Two similarly and equally charged identical metal spheres A and B repel each other with a force of `2xx10^(-5)N`. A third identical uncharged sphere C is touched with A and then placed at the midpoint between A and B. Find the net electric force on C.

Options

A. `2xx10^(-3)N`
B. `2xx10^(-4)N`
C. `2xx10^(-5)N`
D. `2xx10^(-6)N`

Answer

C

Explanation

Let initially, the charge on each sphere be q and separtion between their centres ba r, then according to the given problem

`F=(1)/(4piepsilon_(0))(qxxq)/(r^(2))=2xx10^(-5)N`
When sphere C touches B, the charges B, the charge of B will distribute equally on B and C as sphere are identical conductors, i.e.
`q_(B)=q_(C)=((q)/(2))`
So, sphere C will experience a force
`F_(CA)=(1)/(4piepsilon_(0))(q((q)/(2)))/((r)/(2))^(2)=2F` along AB due to charge on A
and `F_(CB)=(1)/(4piepsilon_(0))((q//2)(q//2))/((r//2)^(2))=F` along BA due to charge on B
So, the net force on C due to charges on A and B,
`F_(C)=F_(CA)-F_(CB)=2F-F`
`F_(C)=F=2xx10^(-5)"N along AB"`