Q.(a) Two insulated charged copper spheres A and B have their centres separated by a distance of $50\,\text{cm}$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7}\,\text{C}$? The radii of A and B are negligible compared to the distance of separation.

The key idea is Coulomb's Law, which quantifies the electrostatic force between two point charges. It is an inverse square law, meaning the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

(a)

1. Given charges are $q_1 = q_2 = 6.5 \times 10^{-7}\,\text{C}$ and separation $r = 50\,\text{cm} = 0.5\,\text{m}$.
2. Substitute these values into Coulomb's Law: $F = (9 \times 10^9\,\text{N m}^2/\text{C}^2) \frac{(6.5 \times 10^{-7}\,\text{C})^2}{(0.5\,\text{m})^2}$
3. Calculate the force: $F = (9 \times 10^9) \frac{42.25 \times 10^{-14}}{0.25} = 9 \times 10^9 \times 169 \times 10^{-14} = 1521 \times 10^{-5}\,\text{N} = 1.521 \times 10^{-2}\,\text{N}$.

(b)

1. The charges are doubled ($q_1' = 2q_1$, $q_2' = 2q_2$) and the distance is halved ($r' = r/2$).
2. The new force $F'$ is: …

The fundamental principle governing the interaction between stationary electric charges is Coulomb's Law. It describes the electrostatic force as directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centres. This "inverse square law" is a recurring theme in physics, appearing in gravitation and light intensity as well.

The reason this law works for spheres with negligible radii is that, for practical purposes, we can treat them as point charges located at their centres. This simplification is crucial because Coulomb's Law is derived for point charges. If the spheres were large compared to their separation, the charge distribution on their surfaces would become non-uniform due to mutual repulsion, making the calculation more complex. However, the problem statement explicitly allows us to treat them as point charges.

Let's apply this law to the given scenarios.

Part (a): Calculating the initial force of repulsion

1. Identify the given quantities:

   • Charge on sphere A, $q_A = 6.5 \times 10^{-7}\,\text{C}$
   • Charge on sphere B, $q_B = 6.5 \times 10^{-7}\,\text{C}$
   • Distance of separation, $r = 50\,\text{cm}$
   • Coulomb's constant, $k = 9 \times 10^9\,\text{N m}^2/\text{C}^2$

2. Convert units to SI standard:

   The distance $r$ is given in centimetres, so we convert it to metres:

   $r = 50\,\text{cm} = 0.50\,\text{m}$

   The charges are already in Coulombs (C).

3. Apply Coulomb's Law:

   Substitute the values into the formula $F = k \frac{q_A q_B}{r^2}$. Since both charges are positive, the force will be repulsive.

   $F = (9 \times 10^9\,\text{N m}^2/\text{C}^2) \frac{(6.5 \times 10^{-7}\,\text{C})(6.5 \times 10^{-7}\,\text{C})}{(0.50\,\text{m})^2}$

   $F = (9 \times 10^9) \frac{(6.5)^2 \times (10^{-7})^2}{(0.50)^2}$

   $F = (9 \times 10^9) \frac{42.25 \times 10^{-14}}{0.25}$

   $F = (9 \times 10^9) \times (169 \times 10^{-14})$

   $F = 1521 \times 10^{-5}\,\text{N}$

   $F = 1.521 \times 10^{-2}\,\text{N}$

Part (b): Calculating the force with changed conditions

1. Identify the new conditions:

   • Each sphere is charged double the amount: $q_A' = 2q_A = 2 \times (6.5 \times 10^{-7}\,\text{C}) = 13 \times 10^{-7}\,\text{C}$
   • $q_B' = 2q_B = 13 \times 10^{-7}\,\text{C}$
   • The distance between them is halved: $r' = \frac{r}{2} = \frac{50\,\text{cm}}{2} = 25\,\text{cm}$

2. Convert units to SI standard:

   $r' = 25\,\text{cm} = 0.25\,\text{m}$

3. Apply Coulomb's Law with new values:

   $F' = k \frac{q_A' q_B'}{(r')^2}$ …

Concept: Inverse Square Law Comparison

1. Initial force: By Coulomb’s law, the repulsion between A and B at $10\,\text{cm}$ is

2. After touching with identical uncharged spheres:

   • When A (charge $q_A$) touches identical uncharged C, each gets $q_A/2$.
   • When B (charge $q_B$) touches identical uncharged D, each gets $q_B/2$. So new charges: $q_A' = q_A/2$, $q_B' = q_B/2$.

3. New force at $5.0\,\text{cm}$: …

🔍 The Core Concept

Coulomb’s Law:

where $k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$.

For part (a):

• $q_1 = q_2 = 6.5 \times 10^{-7} \, \text{C}$
• $r = 0.50 \, \text{m}$

So:

Let’s compute step-by-step:

• $q^2 = (6.5 \times 10^{-7})^2 = 42.25 \times 10^{-14} = 4.225 \times 10^{-13}$
• $r^2 = (0.50)^2 = 0.25$
• $F = 9 \times 10^9 \times \frac{4.225 \times 10^{-13}}{0.25} = 9 \times 10^9 \times 1.69 \times 10^{-12}$

Result (a):

✗ Common Mistake #1: Forgetting to convert cm to m

The error: Using $r = 50$ instead of $r = 0.50$ in the formula.

Why it’s wrong: The SI unit of distance in Coulomb’s Law is metres. Using cm gives an answer that is $10^4$ times too small (since $r^2$ is off by $10^4$).

How to avoid:

• Always write the given distance in metres before plugging in.
• Quick check: $50 \, \text{cm} = 0.50 \, \text{m}$.

✗ Common Mistake #2: Mishandling the “negligible radii” condition

The error: Thinking this condition is irrelevant or ignoring it.

Why it’s important: It tells you the spheres can be treated as point charges. If radii were not negligible, the effective distance between charges would be less than centre-to-centre distance, and the force would be larger.

How to avoid:

• Read the condition carefully. If radii are negligible, use centre-to-centre distance directly.
• If radii are not negligible, you would need to consider charge distribution on surfaces.

🔁 Part (b) — The Ratio Trap

New conditions:

• Each charge is doubled: $q' = 2q$
• Distance is halved: $r' = r/2$

New force:

So:

✗ Common Mistake #3: Only doubling the numerator

The error: Thinking $F' = 2F$ or $4F$ instead of $16F$.

Why it’s wrong:

• Doubling each charge multiplies numerator by $2 \times 2 = 4$.
• Halving the distance multiplies denominator by $(1/2)^2 = 1/4$, which is equivalent to multiplying the whole fraction by $4$.
• Combined effect: $4 \times 4 = 16$.

How to avoid:

• Write the ratio explicitly: …