Q138 / 180 · COMEDK 2025 (10 May Shift 1) · Physics
Two identical conducting balls having unequal 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
- Azero
- Correct answerBMore than before
- CSame as before
- DLess than before
Solution
Initial force: $F_1 = k\dfrac{q_1 q_2}{r^2}$
After touching, each ball gets charge $q' = \dfrac{q_1+q_2}{2}$
New force: $F_2 = k\dfrac{\left(\dfrac{q_1+q_2}{2}\right)^2}{r^2}$
Comparing: since $q_1 \neq q_2$, $(q_1-q_2)^2 > 0$
$\Rightarrow (q_1+q_2)^2 > 4q_1q_2 \Rightarrow \left(\dfrac{q_1+q_2}{2}\right)^2 > q_1q_2$
$\therefore F_2 > F_1$, the force is more than before.
After touching, each ball gets charge $q' = \dfrac{q_1+q_2}{2}$
New force: $F_2 = k\dfrac{\left(\dfrac{q_1+q_2}{2}\right)^2}{r^2}$
Comparing: since $q_1 \neq q_2$, $(q_1-q_2)^2 > 0$
$\Rightarrow (q_1+q_2)^2 > 4q_1q_2 \Rightarrow \left(\dfrac{q_1+q_2}{2}\right)^2 > q_1q_2$
$\therefore F_2 > F_1$, the force is more than before.
Hence, the correct answer is option: 'B'.