Question

A point mass of `-q` mass `m` is released form rest form a distance `R` along the axis of a uniform circular disc of charge `Q` Assuming gravity free situation the velocity attained by point mass when it reaches centre of disc is (Radius of disc-R)
`
A. `{((sqrt(2)Qqsqrt(2)-1))/(piepsilon_(0)mR)}^(1//2)`
B. `{((Qqsqrt(2)-1))/(piepsilon_(0)mR)}^(1//2)`
C. `{((Qqsqrt(2)-1))/(2piepsilon_(0)mR)}^(1//2)`
D. `{((Qqsqrt(3)-1))/(2piepsilon_(0)mR)}^(1//2)`

Answer 1

Correct Answer - A

The charge on the shaded part
`dQ=(Q)/(piR^(2))2pirdr=(2Q)/(R^(2)r(dr)`
Potential due to `dQ` at `P`
`dV=(2Q)/(R^(2))=((r)(dr))/(4piepsilon_(0)sqrt(R^(2)+r^(2)))=(Q)/(2piepsilon_(0)R^(2))(rdr)/(sqrt(R^(2)+r^(2)))`
`V=(Q)/(2piepsilon_(0)R^(2))int_(0)^(R)(rdr)/(sqrt(R^(2)+r^(1)))(Q)/(2piepsilon_(0)R^(2))(sqrt(R^(2)+r^(2)))_(0)^(R)`
`=(Q)/(2piepsilon_(0)R^(2))(sqrt(2)-1)R=(Q(sqrt(2)-1))/(3piepsilon_(0)R)`
`rArr U` of `-q=(-Qq(sqrt(2)-1))/(2piepsilon_(0)R)`
Similarly the potential due to `dQ` at centre of disc is
`dV=(2Q)/(R^(2))(r(dr))/(4piepsilon_(0))=(Q)/(2piepsilon_(0)R^(2))dr`
`V=(Q)/(2piepsilon_(0)R)rArr (1)/(2)mv^(2)-(Qq)/(2piepsilon_(0)R)`
`rArr (1)/(2)mv^(2)=(qQ)/(2piepsilon_(0)R)(1-sqrt(2)+1)`
`rArr (1)/(2)mv^(2)=(Qq)/(2piepsilon_(0)R)(2-sqrt(2))`
`rArr (1)/(2)mv^(2)=((Qq)sqrt(2)(sqrt(2)-1))/(2piepsilon_(0)R)V=[(sqrt(2)Qq(sqrt(2)-1))/(piepsilon_(0)mR)]^(1//2)`