A) \[\frac{{{q}^{2}}}{4\sqrt{2}\pi {{\varepsilon }_{0}}{{l}^{2}}}\]
B) \[\frac{-{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}\]
C) \[\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}\]
D) zero
Correct Answer: D
Solution :
(d) zero From diagram, force on\[{{q}_{1}}\left( =q \right)\] at A, \[{{\overrightarrow{F}}_{1}}{{\overrightarrow{F}}_{12}}+{{\overrightarrow{F}}_{13}}=F{{\hat{r}}_{1}}\] where, \[F=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}\]and \[{{\hat{r}}_{1}}\]is the unit vector along B(C)You need to login to perform this action.
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