Draw equipotential surfaces for an electric dipole.

Question

Q. Draw equipotential surfaces for an electric dipole.

CBSECBSE Class XII Board 2019Subjective· 1mImportance★★★★★

✓ Free question

Equipotential surfaces for an electric dipole are closed, three-dimensional surfaces that are perpendicular to electric field lines at every point. They are not concentric spheres but rather pear-shaped surfaces that bulge outward near the dipole axis and flatten near the equatorial plane. The equatorial plane itself is an equipotential surface at zero potential.

Understanding Electric Potential of a Dipole

Before drawing the surfaces, we need to understand what the potential looks like around a dipole. An electric dipole consists of two equal and opposite charges $+q$ and $-q$ separated by a small distance $2a$ .

The electric potential at any point P due to a dipole is given by:

V = \frac{1}{4\pi\epsilon_0} \cdot \frac{p\cos\theta}{r^2}

where $p = q \cdot 2a$ is the dipole moment, $r$ is the distance from the dipole centre, and $\theta$ is the angle measured from the dipole axis (the line joining the charges).

V(r,\theta) = \frac{1}{4\pi\epsilon_0} \cdot \frac{p\cos\theta}{r^2}

This formula tells us something crucial: the potential depends on both distance $r$ and direction $\theta$ . Unlike a single point charge where equipotentials are concentric spheres, here the surfaces are more interesting.

Step-by-Step Construction

1. Identify the equatorial plane ( $\theta = 90^\circ$ )

When $\theta = 90^\circ$ , $\cos\theta = 0$ , so $V = 0$ everywhere on this plane. The entire plane perpendicular to the dipole axis and passing through its centre is at zero potential. This is the simplest equipotential surface — an infinite plane.

2. Consider points along the dipole axis ( $\theta = 0^\circ$ and $\theta = 180^\circ$ )

Along the axis, $\cos\theta = \pm 1$ , so $|V|$ is maximum for a given $r$ . The potential is positive on the side of the positive charge and negative on the side of the negative charge. For a fixed potential value $V_0$ , the distance $r$ must be smaller here than at other angles — meaning the equipotential surface is pulled in closer to the dipole along the axis.

3. Trace a constant potential surface

For a given $V = V_0$ , the equation becomes:

r^2 = \frac{p\cos\theta}{4\pi\epsilon_0 V_0}

Since $r$ must be positive, $\cos\theta$ must have the same sign as $V_0$ . So:

For $V_0 > 0$ : only angles where $\cos\theta > 0$ ( $-90^\circ < \theta < 90^\circ$ ) are possible — the surface exists only on the positive charge side.
For $V_0 < 0$ : only angles where $\cos\theta < 0$ ( $90^\circ < \theta < 270^\circ$ ) — the surface exists only on the negative charge side.

This is a key difference from a point charge: equipotential surfaces of a dipole are not closed around the dipole centre — each surface encloses only one of the two charges.

4. Shape of the surfaces

As $\theta$ increases from $0^\circ$ to $90^\circ$ , $\cos\theta$ decreases from 1 to 0. To keep $V$ constant, $r$ must increase. So the surface bulges outward as we move away from the axis toward the equatorial plane. Near the equatorial plane, $r$ becomes very large — the surface stretches far out before curving back.

The resulting shape is pear-like or teardrop-shaped, with the narrow end near the charge and the broad end opening toward the equatorial plane.

Watch out

A common mistake is to draw equipotential surfaces of a dipole as concentric spheres or as ellipsoids. Neither is correct. The surfaces are not symmetric about the centre — they are asymmetric, bulging more on one side and flattening near the equator.

5. The complete picture

For a dipole, the equipotential surfaces are:

The equatorial plane ( $V = 0$ ) — an infinite plane
A family of closed, pear-shaped surfaces surrounding each charge separately
Surfaces with positive $V$ surround the positive charge
Surfaces with negative $V$ surround the negative charge
All surfaces are perpendicular to electric field lines at every intersection point

Tip

To visualise this: imagine a balloon that is tied near the positive charge. As you move away from the axis toward the equator, the balloon stretches outward. The surface never crosses the equatorial plane — it approaches it asymptotically. The same happens on the negative side, but inverted.

The Final Visual Description

If you were to draw this in a plane containing the dipole axis (say, the plane of the paper), the equipotential surfaces appear as closed curves around each charge. Near the charge, they are roughly circular. As they extend outward, they become elongated and flattened, approaching the equatorial line (which appears as a straight line in this 2D cross-section) but never crossing it.

In three dimensions, these curves rotate about the dipole axis to form the actual surfaces.

✓Final answer

The equipotential surfaces of an electric dipole consist of the equatorial plane at zero potential and pear-shaped closed surfaces surrounding each charge separately, with positive surfaces around the positive charge and negative surfaces around the negative charge.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set V11 markMCQ
$Q. The equipotential surfaces of an isolated point charge are :$
(a) Coaxial cylindrical surfaces
(b) Plane surfaces parallel to each other
(c) Concentric spherical surfaces centred at the charge
(d) Spherical surfaces but not centred on the charge
›Reveal solutionSolution

(c) Concentric spherical surfaces centred at the charge

✓Final answer
$(c) Concentric spherical surfaces centred at the charge$

The potential due to an isolated point charge is $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ , which depends only on the distance $r$ from the charge. Hence all points at the same $r$ have the same potential, forming spheres centred on the charge. Equipotential surfaces are therefore concentric spheres, and they are everywhere perpendicular to the radial field lines.
CBSE 2026Set ANNUAL1 markMCQ
$Q. The angle between equipotential surface and electric field lines is$
(a) 180°
(b) 90°
(c) 45°
(d) 0°
›Reveal solutionSolution

Electric field lines are normal to equipotential surfaces everywhere, so the angle between them is always 90 degrees.

An equipotential surface is one on which potential V is constant, so no work is done in moving a charge along it (W = q(V2-V1) = 0). If the field had any component along the surface, that component would do work in moving a charge along the surface, contradicting W=0. Hence the electric field can only be perpendicular to the equipotential surface at every point.

✓Final answer
$(b) 90 degrees.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Equipotential surface at a large distance from a collection of charge whose total sum is not zero are -$
(a) ellipsoids
(b) planes
(c) paraboloids
(d) spheres
›Reveal solutionSolution

Far away, any localized charge distribution with nonzero total charge looks like a point charge.

At a large distance from a collection of charges whose total (net) charge is not zero, the finer details of the actual charge distribution become negligible, and the distribution behaves effectively like a single point charge equal to the total charge, located roughly at its "centre". The equipotential surfaces of a point charge are concentric spheres centred on it. So at large distances, the equipotential surfaces of any charge collection with nonzero net charge approach spheres.

✓Final answer
$(d) spheres.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following statement is/are true for equipotential surface ?$
(a) The potential is different for different equipotential surfaces.
(b) Electric field must always be normal to equipotential surface.
(c) Work done to move a charge between any two points is zero.
(d) All of the above
›Reveal solutionSolution
All three listed statements about equipotential surfaces are individually true, so the correct choice is that all of the above hold.

Working

Different equipotential surfaces correspond to different potential values. By definition, each equipotential surface is the locus of points at one particular potential; a different surface (further/closer to the source) is at a different potential value.

The electric field is always normal to an equipotential surface. If $\vec E$ had any component along the surface, it would do work moving a charge along that surface, changing its potential energy -- contradicting the surface being at constant potential. So $\vec E$ must be entirely perpendicular to it.

Work done to move a charge between any two points on the same equipotential surface is zero. Since $W=q(V_A-V_B)$ and $V_A=V_B$ on the same surface, $W=0$ .

All three statements are correct descriptions of equipotential surfaces.

✓Final answer
$The correct option is (d): all of the above statements about equipotential surfaces are true$
CBSE 2025Set X11 markMCQ
$Q. Equipotential surfaces for an isolated point charge are __________ in shape.$
(a) spherical
(b) planar
(c) cylindrical
(d) conical
›Reveal solutionSolution

(a) spherical. The potential of an isolated point charge is $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ , which has the same value at all points a distance $r$ from the charge. The locus of cons

✓Final answer
$V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$
CBSE 2025Set ANNUAL1 mark
$Q. Can two equipotential surfaces cross each other?$
›Reveal solutionSolution

If two equipotential surfaces intersected, the point of intersection would have to be at two different potentials at once, which is impossible; it would also mean the electric field has two directions at that point, which is equally impossible.

Each equipotential surface is, by definition, a surface on which the electric potential has one single fixed value everywhere on it.

If two equipotential surfaces at different potentials (say V1 and V2, with V1 not equal to V2) were to cross at some point P, then P would simultaneously have potential V1 (being on the first surface) and potential V2 (being on the second) - a contradiction, since a point can have only one value of potential.

Additionally, the electric field at any point is always perpendicular to the equipotential surface through that point. If two surfaces crossed, the field at the crossing point would have to be perpendicular to both surfaces at once, i.e. point in two different directions simultaneously, which is physically impossible.

✓Final answer
$No, two equipotential surfaces can never intersect or cross each other.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The angle between the electric lines of force and the equipotential surface is$
(a) 0°
(b) 45°
(c) 90°
(d) 180°
›Reveal solutionSolution

Electric field lines are always perpendicular to equipotential surfaces.

An equipotential surface is one on which the potential V is the same everywhere, so moving along it does no work: $dV = -\vec{E}\cdot d\vec{l} = 0$ for any displacement $d\vec{l}$ on the surface. Since $\vec E$ is generally non-zero, this requires $\vec E \perp d\vec l$ , i.e., the field (and hence the line of force, which is tangent to $\vec E$ ) must be perpendicular to the equipotential surface at every point.

✓Final answer
$The angle is 90° — option (c).$
CBSE 2025Set ANNUAL1 mark
$Q. Figure below shows the equipotential surfaces for two charges q₁ and q₂. Identify the nature of charges q₁ and q₂.$
›Reveal solutionSolution

Only like charges give equipotential surfaces that merge into a single closed peanut/dumbbell-shaped surface enclosing both charges - unlike charges never do.

For unlike (opposite-sign) charges, e.g. a dipole, the equipotential value $V=0$ occurs on the whole perpendicular-bisector plane between them, which is an open, unbounded surface, not a closed loop. Any closed equipotential surface for a non-zero potential value stays wrapped around one charge only - it can never merge with the loop of the other charge, because the sign of the potential contributed by the two charges is opposite.

For like charges (same sign), the potential everywhere between and around them has the same sign, and never passes through zero at a finite distance. Close to each charge the equipotential surfaces are nearly circular loops around that charge alone. As the chosen potential value gets smaller (surface drawn farther away), the two individual loops widen and eventually touch and merge into a single closed surface enclosing both charges - exactly the overlapping peanut/dumbbell shape described, which at even larger distance becomes a single large circle (as though it were one point charge $q_1+q_2$ ).

The given figure shows exactly this merged, closed peanut-shaped outer boundary, which is only possible for two charges of the same sign.

✓Final answer
$q_1$
CBSE 2024Set ANNUAL1 mark
$Q. Draw an equipotential surface for a positive charge (q > 0).$
›Reveal solutionSolution

For an isolated positive point charge, potential depends only on distance from the charge, so every surface of constant distance (a sphere centred on the charge) is an equipotential surface.

The potential due to a point charge q at a distance r is:

$V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$

Since V depends only on r, every point at the same distance r from the charge is at the same potential. So the equipotential surfaces are concentric spheres centred on the charge.

To sketch: draw the positive charge q as a dot, then draw several concentric circles (representing spheres in 3D) around it, spaced progressively farther apart as r increases (since V falls off as 1/r, equal potential steps correspond to increasingly widely-spaced spheres). Each circle is labelled as an equipotential surface, and by definition the electric field lines (radially outward from q) are everywhere perpendicular to these spherical surfaces.

✓Final answer
$Concentric spheres centred on the charge q, spaced farther apart with increasing distance; field lines are perpendicular to them.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. A charge of 100 μC is placed at the centre of a circle of radius 5 m. Work done in moving a unit positive charge around the circumference of the circle once is$
(a) 500 J
(b) 20 J
(c) 0.05 J
(d) 0 J.
›Reveal solutionSolution

Work done by a conservative (electrostatic) field around any closed path is always zero — the charge returns to its starting point, so its potential energy is unchanged.

The charge of 100 μC at the centre sets up a radially symmetric electric field. As the unit positive charge is carried once around the circumference, at every point on the circle the potential due to the central charge is the same, $V = kq/r$ (constant, since $r$ = 5 m throughout).

Work done in moving a charge $q_0$ between two points equals $W = q_0(V_i - V_f)$ . Here the path starts and ends at the SAME point on the circle, so $V_i = V_f$ , giving $W = 0$ regardless of the path taken (straight line, circle, or any curve) — this is the defining property of a conservative field: the work done around any closed loop is zero.

✓Final answer
$Work done = 0 J. Choice (d).$
CBSE 2022Set ANNUAL1 mark
$Q. What is an equipotential surface?$
›Reveal solutionSolution

A surface where potential is constant everywhere; no work is done moving a charge along it.

An equipotential surface is defined as a surface at every point of which the electric potential is the same. Since the potential does not change as we move from one point to another on such a surface, no work is done in moving a test charge along it ( $W = q(V_A - V_B) = 0$ since $V_A = V_B$ ). This also means the electric field is always perpendicular to an equipotential surface at every point — if it had a component along the surface, work would be done moving a charge along that component, contradicting $V$ being constant. For a single point charge, equipotential surfaces are concentric spheres centred on the charge.

✓Final answer
$An equipotential surface is one on which the electric potential is the same at every point; the electric field is always normal to it, and no work is done moving a charge along it.$
CBSE 2020Set ANNUAL1 mark
$Q. The angle between electric line of force and equipotential surface is ____. (Fill in the blank.)$
›Reveal solutionSolution

Electric field lines always cut equipotential surfaces at right angles (90°).

An equipotential surface is one on which every point is at the same electric potential; no work is done in moving a charge along such a surface. Since the electric field $\vec{E} = -\nabla V$ points in the direction of the steepest decrease of potential, and there is no change in potential along an equipotential surface, the field must have zero component along the surface — i.e., it can only point perpendicular to it. If the field had any component along the surface, moving a charge along that component would require or release work, contradicting the surface being equipotential.

✓Final answer
$90° — the angle between an electric line of force and an equipotential surface is always a right angle.$

Q.Draw equipotential surfaces for an electric dipole.

Concept understanding — Equipotential Surfaces

Understanding Electric Potential of a Dipole

Step-by-Step Construction

The Final Visual Description

Q.Draw equipotential surfaces for an electric dipole.

Understanding Electric Potential of a Dipole

Step-by-Step Construction

The Final Visual Description

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