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

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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 $\;E_{x}=E_{x_{1}}$

$\;E_{y}=E_{y_{1}}$

$\overrightarrow{E}_{net}\;on\;p=(E_{x}+E_{x_{1}}) \hat{i}+(E_{y_{1}}-E_{y}) \hat{j}$

$i\;.\;e\;.\;E\;is\;along\;x\;axis$

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