Animated Solution for Physics - Electrostatics: Two point charges $+q$ and $-q$ are held fixed at $(-d,0)$ and $(d,0)$ respectively of a $x$-$y$ co-ordinate system. Then

The problem presents us with a classic electrostatic setup: an electric dipole. We have a positive charge $+q$ located at $(-d,0)$ and a negative charge $-q$ located at $(d,0)$. Our goal is to evaluate four different statements regarding the electric field, potential, and dipole moment of this configuration. Let's break them down one by one.

The first statement claims that the electric field $E$ at all points on the $x$-axis has the same direction. Let's test this hypothesis.

Imagine placing a positive test charge anywhere between the two charges (i.e., $-d<x<d$). The positive charge $+q$ will repel it towards the right (positive $x$-direction), and the negative charge $-q$ will attract it towards the right. Thus, the net electric field in this region points strictly in the positive $x$-direction.

Now, let's move our test charge outside this region, say to the right of the negative charge ($x>d$). Here, the test charge is attracted by the closer negative charge (towards the left) and repelled by the farther positive charge (towards the right). Because the negative charge is closer, its attractive pull dominates. Therefore, the net electric field in this region points in the negative $x$-direction.

Since the direction of the electric field flips depending on where you are on the $x$-axis, the first statement is incorrect.

The second statement suggests that work has to be done to bring a test charge from infinity to the origin. To verify this, we need to calculate the electric potential at the origin.

The electric potential $V$ is a scalar quantity. At the origin $(0,0)$, the potential is the sum of the potentials due to each charge:

The potential at infinity is also zero ($V_{\infty }=0$). The work done $W$ by an external agent in moving a test charge $q_0$ is given by:

Since the potential difference is zero, no work is required. Thus, the second statement is also incorrect.

The third statement claims that the electric field at all points on the $y$-axis is along the $x$-axis. Let's pick an arbitrary point $P(0,y)$ on the $y$-axis.

At point $P$, the electric field vector ${\vec{E}}_{+q}$ due to the positive charge points radially outward (up and to the right). The electric field vector ${\vec{E}}_{-q}$ due to the negative charge points radially inward (down and to the right).

Because point $P$ is equidistant from both charges, the magnitudes of these two field vectors are identical. When we resolve these vectors into their components, a beautiful symmetry emerges. The vertical ($y$) components are equal and opposite, perfectly canceling each other out. The horizontal ($x$) components, however, both point in the positive $x$-direction and add up.

Therefore, the net electric field ${\vec{E}}_{\text{net}}$ at any point on the $y$-axis is strictly parallel to the $x$-axis. This makes the third statement absolutely correct!

Finally, let's look at the fourth statement, which claims the dipole moment is $2qd$ along the $x$-axis.

By standard convention in physics, the electric dipole moment vector $\vec{p}$ is directed from the negative charge to the positive charge. In our coordinate system, the negative charge is at $(d,0)$ and the positive charge is at $(-d,0)$.

This means the dipole moment vector points from right to left, which is along the negative $x$-axis (or $-\hat{i}$ direction). While the magnitude is indeed $q\times 2d=2qd$, the direction stated in the option is wrong. Thus, the fourth statement is incorrect.