Q.A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is $10\,\text{cm}$, as shown in Fig. 1.4(a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres A and B are touched by uncharged spheres C and D respectively, as shown in Fig. 1.4(b). C and D are then removed and B is brought closer to A to a distance of $5.0\,\text{cm}$ between their centres, as shown in Fig. 1.4(c).

Concept: Inverse square law comparison with charge redistribution.

Initially at $r_1 = 10\,\text{cm}$, the force is $F_1 = k\frac{q_A q_B}{r_1^2}$.

When identical conducting spheres touch, charge distributes equally. After A touches C, each has $q_A/2$; after B touches D, each has $q_B/2$. C and D are removed, leaving A with charge $q_A/2$ and B with charge $q_B/2$.

At the new separation $r_2 = 5.0\,\text{cm}$, the force becomes

Taking the ratio: …

The heart of this problem is understanding how charge redistributes when conductors touch, then applying Coulomb's law to compare forces at different configurations.

When two identical conducting spheres touch, they reach the same potential — which for identical spheres means they share the total charge equally. If sphere A carries charge $q_A$ and touches an uncharged identical sphere C, each ends up with $\frac{q_A}{2}$. The same happens to B and D.

Coulomb's law tells us the electrostatic force between two point charges (or spherically symmetric charge distributions) is

where $k = \frac{1}{4\pi\varepsilon_0}$ and $r$ is the centre-to-centre separation. The force scales as the product of charges and inversely as the square of distance.

Step-by-step comparison

1. Initial configuration (a): distance $10\,\text{cm}$, original charges

   Let the initial charges be $q_A$ and $q_B$. The repulsive force is

2. Charge redistribution (b): touching with uncharged spheres

   • Sphere A touches uncharged sphere C (identical size). Total charge $q_A + 0 = q_A$ splits equally: each gets $\frac{q_A}{2}$.
   • Sphere B touches uncharged sphere D. Total charge $q_B + 0 = q_B$ splits equally: each gets $\frac{q_B}{2}$.

   After C and D are removed, A carries $\frac{q_A}{2}$ and B carries $\frac{q_B}{2}$.

3. Final configuration (c): distance $5.0\,\text{cm}$, halved charges

   The new force is

4. Comparison …

Method: Coulomb's Law with Charge Sharing by Contact

This problem uses Coulomb's inverse square law combined with the principle of charge redistribution when identical conducting spheres are touched.

Step-by-step solution

Step 1: Initial situation (Fig. 1.4a)

• Let the initial charges on A and B be $q_A$ and $q_B$.
• Distance between centres: $r_1 = 10\ \text{cm} = 0.1\ \text{m}$
• Initial repulsive force: $F_1 = k \dfrac{q_A q_B}{r_1^2}$

Step 2: Charge sharing process (Fig. 1.4b)

• Sphere A (charge $q_A$) is touched by identical uncharged sphere C.

• When identical conductors touch, total charge distributes equally.

• After separation: $q_A' = q_C = \dfrac{q_A}{2}$

• Similarly, sphere B (charge $q_B$) is touched by identical uncharged sphere D.

• After separation: $q_B' = q_D = \dfrac{q_B}{2}$

Step 3: Final situation (Fig. 1.4c)

• New charges: $q_A' = \dfrac{q_A}{2}$ and $q_B' = \dfrac{q_B}{2}$
• New distance: $r_2 = 5.0\ \text{cm} = 0.05\ \text{m}$
• New repulsive force: $F_2 = k \dfrac{q_A' q_B'}{r_2^2} = k \dfrac{(q_A/2)(q_B/2)}{(0.05)^2}$

Step 4: Compare forces

• Ratio of forces: $\frac{F_2}{F_1} = \frac{k \cdot \frac{q_A q_B}{4} / (0.05)^2}{k \cdot q_A q_B / (0.1)^2} = \frac{1}{4} \times \frac{(0.1)^2}{(0.05)^2}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting that charge is shared equally between identical spheres

Many students assume the charge on A and B remains unchanged after touching C and D.

Reality: When two identical metallic spheres touch, the total charge redistributes equally.

How to avoid:

• Whenever identical conductors touch, each gets half of the total charge.
• Let initial charges be $q_A$ and $q_B$. After A touches C (identical, uncharged):

• Similarly, after B touches D:

Mistake 2: Using the wrong distance in Coulomb’s law

Students sometimes plug in $10\,\text{cm}$ again, forgetting the distance changes to $5.0\,\text{cm}$.

How to avoid:

• Read the problem carefully: the second separation is half the first.
• Always write the two force expressions explicitly:

Initial force (distance $r_1 = 10\,\text{cm}$):

Final force (distance $r_2 = 5.0\,\text{cm}$, charges halved):

Mistake 3: Ignoring the inverse square effect when distance halves

Some students think halving the distance doubles the force — but Coulomb’s law says force quadruples when distance halves (if charges were constant).

How to avoid:

• Remember: $F \propto 1/r^2$. If $r$ becomes $r/2$, then $F$ becomes $F/(1/4) = 4F$ (for same charges).
• Combine with charge reduction:

Better: compute ratio directly:

So $F_2 = F_1$ — the repulsion is unchanged.

Mistake 4: Not accounting for identical sizes of spheres …