A uniformly charged solid (non-conducting) sphere of radius R carrying positive charge, has a volume charge density equal to `rho`. A tunnel of very small radius is made along the diameter of the sphere. A particle of mass m and charge -q is released from rest near the opening of tunnel described above. Calculate the speed of particle as it passes through the center of the sphere.

Explanation

Electric potential at a point lying inside a charged solid non-conducting sphere of radius R is given by the following relation:
`V+(rho)/(6epsilon_(0))[3R^(2)-x^(2)]" for " x le R`
here x is the distance of point from the centre of sphere

Initial location of particle is at the surface of the sphere.
Let the potential at this point be `V_(1)`.
Using x = R, in above relation, we have, `V_(1)=(rhoR^(2))/(3epsilon_(0))`
The final location is centre where we need to calculate the speed. Let electric potential at the centre be `V_(2)`.
Then we can use x = 0 in the formula of electric potential: `V_(2)=(rho R^(2))/(2epsilon_(0))`
By definition we have:
`V_(2)-V_(1)=(-W_(el))/(q) rArr W_(el)=q(V_(1)-V_(2))`
`rArr W_(el)=(-q) ((rhoR^(2))/(3epsilon_(0))-(rhoR^(2))/(2epsilon_(0)))`, (note that we have used negative sign of charge in this step).
`rArr W_(el)=(rho q R^(2))/(6epsilon_(0))`
For a negatively charged particle
`W_(el)=K_(2)-K_(1)`
`rArr (pqR^(2))/(6epsilon_(0)) =(1)/(2)mv^(2)-0`
`rArr v= sqrt((rho qR^(2))/(3mepsilon_(0)))`