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Electric field at P due to charge at A is,
EA = kq/(a2/2) along AP
Electric field at P due to charge at B is,
EB = k(2q)/(a2/2) along BP
Electric field at P due to charge at C is,
EC = k(3q)/(a2/2) along CP
Electric field at P due to charge at D is,
ED = k(4q)/(a2/2) along DP
Electric field EA and EC are in opposite directions, so, their net field is,
ECA = k(3q)/(a2/2) - kq/(a2/2) = k(2q)/(a2/2) along CA
Electric field ED and EB are in opposite directions, so, their net field is,
EDB = k(4q)/(a2/2) - k(2q)/(a2/2) = k(2q)/(a2/2) along DB
Since, ECA and EDB are equal in magnitude and they are perpendicular to each other, so, their resultant E will bisect angle APB.
Or we can say it is along CB
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