A) \[\frac{3{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]
B) \[\frac{4{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]
C) \[\left( \frac{1+2\sqrt{2}}{2} \right)\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]
D) \[\left( 2+\frac{1}{2} \right)\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]
Correct Answer: C
Solution :
\[{{F}_{net}}={{F}_{AC}}+{{F}_{D}}=\sqrt{F_{A}^{2}+F_{C}^{2}}+{{F}_{D}}\] |
Since \[{{F}_{A}}={{F}_{C}}=\frac{k{{q}^{2}}}{{{a}^{2}}}\]and \[{{F}_{D}}=\frac{k{{q}^{2}}}{{{\left( a\sqrt{2} \right)}^{2}}}\] |
\[{{F}_{net}}=\frac{\sqrt{2}k{{q}^{2}}}{{{a}^{2}}}+\frac{k{{q}^{2}}}{2{{a}^{2}}}=\frac{k{{q}^{2}}}{{{a}^{2}}}\left( \sqrt{2}+\frac{1}{2} \right)\] |
\[=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\,\left( \frac{1+\sqrt{2}}{2} \right)\] |
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