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(A) The potential at a distance $r$ from the center of a uniformly charged non-conducting sphere is given by $V(r) = \frac{\rho}{6\varepsilon_0} (3R^2

Vedclass · Accepted Answer

$[\rho R^2 / 4m\varepsilon_0]^{1/2}$

Vedclass · Answer

(A) The potential at a distance $r$ from the center of a uniformly charged non-conducting sphere is given by $V(r) = \frac{\rho}{6\varepsilon_0} (3R^2 - r^2)$.For the tunnel at distance $d = R/2$ from the center,the potential at the surface $(r=R)$ is $V_s = \frac{\rho}{6\varepsilon_0} (3R^2 - R^2) = \frac{\rho R^2}{3\varepsilon_0}$.The potential at the center of the tunnel $(r=R/2)$ is $V_c = \frac{\rho}{6\varepsilon_0} (3R^2 - (R/2)^2) = \frac{\rho}{6\varepsilon_0} (3R^2 - R^2/4) = \frac{\rho}{6\varepsilon_0} (11R^2/4) = \frac{11\rho R^2}{24\varepsilon_0}$.To reach the opposite end,the particle must at least reach the center of the tunnel where the potential is maximum. By conservation of energy: $\frac{1}{2}mv^2 + qV_s = 0 + qV_c$.Given $q=1$,$\frac{1}{2}mv^2 = V_c - V_s = \frac{11\rho R^2}{24\varepsilon_0} - \frac{\rho R^2}{3\varepsilon_0} = \frac{\rho R^2}{24\varepsilon_0} (11 - 8) = \frac{3\rho R^2}{24\varepsilon_0} = \frac{\rho R^2}{8\varepsilon_0}$.Thus,$v^2 = \frac{\rho R^2}{4m\varepsilon_0}$,which gives $v = [\rho R^2 / 4m\varepsilon_0]^{1/2}$.

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