Two point charges +q and -q are held fixed at | vedclass

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(C) The charges $+q$ and $-q$ are placed at $(-d, 0)$ and $(d, 0)$ respectively.For option $(a)$: On the $x$-axis,the electric field direction changes

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electric field at all points on the $y$-axis is along the $x$-axis

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(C) The charges $+q$ and $-q$ are placed at $(-d, 0)$ and $(d, 0)$ respectively.For option $(a)$: On the $x$-axis,the electric field direction changes depending on whether the point is between the charges or outside them. Thus,it is not the same everywhere.For option $(b)$: The potential at the origin $(0, 0)$ is $V = k(q/d) + k(-q/d) = 0$. The potential at $\infty$ is also $0$. Thus,the work done $W = q(V_{origin} - V_{\infty}) = 0$. So,no work is required.For option $(c)$: At any point $(0, y)$ on the $y$-axis,the electric field due to $+q$ points away and the field due to $-q$ points towards the charge. By symmetry,the vertical components cancel out,and the resultant electric field is directed along the negative $x$-axis. Thus,the field is along the $x$-axis.For option $(d)$: The dipole moment vector $\vec{p}$ is directed from $-q$ to $+q$,which is from $(d, 0)$ to $(-d, 0)$,i.e.,along the negative $x$-axis. The magnitude is $2qd$.

Two point charges $+q$ and $-q$ are held fixed at $(-d, 0)$ and $(d, 0)$ respectively of an $x-y$ coordinate system. Then:

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