question_answer

Four point charges \[q,+q,+q\]and -q are placed on y-axis at \[y=2d,y=d,y=+d\] and \[y=+2d,\]respectively. The magnitude of the electric field E at a point on the x-axis at \[x=D,\] with \[D>>d,\]will behave as :- [JEE Main 9-4-2019 Afternoon]

A) \[E\propto \frac{1}{D}\]

B) \[E\propto \frac{1}{{{D}^{3}}}\]

C) \[E\propto \frac{1}{{{D}^{2}}}\]

D) \[E\propto \frac{1}{{{D}^{4}}}\]

Correct Answer: D

Solution :

[d]

Electric field at \[p=2{{E}_{1}}\cos {{\theta }_{1}}-2{{E}_{1}}\cos {{\theta }_{2}}\]

\[=\frac{2{{K}_{q}}}{({{d}^{2}}+{{D}^{2}})}\times \frac{D}{{{({{d}^{2}}+{{D}^{2}})}^{1/2}}}-\frac{2{{K}_{q}}}{[{{(2d)}^{2}}+{{D}^{2}}]}\times \frac{D}{{{[{{(2d)}^{2}}+{{D}^{2}}]}^{1/2}}}\]\[=2KqD\left[ {{({{d}^{2}}+{{D}^{2}})}^{-3/2}}-{{(4{{d}^{2}}+{{D}^{2}})}^{-3/2}} \right]\]

\[=\frac{2KqD}{{{D}^{3}}}\left[ {{\left( 1+\frac{{{d}^{2}}}{{{D}^{2}}} \right)}^{-3/2}}-{{\left( 1+\frac{4{{d}^{2}}}{{{D}^{2}}} \right)}^{-3/2}} \right]\]

Applying binomial approximation \[\because d<<D\]

\[=\frac{2KqD}{{{D}^{3}}}\left[ 1-\frac{3}{2}\frac{{{d}^{2}}}{{{D}^{2}}}-\left( 1-\frac{3\times 4{{d}^{2}}}{2{{D}^{2}}} \right) \right]\]

\[=\frac{2KqD}{{{D}^{3}}}\left[ \frac{12}{2}\frac{{{d}^{2}}}{{{D}^{2}}}-\frac{3}{2}\frac{{{d}^{2}}}{{{D}^{2}}} \right]\]

\[=\frac{9Kq{{d}^{2}}}{{{D}^{4}}}\]