question_answer

Equal charges \[q\] are placed at the four corners \[A,\,B,\,C,\,D\] of a square of length \[a\]. The magnitude of the force on the charge at B will be                                                                  [MP PMT 1994; DPMT 2001]

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)                    \[\left( 2+\frac{1}{\sqrt{2}} \right)\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]

Correct Answer: D

Solution :

                     After following the guidelines mentioned above \[{{F}_{net}}={{F}_{AC}}+{{F}_{D}}=\sqrt{F_{A}^{2}+F_{C}^{2}+}{{F}_{D}}\] Since \[{{F}_{A}}={{F}_{C}}=\frac{k{{q}^{2}}}{{{a}^{2}}}\]and \[{{F}_{D}}=\frac{k{{q}^{2}}}{{{(a\sqrt{2})}^{2}}}\] \[{{F}_{net}}=\frac{\sqrt{2}k{{q}^{2}}}{{{a}^{2}}}+\frac{k{{q}^{2}}}{2{{a}^{2}}}=\frac{k{{q}^{2}}}{{{a}^{2}}}\left( \sqrt{2}+\frac{1}{2} \right)\]\[=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\left( \frac{1+2\sqrt{2}}{2} \right)\]