A) \[{{[\rho {{R}^{2}}/4m{{\varepsilon }_{0}}]}^{1/2}}\]
B) \[{{[\rho {{R}^{2}}/24m{{\varepsilon }_{0}}]}^{1/2}}\]
C) \[{{[\rho {{R}^{2}}/6m{{\varepsilon }_{0}}]}^{1/2}}\]
D) zero because the initial and the final points are at some potential
Correct Answer: A
Solution :
[a] If we thraw the charged particle just right the tunnel. Hence, applying conservation of momentum between start point and center of tunnel we get \[\Delta K+\Delta U=0\] \[\text{or }\left( 0+\frac{1}{2}m{{v}^{2}} \right)+q\left( {{V}_{f}}-{{V}_{i}} \right)=0\] \[\text{or }{{V}_{f}}=\frac{{{V}_{s}}}{2}\left( 3-\frac{{{r}^{2}}}{{{R}^{2}}} \right)=\frac{\rho {{R}^{2}}}{6{{\varepsilon }_{0}}}\left( 3-\frac{{{r}^{2}}}{{{R}^{2}}} \right)\] \[\text{Hence }r=\frac{R}{2}\] \[{{V}_{f}}=\frac{\rho {{R}^{2}}}{6{{\varepsilon }_{0}}}\left( 3-\frac{{{R}^{2}}}{4{{R}^{2}}} \right)=\frac{11\rho {{R}^{2}}}{24{{\varepsilon }_{0}}};{{V}_{i}}=\frac{\rho {{R}^{2}}}{3{{\varepsilon }_{0}}}\]\[\frac{1}{2}m{{v}^{2}}=1\left[ \frac{11\rho {{R}^{2}}}{24{{\varepsilon }_{0}}}-\frac{\rho {{R}^{2}}}{3{{\varepsilon }_{0}}} \right]\] \[=\frac{\rho {{R}^{2}}}{{{\varepsilon }_{0}}}\left[ \frac{11}{24}-\frac{1}{3} \right]=\frac{\rho {{R}^{2}}}{3{{\varepsilon }_{0}}}\text{ or }V={{\left( \frac{\rho {{R}^{2}}}{4m{{\varepsilon }_{0}}} \right)}^{1/2}}\] Hence velocity should be slightly greater than V.
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