Question 71: Electrostats & Capacitance

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

Accepted Answer

Let the Coulomb constant be $k$.
Before touching, the repulsive force is
$      F_{m initial}      = \frac{k\,q_1\,q_2}{r^2}.    $
After touching, the total charge $q_1+q_2$ redistributes equally, so each ball carries
$      q' = \frac{q_1 + q_2}{2}.    $
The force on separating them again by $r$ is
$      F_{m final}      = \frac{k\,(q')^2}{r^2}      = \frac{k\,(q_1 + q_2)^2}{4\,r^2}.    $
Compare $F_{m final}$ with $F_{m initial}$:
$      F_{m final} - F_{m initial}      = \frac{k}{4r^2}\bigl((q_1+q_2)^2 - 4\,q_1q_2\bigr)      = \frac{k\,(q_1 - q_2)^2}{4r^2}      \;>\;0      \quad	ext{(for }q_1
eq q_2	ext{).}    $
Hence $F_{m final} > F_{m initial}$.
Answer: More than before (Option D).

Answer

Same as before

Answer

zero

Answer

Less than before

Answer

More than before

Electrostats & Capacitance

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

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