There is a uniformly charged ring having radius R. An infinite line charge (charge per unit length lambda) is placed along a diameter of the ring (in gravity free space). Total charge on the ring `Q=4sqrt(2lambda)R`. An electron of mass m is released from rest on the axis of the ring at a distance `x=sqrt(3) R` from the centre.

Magnitude of initial acceleration of the electron.

Options

A. `(elambda)/(pivarepsilon_(0)mR)((3-2sqrt(2))/(4sqrt(6)))`
B. `(elambda)/(pivarepsilon_(0)mR) ((3+2sqrt(2))/(4sqrt(6)))`
C. `(elambda)/(pivarepsilon_(0)mR)((3+2sqrt(2))/(4sqrt(3)))`
D. none

Answer

A

Explanation

Magnitude of initial acceleration of the electron.
Electricfield `E=-lambda/(2pivarepsilon_(0)x)+Q/(4pivarepsilon_(0)) x/((R^(2)+x^(2))^(3//2))` (considering right direction as positive)
`=lambda/(2pivarepsilon_(0))[-1/x+(x4sqrt(2)R)/(2(R^(2)+x^(2))^(3//2))]`
`=lambda/(2pivarepsilon_(0))[-1/x+(2sqrt(2)xxR)/((R^(2)+x^(2))^(3//2))]`
Initially`x=sqrt(3)R`
`E=lambda/(2pivarepsilon_(0)R)[-1/sqrt(3)+(2sqrt(2)sqrt(3))/8]`
`lambda/(2pivarepsilon_(0)R)[(-2sqrt(2)+3)/(sqrt(3)(2sqrt(2)))]=lambda/(2pivarepsilon_(0)R)((3-2sqrt(2))/(2sqrt(6)))`
a(acceleration)`=((-e)(E))/m=-(elambda)/(pivarepsilon_(0) mR) ((3-2sqrt(2))/(4sqrt(6)))`
Sol(2) Force on electron is zero at point where `E=0Rightarrow x=R`
Sol(3) Potential difference between two points
`DeltaV=-E dx`
P.d. due to line charge between
`x=R& x=sqrt(3)R`
`V_(A)-V_(B)=-underset(R)oversetsqrt((3)R)(int) -(lambda dx)/(2pivarepsilon_(0)x)=lambda/(pivarepsilon_(0))((l n3)/4)`
Potential difference due to ring between `x=sqrt(3) R and x=R`
`V_(A)-V_(B)=1/(4pivarepsilon_(0))((4sqrt(2)lambda R)/(2R)-(4sqrt(2)lambdaR)/(sqrt(2)R))=(-lambda)/(pivarepsilon_(0))(1-1/sqrt(2))`
Net `V_(A)-V_(B)=-lambda/(pivarepsilon_(0))[(1-1/sqrt(2))-(l n3)/4]`