \frac{|q_1 q_2|}{F}$
2. Substitute carefully:

3. Simplify powers of 10 separately:

4. Take square root: $r = \sqrt{1.44 \times 10^{-2}} = 1.2 \times 10^{-1} = 0.12\,\text{m}$

Check: $0.12\,\text{m} = 12\,\text{cm}$ — a reasonable distance for these charges and force.

Mistake 5: Answering part (b) with a different value than $0.2\,\text{N}$

The error: Students recalculate the force using the distance found in (a), but make an arithmetic slip, or think the force on the second sphere is somehow different.

Why it happens: They forget Newton's Third Law — electrostatic forces are action-reaction pairs.

How to avoid: Remember: The force on sphere 2 due to sphere 1 is equal in magnitude and opposite in direction to the force on sphere 1 due to sphere 2.

So part (b) answer is simply:

Direction: Attractive (since charges are opposite), but the question only asks for force magnitude.

Mistake 6: Giving the distance in wrong units

The error: After calculating $r = 0.12\,\text{m}$, students write the answer as $0.12$ without units, or convert unnecessarily to cm without being asked.

How to avoid: Always state the unit. The standard SI unit for distance is metres. Write:

If you prefer, you can add $(12\,\text{cm})$ in brackets, but the primary answer should be in metres unless the question specifies otherwise.

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