Question

Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

• A. attract with a force \( \frac{F}{2} \)
• B. repel with a force \( \frac{F}{2} \)
• C. repel with a force \( F \)
• D. attract with a force \( F \)

Hint:

For identical conducting spheres: - Charges equalize after contact. - Always compare forces using Coulomb’s law carefully (charge and distance both matter).

Answer 1

The Correct Option is B

To analyze the problem, let's break it down step-by-step:

1. Initially, we have two metallic balls with charges \(q\) and \(-2q\). The force \(F\) between them when they are at a distance \(r\) is given by Coulomb's Law:

\(F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} = \frac{k \cdot |q \cdot (-2q)|}{r^2} = \frac{2kq^2}{r^2}\)

1. When the balls are brought into contact, they share their charge equally due to conduction. The total initial charge is \((q) + (-2q) = -q\). Thus, each ball will have a charge of \(\frac{-q}{2}\) after separation.
2. Once separated by a distance of \(\frac{r}{2}\), we calculate the new force \(F'\) using Coulomb's Law:
3. The charges are now \(\frac{-q}{2}\) on both balls. The new force \(F'\) is:

\(F' = \frac{k \cdot |\left(\frac{-q}{2}\right) \cdot \left(\frac{-q}{2}\right)|}{\left(\frac{r}{2}\right)^2}\)

Solving for \(F'\) gives:

\(F' = \frac{k \cdot \left(\frac{q^2}{4}\right)}{\frac{r^2}{4}} = \frac{k \cdot q^2}{r^2}\)

1. Comparing \(F'\) with \(F\):
2. \(F' = \frac{1}{2}F\)

Hence, the force becomes half the magnitude of the initial force \(F\), and since the charges are both negative, they repel each other.

The correct answer is: They will repel with a force \(\frac{F}{2}\).

Answer 2

This can be solved quickly using proportional reasoning instead of recomputing the force formula from scratch each time.

Initially the spheres carry \( q \) and \( -2q \), separated by \( r \), with force magnitude \( F = k\dfrac{2q^2}{r^2} \), attractive because the charges have opposite signs.

When identical spheres touch, charge is shared equally. The combined charge is \( q - 2q = -q \), split evenly gives \( -\dfrac{q}{2} \) on each sphere. Both are now negative, so the interaction becomes repulsive going forward, regardless of the exact numbers.

Now compare the new setup to the old one using ratios. The product of the charge magnitudes changes from \( |q \cdot (-2q)| = 2q^2 \) to \( \left|\left(-\dfrac{q}{2}\right)\left(-\dfrac{q}{2}\right)\right| = \dfrac{q^2}{4} \), a factor of \( \dfrac{1}{8} \) of the original product. The separation changes from \( r \) to \( \dfrac{r}{2} \), so \( r^2 \) becomes \( \dfrac{r^2}{4} \), a factor of \( \dfrac{1}{4} \).

Since force is proportional to (charge product) divided by (distance squared), the new force relative to the old one scales by:

\[ \frac{F'}{F} = \frac{1/8}{1/4} = \frac{1}{2} \]

So \( F' = \dfrac{F}{2} \), and because both final charges are negative, the force is repulsive.

The correct answer is that they will repel with a force \( \frac{F}{2} \).