question_answer

Two point charges \[{{q}_{1}}\,(\sqrt{10}\,\mu C)\]and \[{{q}_{2}}(-25\,\,\mu C)\] are placed on the x-axis at \[x=1\,m\] and \[x=4\,m\]respectively. The electric field (in V/m) at a point \[y=3\]m on y-axis is, [take\[\frac{1}{4\pi {{\in }_{0}}}\,\,=\,\,9\,\times {{10}^{9}}\,N{{m}^{2}}{{C}^{-}}^{2}\]] [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

A) \[(-81\widehat{i}\,\,+\,\,81\widehat{j})\,\,\times \,{{10}^{2}}\]                

B) \[(81\widehat{i}-81\widehat{j})\times {{10}^{2}}\]

C) \[(63\widehat{i}\,\,-\,\,27\widehat{j})\,\,\times \,{{10}^{2}}\]

D) \[(-63\widehat{i}+27\widehat{j})\times {{10}^{2}}\]

Correct Answer: C

Solution :

  Electric field due to \[\sqrt{10}\,\mu C\] \[\left| {{E}_{1}} \right|\,\,=\,\,\frac{k\,\times \,\sqrt{10}\,\times \,{{10}^{-6}}}{10}\,\,=\,\,\frac{9\,\times \,{{10}^{3}}}{\sqrt{10}}\] \[{{\overrightarrow{E}}_{1\,\,}}=\,\,{{E}_{1}}\sin \theta \,i\,\,+\,\,\,{{E}_{1}}\cos \theta j\] \[{{\overrightarrow{E}}_{1\,\,}}=\,\,\frac{9\,\times \,{{10}^{3}}}{\sqrt{10}}\,\left[ \frac{-3}{\sqrt{10}}\,i\,+\,\,\frac{1}{\sqrt{10}}\,j \right]\,\,=\,\,9\,\times \,{{10}^{2}}\] \[~\left( -3i+j \right)\] Electric field due to \[-25\,\mu C\] charge \[\left| {{E}_{2}} \right|\,\,\,=\,\,\frac{K\,\,\times \,\,25\,\,\times {{10}^{-6}}}{25}\,=\,9\,\times \,{{10}^{3}}\,\] \[{{E}_{2}}={{E}_{2}}sin\theta i-{{E}_{2}}cos\theta j\] \[{{\overrightarrow{E}}_{2}}=9\,\times \,{{10}^{3}}\,\left[ \frac{4}{5}i\,\,-\,\,\frac{3}{5}\,j \right]\] \[=\,\,18\,\times \,{{10}^{2}}\,[4i\,-\,3j]\] \[\overrightarrow{{{E}_{2}}}={{\overrightarrow{E}}_{1}}+{{\overrightarrow{E}}_{2}}={{10}_{2}}[63i-27j]\] Option (c)