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

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(A) The electric dipole is formed by charges $+q$ at $(-d, 0)$ and $-q$ at $(d, 0)$.$1$. The dipole moment $\overrightarrow{p}$ is directed from $-q$

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$E$ at all points on the $Y$-axis is along $\hat{i}$

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(A) The electric dipole is formed by charges $+q$ at $(-d, 0)$ and $-q$ at $(d, 0)$.$1$. The dipole moment $\overrightarrow{p}$ is directed from $-q$ to $+q$,which is along the negative $X$-axis ($-\hat{i}$ direction).$2$. For any point on the $Y$-axis (equatorial plane),the electric field $\overrightarrow{E}$ is antiparallel to the dipole moment $\overrightarrow{p}$. Since $\overrightarrow{p}$ is along $-\hat{i}$,$\overrightarrow{E}$ at all points on the $Y$-axis is along $+\hat{i}$. Thus,option $(a)$ is correct.$3$. On the $X$-axis,the electric field direction changes depending on whether the point is between the charges or outside them. Thus,option $(b)$ is incorrect.$4$. The dipole moment is $\overrightarrow{p} = q(2d)$ directed from $-q$ to $+q$,which is along $-\hat{i}$. Thus,option $(c)$ is incorrect.$5$. The electric potential $V$ at the origin $(0, 0)$ is $V = k(+q)/d + k(-q)/d = 0$. The work done in bringing a test charge $q_0$ from infinity to the origin is $W = q_0 \Delta V = q_0(V_{origin} - V_{\infty}) = q_0(0 - 0) = 0$. Thus,option $(d)$ is incorrect.

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

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