Two identical conducting balls A and B have positive charges `q_(1)` and `q_(2)` respectively. But `q_(1)!=q_(2)`. The balles are brought together so that they touch each other and then kept in their original positions. The force between them is

Options

A. less than that before the balls touched
B. greater than that before the balls touched
C. same as that before the balls touched
D. zero

Answer

B

Explanation

According to Coulomb.s law, the force of repulsion between them is
`F=(q_(1)q_(2))/(4piepsilon_(0)r^(2))`
When the charged spheres A and B are brought in contact, each sphere will attain equal charge `q.`
`q.=(q_(1)+q_(2))/(2)`
Now, the force of repulsion between then at the same distance r is

`F.=(q.xxq.)/(4piepsilon_(0)r^(2))=(1)/(4piepsilon_(0))(((q_(1)+q_(2))/(2))((q_(1)+q_(2))/(2)))/(r^(2))`
`=(((q_(1)+q_(2))/(2))^(2))/(4piepsilonr^(2))`
As `((q_(1)+q_(2))/(2))^(2)gtq_(1)q_(2)`
`:.F.gtF`