Two identical thin rings, each of radius 10 cm carrying charges 10 C and 5 C are coaxially placed at a distance 10 cm apart. The work done in moving a charge q from the centre of the first ring to that of the second is

Options

A. `(q)/(8piepsilon_(0))((sqrt(2)+1)/(sqrt(2)))`
B. `(q)/(8piepsilon_(0))((sqrt(2)-1)/(sqrt(2)))`
C. `(q)/(4piepsilon_(0))((sqrt(2)+1)/(sqrt(2)))`
D. `(q)/(4piepsilon_(0))((sqrt(2)-1)/(sqrt(2)))`

Answer

B

Explanation

Work done `W=q(V_(2)-V_(1))`

Now, `V_(1)=(Q_(1))/(4piepsilon_(0)R_(1))+(Q_(2))/(4piepsilon_(0)Rsqrt(2))`
`=(10)/(4piepsilon_(0)xx10)+(5)/(4piepsilon_(0)10sqrt(2))`
`V_(2)=(Q_(2))/(4piepsilon_(0)R)+(Q_(1))/(4piepsilon_(0)Rsqrt(2))`
`=(5)/(4piepsilon_(0)xx10)+(10)/(4piepsilon10sqrt(2))`
`V_(2)-V_(1)=(5)/(4piepsilon_(0)10sqrt(2))-(5)/(4piepsilon_(0)xx10)`
`=(5)/(4piepsilon_(0)10)[(1)/(sqrt(2))-1]=(1)/(8piepsilon_(0))[(sqrt(2)-1)/(sqrt(2))]`
`therefore W=(q)/(8piepsilon_(0))[(sqrt(2)-1)/(sqrt(2))]`