The work done to move a charge along an equipotential from A to B (a) cannot be defined as -∫AB vecE· dvecl. (

NCERT Exemplar · Q9

Q. The work done to move a charge along an equipotential from A to B

(a) cannot be defined as

-\int_A^B \vec{E}\cdot d\vec{l}

(b) must be defined as

-\int_A^B \vec{E}\cdot d\vec{l}

(d) can have a non-zero value.

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Concept understanding — Electric Potential

Electric Potential

The Idea

When you lift a book onto a shelf you do work against gravity, and the book gains gravitational potential energy that depends on where it sits. Electric charges behave the same way in an electric field. Electric potential is the electrical analogue of "height" in a gravitational field: it tells you the potential energy a unit charge would have at a given point.

Formally, the electric potential at a point is the work done in bringing a unit positive charge from infinity to that point, slowly (without giving it kinetic energy), against the electric field.

V = \frac{W}{q_0}

Here $W$ is the work done to move a small test charge $q_0$ from infinity to the point. Because both $W$ and $q_0$ are scalars, electric potential is a scalar quantity — it has magnitude but no direction. Its SI unit is the volt:

1\ \text{V} = 1\ \text{J/C}

Potential Due to a Point Charge

For a single point charge $q$ , the potential at a distance $r$ from it is

V = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{r}

Notice it falls off as $1/r$ , whereas the electric field of a point charge falls off as $1/r^2$ . The potential is taken as zero at infinity, our chosen reference. A positive charge makes the potential around it positive; a negative charge makes it negative.

Potential Difference

Usually we care about the potential difference between two points $A$ and $B$ :

V_A - V_B = \frac{W_{B\to A}}{q_0}

the work per unit charge needed to move a charge from $B$ to $A$ . This is the quantity a voltmeter reads and the "voltage" that drives current in a circuit.

Relation Between Field and Potential

Field and potential are two views of the same thing. The electric field is the negative rate of change of potential with distance:

E = -\frac{dV}{dr}

The minus sign says the field points from high potential toward low potential — a positive charge, left free, rolls "downhill" in potential. Where the potential changes steeply, the field is strong.

Superposition

Because potential is a scalar, the potential due to several charges is just the algebraic sum of their individual potentials — no vectors, no angles:

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots\right)

This makes potential far easier to compute than the field, which needs vector addition. Once you have $V$ everywhere, you can get the field by differentiating.

Equipotential Surfaces

An equipotential surface is a surface on which the potential has the same value everywhere.

No work is done in moving a charge along an equipotential surface (since $\Delta V = 0$ ).
The electric field is always perpendicular to an equipotential surface. …

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The work done moving a charge from A to B is, by definition, always $W=-q\int_A^B \vec E\cdot d\vec l$ — this formula is universally valid, so option (b) is correct — and because $\vec E$ is perpendicular to $d\vec l$ everywhere on an equipotential, that integral evaluates to exactly zero, so option (c) is also correct. Both (b) and (c) hold; (a) and (d) are false.

The general definition of work, first

For any path from A to B in an electrostatic field, the work done by the field on a charge $q$ is

W = q\int_A^B \vec E\cdot d\vec l,

or, equivalently, the work required by an external agent (against the field) is $W_{\text{ext}}=-q\int_A^B\vec E\cdot d\vec l = q(V_B-V_A)$ . This line-integral formula for work is not a special rule that applies only away from equipotentials — it is the general definition of electrostatic work along any path, equipotential or not. So option (b), "must be defined as $-\int_A^B \vec E\cdot d\vec l$ " (up to the constant $q$ ), is simply true — this is always how the work is defined, and option (a), which claims the opposite (that it cannot be defined this way), is therefore false.

Now apply it to a path along an equipotential

An equipotential surface, by definition, is a surface where $V$ has the same value at every point. Two known facts about equipotentials:

$V_A = V_B$ if A and B lie on the same equipotential.
The electric field $\vec E$ is always perpendicular to an equipotential surface at every point on it.

Step 1 — Using the potential-difference form. Since $V_A=V_B$ on the same equipotential, $W=q(V_B-V_A)=q\cdot 0=0$ immediately.

Step 2 — Confirming it via the line integral (option b's formula). Because $\vec E$ is everywhere perpendicular to the equipotential surface, and the displacement $d\vec l$ along a path on the surface is everywhere tangent to it, $\vec E\cdot d\vec l=0$ at every point of the path. Integrating zero along the whole path gives

$\int_A^B \vec E\cdot d\vec l = 0 \implies W=0.$ …

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Q.The work done to move a charge along an equipotential from A to B

Concept understanding — Electric Potential

Electric Potential

The Idea

Potential Due to a Point Charge

Potential Difference

Relation Between Field and Potential

Superposition

Equipotential Surfaces

The general definition of work, first

Now apply it to a path along an equipotential

Method: Computing Electrostatic Work — General Formula vs. Equipotential Shortcut

Steps

Q.The work done to move a charge along an equipotential from A to B

The general definition of work, first

Now apply it to a path along an equipotential

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