A) 0
B) v
C) \[\sqrt{2}v\]
D) \[\sqrt{1.5}v\]
Correct Answer: D
Solution :
[d]: Potential at \[\infty ,V\infty =0\]. Potential at the surface of the sphere,\[{{V}_{s}}=k\frac{Q}{R}\]. \[\left( \text{where}\,k=\frac{1}{4\pi {{\varepsilon }_{0}}} \right)\] Potential at the centre of the sphere,\[{{V}_{c}}=\frac{3}{2}k\frac{Q}{R}\]. Let m and -q be the mass and the charge of the particle respectively. Let \[{{v}_{0}}=\]speed of the particle at the centre of the sphere. \[\frac{1}{2}m{{v}^{2}}=-q[{{V}_{\infty }}-{{V}_{s}}]=qk\frac{Q}{R}\] ...(i) \[\frac{1}{2}mv_{0}^{2}=-q[{{V}_{\infty }}-{{V}_{c}}]=q.\frac{3}{2}k\frac{Q}{R}\] ?(ii) Dividing (ii) by (i),\[\frac{v_{0}^{2}}{{{v}^{2}}}=\frac{3}{2}=1.5\]or\[{{v}_{0}}=\sqrt{1.5}v\]
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