Two identical conducting balls having positive charges q₁ and q₂ 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) less than before(b) same as before(c) more than before(d) zero
Answers
Answered by
4
Answer:
C) More than before
Explanation:
According to the Coulomb's law, the force of repulsion between them is given as - q1q2/4π0r²
when the charged spheres A and B are bought in contac, each sphere will attain equal charge q
q = q1+q2/2
The force of repulsion between them at same distance r will be -
F = q1q'/4π0r² = 1/4π0 [(q1+q2/2) (q1+q2/2)]/ r²
F = (q1+q2/2)²/ 4π0r²
As (q1+q2/2)²≥ q1q2
Therefore F'≥F
Thus, the force between the conducting balls will be greater than that before the balls touched.
Similar questions