A) \[\sqrt{\frac{2}{m}}{{\left( \frac{1}{15}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]
B) \[\sqrt{\frac{2}{m}}{{\left( \frac{2}{15}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]
C) \[\sqrt{\frac{2}{m}}{{\left( \frac{4}{15}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]
D) \[\sqrt{\frac{2}{m}}{{\left( \frac{1}{5}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]
Correct Answer: B
Solution :
\[{{U}_{i}}+{{K}_{i}}={{U}_{f}}+{{K}_{f}}\] \[\frac{k{{q}^{2}}}{\sqrt{16{{a}^{2}}+9{{a}^{2}}}}+\frac{1}{2}m{{\text{v}}^{2}}=\frac{k{{q}^{2}}}{3a}\] \[\frac{1}{2}m{{\text{v}}^{2}}=\frac{k{{q}^{2}}}{a}\left( \frac{1}{3}-\frac{1}{5} \right)=\frac{2k{{q}^{2}}}{15a}\] \[\text{v}=\sqrt{\frac{4k{{q}^{2}}}{15ma}}\]
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