Two identical conducting balls having positive charges q1 and q2 are s...
Answer:
Explanation:
When two identical conducting balls having positive charges q1 and q2 are separated by a centre to centre distance r, the force between them is given by Coulomb's law as:
F = (1/4πε) * (q1q2/r²)
where ε is the permittivity of free space.
When the two balls are made to touch each other, their charges get redistributed and they acquire a common potential. Let this potential be V. Then, the charges on the two balls will be q1' = q2' = q1 + q2.
When the balls are separated to the same distance r, their charges remain the same but the distance between them decreases to r/2. Therefore, the force between them is given by:
F' = (1/4πε) * [(q1+q2)²/(r/2)²] = (1/4πε) * 4(q1+q2)²/r² = (1/4πε) * 4(q1²+2q1q2+q2²)/r²
= (1/4πε) * (q1q2/r²) * 4(q1/q2 + 2 + q2/q1 + 2) = 4F * (q1/q2 + 2 + q2/q1 + 2)
From the above equation, we can see that:
- If q1 = q2, then F' = 8F, which means the force between the balls is more than before.
- If q1 > q2, then q1/q2 > 1 and q2/q1 < 1,="" which="" means="" f'="" /> F, and the force is more than before.
- If q1 < q2,="" then="" q1/q2="" />< 1="" and="" q2/q1="" /> 1, which means F' < f,="" and="" the="" force="" is="" less="" than="" />
Therefore, the correct answer is option A) less than before.
Two identical conducting balls having positive charges q1 and q2 are s...
More than before
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