Question

Two identical conducting balls having positive charges \(q_1\) and \(q_2\) are separated by a distance \(r\). If they are made to touch each other and then separated to the same distance, the force between them will be

• A. Same as before
• B. zero
• C. Less than before
• D. More than before

Hint:

When identical charged conductors touch, charges equalize → usually increases force unless charges were equal initially.

Answer 1

The Correct Option is D

Step 1: Initial force between charges.
\[ F_1 = \frac{k q_1 q_2}{r^2} \]

Step 2: Charge distribution after touching.
Since balls are identical, charge gets equally distributed:
\[ q' = \frac{q_1 + q_2}{2} \]

Step 3: New force after separation.
\[ F_2 = \frac{k (q')^2}{r^2} \]
\[ F_2 = \frac{k}{r^2} \left(\frac{q_1 + q_2}{2}\right)^2 \]

Step 4: Compare \(F_1\) and \(F_2\).
\[ F_1 = \frac{k q_1 q_2}{r^2}, \quad F_2 = \frac{k (q_1 + q_2)^2}{4r^2} \]

Step 5: Expand \(F_2\).
\[ F_2 = \frac{k}{4r^2}(q_1^2 + 2q_1q_2 + q_2^2) \]

Step 6: Compare magnitudes.
Since \(q_1^2 + q_2^2>2q_1 q_2\), we get:
\[ F_2>F_1 \]

Step 7: Final conclusion.
\[ \boxed{\text{Force increases}} \] Hence, correct answer is option (D).