question_answer

Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance from centre will experience maximum coulomb force, when

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

B) \[x=d/2\]

C)  \[x=d/2\sqrt{2}\]           

D)         \[x=d/2\sqrt{3}\]

Correct Answer: C

Solution :

From Coulombs law, the force of attraction/repulsion acting between two stationary point charges\[({{q}_{1}},\,\,{{q}_{2}})\]separated by a distance\[(r)\] is \[F=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}\,\frac{{{q}_{1}}\,{{q}_{2}}}{{{r}^{2}}}\] Taking the net force, we have \[{{F}_{net}}=2F\,\,\cos \theta =2\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qqx}{{{\left( {{x}^{2}}+\frac{{{d}^{2}}}{4} \right)}^{3/2}}} \right]\] For maximum,                 \[\frac{d\,\,{{F}_{net}}}{dx}=0\] \[\therefore \] \[{{\left( {{x}^{2}}+\frac{{{d}^{2}}}{4} \right)}^{3/2}}-\frac{3}{2}x{{\left[ {{x}^{2}}+\frac{{{d}^{2}}}{4} \right]}^{1/2}}(2x)=0\] \[\therefore \] \[{{\left( {{x}^{2}}+\frac{{{d}^{2}}}{4} \right)}^{1/2}}\left( {{x}^{2}}+\frac{{{d}^{2}}}{4}-3{{x}^{2}} \right)=0\] or            \[2{{x}^{2}}=\frac{{{d}^{2}}}{4}\] \[\Rightarrow \]               \[x=\frac{d}{2\sqrt{2}}\]