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 :
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 \[\because d<<D\] \[=\frac{2KqD}{{{D}^{3}}}\left[ 1-\frac{3}{2}\frac{{{d}^{2}}}{{{D}^{2}}}-\left( 1-\frac{3\times 4{{d}^{2}}}{2{{D}^{2}}} \right) \right]\] \[=\frac{2KqD}{{{D}^{3}}}\left[ \frac{12}{2}\frac{{{d}^{2}}}{{{D}^{2}}}-\frac{3}{2}\frac{{{d}^{2}}}{{{D}^{2}}} \right]\] \[=\frac{9Kq{{d}^{2}}}{{{D}^{4}}}\]You need to login to perform this action.
You will be redirected in
3 sec