A cavity of radius r is made inside a solid sphere. The volume charge density of the remaining sphere is `rho`. An electron (charge e, mass m) is released inside the cavity from point P as shown in figure. The centre of sphere and center of cavity are separated by a distance a. The time after which the electron again touches the sphere is

Options

A. `sqrt((6 sqrt (2) r epsi _(0) m )/(e rho a ))`
B. `sqrt ((sqrt2 r epsi _(0) m )/(e rho a ))`
C. `sqrt (( 6 r epsi_(0) m )/( e rho a ))`
D. `sqrt (( r epsi _(0) m )/( e rho a ))`

Answer

A

Explanation

`E = ( rho a )/( 3 epsi _(0)) F = e E = (e rho a )/( 3 m epsi _(0))`
Acceleation `= (rho ea )/( 3 m epsi _(0))`

`= (1)/(2) xx ( rho ea)/ ( 3m epsi _(0)) xx t ^(2) = sqrt2 r`
Electron will move opposite electronic field with constant acceleration along show horizontal line of length `= sqrt2 r`