Two charges +q and -q are placed at a distanc | vedclass

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Question

(A) The electric field due to a positive charge $+q$ at point $P$ is directed away from the charge, represented by vector $E_1$.The electric field due

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along vector $A$

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(A) The electric field due to a positive charge $+q$ at point $P$ is directed away from the charge, represented by vector $E_1$.The electric field due to a negative charge $-q$ at point $P$ is directed towards the charge, represented by vector $E_2$.Since the point $P$ is on the perpendicular bisector, the magnitudes of the electric fields are equal, i.e., $|E_1| = |E_2|$.When resolving these vectors into components, the vertical components ($E_1 \sin 	heta$ and $E_2 \sin 	heta$) cancel each other out because they are in opposite directions.The horizontal components ($E_1 \cos 	heta$ and $E_2 \cos 	heta$) add up in the direction parallel to the line joining the charges, pointing from the positive charge towards the negative charge.Looking at the provided figure, vector $A$ is parallel to the line joining the charges and points from $+q$ to $-q$.Therefore, the resultant electric field is along vector $A$.

Two charges $+q$ and $-q$ are placed at a distance $b$ apart as shown in the figure. The electric field at a point $P$ on the perpendicular bisector as shown is

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