A) y
B) − y
C) \[\frac{1}{y}\]
D) \[-\frac{1}{y}\]
Correct Answer: A
Solution :
\[{{F}_{net}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2q{{q}_{0}}}{{{y}^{2}}+{{a}^{2}}}\cos \theta \] \[=\frac{2q{{q}_{0}}y}{4\pi {{\varepsilon }_{0}}{{\left( {{y}^{2}}+{{a}^{2}} \right)}^{3/2}}}=\frac{2q{{q}_{0}}}{4\pi {{\varepsilon }_{0}}}\times \frac{y}{{{a}^{3}}}\]as \[(y<<a)\] \[\therefore \] \[F\propto y\]You need to login to perform this action.
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