# Find the direction of electric field at point P for the charge distribution as shown in figure

$(a)\;towards\;x\;axis\qquad(b)\;towards\;y\;axis\qquad(c)\;towards\;-x\;axis\qquad(d)\;towards\;-y\;axis$

Answer : (a) towards x axis
Explanation : Field due to upper charge Q at P
$=E_{x}\hat{i} + E_{y} \hat{j}\quad\;(E_{x},E_{y} > 0)$
Field due to lower charge Q at P
$=E_{x_{1}}\hat{i} + E_{y_{1}} \hat{j}\quad\;(E_{x_{1}},E_{y_{1}} > 0)$
By symmetry $\;E_{x}=E_{x_{1}}$
$\;E_{y}=E_{y_{1}}$
$\overrightarrow{E}_{net}\;on\;p=(E_{x}+E_{x_{1}}) \hat{i}+(E_{y_{1}}-E_{y}) \hat{j}$
$i\;.\;e\;.\;E\;is\;along\;x\;axis$

answered Jan 29, 2014 by
edited Aug 11, 2014