A point mass \['m'\] and charge \['q'\]is projected with a velocity v towards a stationary charge \[{{Q}_{0}}\] from a distance of \[2\text{ }m.\] The closest distance that q can approach is: \[\left[ k=\frac{1}{4\pi {{\varepsilon }_{0}}} \right]\] |
A) \[\frac{m{{v}^{2}}+kq{{Q}_{0}}}{k{{Q}_{0}}q}\]
B) \[\frac{K{{Q}_{0}}}{m{{v}^{2}}+kq{{Q}_{0}}}\]
C) \[\frac{k{{Q}_{0}}+m{{v}^{2}}}{kq{{Q}_{0}}}\]
D) \[\frac{k{{Q}_{0}}-m{{v}^{2}}}{kq{{Q}_{0}}}\]
Correct Answer: B
Solution :
\[\frac{K{{Q}_{0}}}{m{{v}^{2}}+Kq{{Q}_{0}}}\]You need to login to perform this action.
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