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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. |
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Answer» Solution :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)))` |
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