Electrostatics Question 819
A unit positive point charge of mass m is projected with a velocity V inside the tunnel as shown. The tunnel has been made inside a uniformly charged nonconducting sphere (charge density $\rho$ ). The minimum velocity with which the point charge should be projected such that it can reach the opposite end of the tunnel is equal to
Options:
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 the same potential
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Answer:
Correct Answer: A
Solution:
If we throw the charged particle just right into the tunnel. Hence, applying conservation of energy between start point and center of tunnel we get $ \Delta K+\Delta U=0 $
$ \text{or }( 0+\frac{1}{2}mv^{2} )+q( V _{f}-V _{i} )=0 $
$ \text{or }V _{f}=\frac{V _{s}}{2}( 3-\frac{r^{2}}{R^{2}} )=\frac{\rho R^{2}}{6{\varepsilon _{0}}}( 3-\frac{r^{2}}{R^{2}} ) $
$ \text{Hence }r=\frac{R}{2} $
$ V _{f}=\frac{\rho R^{2}}{6{\varepsilon _{0}}}( 3-\frac{1}{4} )=\frac{11\rho R^{2}}{24{\varepsilon _{0}}};V _{i}=\frac{\rho R^{2}}{3{\varepsilon _{0}}} $
$ \frac{1}{2}mv^{2}=1[ \frac{11\rho R^{2}}{24{\varepsilon _{0}}}-\frac{\rho R^{2}}{3{\varepsilon _{0}}} ] $
$ =\frac{\rho R^{2}}{{\varepsilon _{0}}}[ \frac{11}{24}-\frac{1}{3} ]=\frac{\rho R^{2}}{3{\varepsilon _{0}}}\text{ or }V={{( \frac{\rho R^{2}}{4{\varepsilon _{0}}} )}^{1/2}} $ Hence velocity should be slightly greater than V.
BETA
