question_answer

Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance will experience maximum coulomb force when   [MP PET 2002]

A)            \[x=\frac{d}{\sqrt{2}}\]

B)                                      \[x=\frac{d}{2}\]

C)            \[x=\frac{d}{2\sqrt{2}}\] 

D)            \[x=\frac{d}{2\sqrt{3}}\]

Correct Answer: C

Solution :

                   Suppose third charge is similar to Q and it is q So net force on it \[{{F}_{net}}=\text{ 2}F\text{cos}\theta \] Where \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Qq}{\left( {{x}^{2}}+\frac{{{d}^{2}}}{4} \right)}\]and \[\cos \theta =\frac{x}{\sqrt{{{x}^{2}}+\frac{{{d}^{2}}}{4}}}\] \\[{{F}_{net}}=2\times \frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Qq}{\left( {{x}^{2}}+\frac{{{d}^{2}}}{4} \right)}\times \frac{x}{{{\left( {{x}^{2}}+\frac{{{d}^{2}}}{4} \right)}^{1/2}}}\]                    \[=\frac{2Qqx}{4\pi {{\varepsilon }_{0}}{{\left( {{x}^{2}}+\frac{{{d}^{2}}}{4} \right)}^{3/2}}}\] For \[{{F}_{net}}\] to be maximum \[\frac{d{{F}_{net}}}{dx}=0\] i.e. \[\frac{d}{dx}\left[ \frac{2Qqx}{4\pi {{\varepsilon }_{0}}{{\left( {{x}^{2}}+\frac{{{d}^{2}}}{4} \right)}^{3/2}}} \right]=0\] or  \[\left[ {{\left( {{x}^{2}}+\frac{{{d}^{2}}}{4} \right)}^{-3/2}}-3{{x}^{2}}{{\left( {{x}^{2}}+\frac{{{d}^{2}}}{4} \right)}^{-5/2}} \right]=0\] i.e.  \[x=\pm \frac{d}{2\sqrt{2}}\]