Question

A point charge having charge 'q' and mass 'm' is released at rest on the axis of uniformly charged fix ring of total charge 'Q' and radius 'R' from a distance √3 R. find out it's velocity when it reaches to centre of ring.

Answer 1

$$\begin{gathered} Whenchargeqisreleasedatdis\tan ce\sqrt{3}Ratrestfromauniformly \\ chargedringwithfixedchargesQthenkineticenergyatthispointiszero. \\ Thepotentialenergyattispointisgivenby \\ U=\frac{Qq}{4\pi {\epsilon }_0\times \sqrt{3}R} \\ Accordingtotheprincipleofconservationofenergy,netenergyat \\ thecentreoftheringmustbezero,so \\ \frac{1}{2}m{v}^2-U=0 \\ Atthedis\tan cex=R,thepotentialenrgyis \\ U=\frac{Qq}{4\pi {\epsilon }_0\times R},so \\ netenergy\frac{1}{2}m{v}^2-\frac{Qq}{4\pi {\epsilon }_0\times R}=0 \\ or{v}^2=\frac{Qq}{2\pi {\epsilon }_{0}m\times R} \\ orv=\sqrt{\frac{Qq}{2\pi {\epsilon }_{0}m\times R}} \\ Thusthevelocityofchargeqatthecentreoftheringis\sqrt{\frac{Qq}{2\pi {\epsilon }_{0}m\times R}}. \end{gathered}$$