Question

I am trying to give the magnitude and direction of the electric field produced at point A. However that angle seems small. Where did I go wrong?

Answer 1

$Calculated\\ E_{A_1}=1.25\times10^6\,\,N/C\\ E_{A_2}=1.66\times10^6\,\,N/C\\\\ Now,\,\,E_{A_1}\,\,will\,\,have\,\,two\,\,component\,\,that\,\,is\,\,E_{A_1}\cos30^{\circ} \,\,and\,\,E_{A_1}\sin30^{\circ}\\ in\,\,-ve\,X\,\,and+ve\,\,\,Y\,\,direction\,\,respectively.\\ In\,\,other\,\,hand\,\,E_{A_2}\,\,will\,\,have\,\,only\,\,one\,\,component\,\,E_{A_2} \,\,in\,\,+ve\,\,X\,\,direction.\\\\$

$E_{A_X}=E_{A_2}-E_{A_1}\cos 30^{\circ}$
$\\ =\left (1.66\times10^6-1.25\times10^6\cos 30^{\circ} \right )\,\,N/C\\ =2.74\times10^6\,\,N/C$

$E_{A_Y}=E_{A_1}\sin 30^{\circ}$
$\\ =1.25\times10^6\sin30^{\circ}\,\,N/C\\ =0.625\times10^6\,\,N/C$

$\therefore E_{A}=\sqrt{E_{A_X}^2+E_{A_Y}^2}$
  $\\ =\sqrt{(2.75\times10^6)^2+(0.625\times10^6)^2}\,\,N/C\\ =2.82\times10^6\,\,N/C$

$Suppose\,\,it\,\,makes\,\,an\,\,angle\,\,\theta \,\,with\,\,positive\,\,X\,\,direction\,\,then,\\\\ \cos\theta =\frac{E_{A_X}}{E_A}\\\\ \Rightarrow \cos\theta=\frac{2.75\times10^6}{2.82\times10^6}\\\\ \Rightarrow \,\,\theta=\cos^{-1}\left (\frac {2.75}{2.82} \right )\\\\ \Rightarrow \,\,\theta =12.8^{\circ}$

$.\\\\\\\ [For\,\,any\,\,doubt\,\,please\,\,leave\,\,a\,\,comment]$