The diagram shows a small bead of mass m carr | vedclass

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Question

(A) Let the radius of the ring be $R$. From the diagram,the distance of $P$ from $+Q$ is $4a$. Thus,$V = \frac{Q}{4\pi\epsilon_0(4a)}$.The bead must r

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$\sqrt{\frac{6qV}{m}}$

Vedclass · Answer

(A) Let the radius of the ring be $R$. From the diagram,the distance of $P$ from $+Q$ is $4a$. Thus,$V = \frac{Q}{4\pi\epsilon_0(4a)}$.The bead must reach the point $B$ (the point closest to $+Q$) to complete the circle. The distance of $B$ from $+Q$ is $a$. The potential at $B$ is $V_B = \frac{Q}{4\pi\epsilon_0 a} = 4V$.Using the principle of conservation of mechanical energy between point $P$ and point $B$:$U_P + K_P = U_B + K_B$$qV + \frac{1}{2}mv_0^2 = qV_B + K_B$$qV + \frac{1}{2}mv_0^2 = q(4V) + K_B$For the bead to complete the circle,it must have a non-zero kinetic energy at point $B$ $(K_B > 0)$.$\frac{1}{2}mv_0^2 = 3qV + K_B$Since $K_B > 0$,we have $\frac{1}{2}mv_0^2 > 3qV$.$v_0^2 > \frac{6qV}{m}$$v_0 > \sqrt{\frac{6qV}{m}}$.

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