Q.Two charges $2\ \mu\text{C}$ and $-2\ \mu\text{C}$ are placed at points A and B $6\ \text{cm}$ apart.

Concept: Electric Potential — The potential at any point due to a dipole is the algebraic sum of potentials from each charge. An equipotential surface is where this sum is constant.

Reasoning:

1. For two equal and opposite charges separated by $6\ \text{cm}$, the net potential at any point is $V = \frac{kq}{r_1} + \frac{k(-q)}{r_2}$.
2. The perpendicular bisector of the line joining A and B has $r_1 = r_2$ for every point on it, so $V = 0$ everywhere on that plane.
3. Hence, the entire plane perpendicular to AB through its midpoint is an equipotential surface (potential zero).
4. The electric field at any point on this plane is perpendicular to the surface (since field lines are always normal to equipotential surfaces). For a dipole, the field on the equatorial plane points from the positive charge toward the negative charge — i.e., parallel to AB, from A to B.

Concept and Intuition

When you have two equal and opposite charges separated by a distance, you have an electric dipole. The key property of a dipole is that the potential due to the two charges cancels exactly at all points that are equidistant from the two charges. Why? Because potential is a scalar quantity — it adds algebraically. For a point charge, $V = \frac{kq}{r}$. If you are at a point that is the same distance $r$ from $+q$ and $-q$, the potentials are $+\frac{kq}{r}$ and $-\frac{kq}{r}$, which sum to zero.

The set of all points equidistant from A and B is the perpendicular bisector plane of the line AB. This is the natural equipotential surface for a dipole — and it's the only one you can identify without doing heavy calculation.

For part (b), recall that the electric field is always perpendicular to an equipotential surface. So the field direction on this plane must be perpendicular to the plane itself. But is it exactly perpendicular? Yes — and the direction is from the positive charge toward the negative charge, i.e., along the dipole axis.

Step-by-step solution

1. Identify the nature of the charge system.

We have $q_1 = +2\ \mu\text{C}$ at A and $q_2 = -2\ \mu\text{C}$ at B, separated by $d = 6\ \text{cm}$. This is a pure electric dipole (equal magnitude, opposite sign). The midpoint O of AB is the centre of the dipole.

2. Condition for an equipotential surface.

An equipotential surface is a surface on which the electric potential $V$ is constant everywhere. For two point charges, the potential at any point P is

where $r_1$ and $r_2$ are distances from P to A and B respectively.

3. Find where the potential is constant (and simple).

The simplest constant value is $V = 0$. Set

So $r_1 = r_2$. Every point equidistant from A and B has zero potential.

4. Describe the surface geometrically.

The set of points equidistant from two fixed points A and B is the plane that is perpendicular to AB and passes through its midpoint O. This is the perpendicular bisector plane of AB.

Since A and B are 6 cm apart, this plane is located 3 cm from each charge, perpendicular to the line joining them.

5. Determine the direction of the electric field on this surface.

The electric field $\vec{E}$ is always perpendicular to an equipotential surface. Since our surface is a plane, $\vec{E}$ must be perpendicular to that plane. But perpendicular to the plane means parallel to the line AB (the normal to the plane).

Now, what is the sense of the field? On the perpendicular bisector of a dipole, the field points from the positive charge toward the negative charge — that is, from A to B. You can verify this by considering a test point just above the midpoint: the field due to $+q$ points away from A, and due to $-q$ points toward B; their horizontal components add, and vertical components cancel, giving a net field along AB toward the negative charge.

6. Summarise the answer.

1. The equipotential surface is the perpendicular bisector plane of AB (the plane through the midpoint O, perpendicular to AB).
2. The electric field at every point on this surface is parallel to AB, directed from the positive charge ($2\ \mu\text{C}$ at A) toward the negative charge ($-2\ \mu\text{C}$ at B).

Method: Identifying an Equipotential Surface by Symmetry

This method finds an equipotential surface (and the field direction on it) for a set of point charges by locating the symmetry that makes the total potential constant.

Steps

Step 1: Write the general potential at a point due to all the charges

For a point P at distances $r_1, r_2, \dots$ from charges $q_1, q_2, \dots$, potential adds algebraically, since $V$ is a scalar:

Step 2: Look for a geometric locus that makes this sum constant

Try the simplest symmetric locus available for the configuration. For two equal-and-opposite charges, points equidistant from both give $r_1=r_2$, so the $+q/r_1$ and $-q/r_2$ terms cancel and $V=0$ everywhere on that locus. For two equal like charges, the same equidistant locus instead gives a fixed nonzero $V$. You only need one geometrically describable surface, not every equipotential surface of the system.

Step 3: Get the field direction from the surface, not by re-deriving it from scratch

Once a surface is identified, use the general rule that $\vec{E}$ is always perpendicular to an equipotential surface and points from higher to lower potential. Find the normal direction at any point on the surface, then fix the sense (toward the lower-potential side) from the sign of the nearby charges.

Step 4: State the surface and the field direction together

Combine Steps 2–3 into the final description: the shape/location of the surface, and the direction (not the magnitude) of the field on it.