The distance between two equal balls having unlike charges is `2cm`. The radii of the balls are much less than the distance between them. The balls attract each other with a force of `36xx10^(-5)`N. After the ball have been connected by a wire and latter has been removed, the balls repel each other with a force of `20.25xx10^(-5)`N. Determine the original charges on the balls

Options

A. `+8xx10^(-9)` and `-2xx10^(-9)C`
B. `-6xx10^(-9)` and `+1.5xx10^(-9)C`
C. `+4xx10^(-9)C` and `-1xx10^(-9)C`
D. `-8xx10^(-9)C` and `+2xx10^(-9)C`

Answer

a

Explanation

Let `q_(1)` and `-q_(2)` be the charges on the two balls. Before connecting with a wire:
`F=(1)/(4piepsilon_(0)).(q_(1)q_(2))/(r^(2))` (attractive)
Or `q_(1)q_(2)= 4pi epsilon_(0)F r^(2)`
Setting `F= 36xx10^(-5) N` and `r= 2 cm= 0.02m`, we get
`q_(1)q_(2)= 16xx10^(-18)`....(i)
After connecting with wire: Let `q'` be the charge on each ball. Then,
`q'(q_(1)+(-q_(2)))/(2)=(q_(1)-q_(2))/(2)`
`F' =(1)/(4piepsilon_(0)).((q_(1)-q_(2))//2xx(q_(1)-q_(2))//2)/(r^(2))`(repulsive)
Or `q_(1)-q_(2)= 2xx sqrt(4piepsilon_(0)F'r)`
Setting `F' = 20.25xx10^(-5)N` and `r= 0.2m`, We get
`q_(1)-q_(2)= 6xx10^(-9)`....(ii)
Solving the equations (i) and (ii), we obtain
`q_(1)= 8xx10^(-9)C`
And `q_(2)= 2xx10^(-9)C`(negative)