(a) An electric field $\vec{E} = E_0\,\hat{i}$ exists in a region of space. Draw three equipotential surfaces in the region. (b) Two point charges $-q$ and $+q$ are located at points $(-a, 0, 0)$ and $(a, 0, 0)$ respectively. Find the electrostatic potential at the point $(x, 0, 0)$ where $x \gg a$.

(a) An electric field vecE = E₀hati exists in a region of space. Draw three equipotential surfaces in the regi

Question

\vec{E} = E_0\,\hat{i}

(b) Two point charges

-q

and

+q

are located at points

(-a, 0, 0)

and

(a, 0, 0)

respectively. Find the electrostatic potential at the point

(x, 0, 0)

where

x \gg a

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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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Figure — Stem (a) 'Draw three equipotential surfaces' for E=E0 i-hat is an explicit draw instruction; a sketch of plane

The key idea is that equipotential surfaces are always perpendicular to the electric field lines, and for a dipole at large distances, the potential falls off as $1/r^2$ and depends on the dipole moment. For part (a), the surfaces are planes perpendicular to the x-axis. For part (b), the potential at $(x,0,0)$ for $x \gg a$ is $V \approx \frac{1}{4\pi\epsilon_0} \frac{2qa}{x^2}$ .

Part (a): Equipotential Surfaces for a Uniform Field

Concept first: An equipotential surface is a surface where the electric potential is constant everywhere. The electric field always points in the direction of the steepest decrease of potential, and it is always perpendicular to equipotential surfaces. If you move along an equipotential, you do no work against the field — the field does no work because your displacement is perpendicular to it.

Here, $\vec{E} = E_0 \hat{i}$ is uniform and points along the positive x-axis. That means the potential decreases as you move in the $+x$ direction. Surfaces of constant potential must be perpendicular to the x-axis — they are planes parallel to the yz-plane.

Choose three different potentials. Let’s pick $V = V_1$ , $V = V_2$ , and $V = V_3$ , with $V_1 > V_2 > V_3$ . Since $E = -\frac{dV}{dx}$ , a larger $x$ gives a smaller $V$ .
Draw them. In a 3D sketch (or even a 2D cross-section in the xy-plane), these are three parallel planes. In the xy-plane, they appear as three vertical lines (parallel to the y-axis) spaced equally apart if $E_0$ is constant. The plane with the highest potential is on the left (smaller $x$ ), and the lowest on the right.

Tip

A common shortcut: for a uniform field, the equipotential surfaces are equally spaced parallel planes. The spacing $\Delta x$ between surfaces differing by $\Delta V$ is $\Delta x = \Delta V / E_0$ .

Part (b): Potential of a Dipole at Large Distance

Concept first: Two equal and opposite charges separated by a small distance form an electric dipole. At points far away compared to the separation ( $x \gg a$ ), the individual charges look almost like they’re at the same location, but their opposite signs cause a partial cancellation of the $1/r$ terms. The leading term in the potential is not $1/x$ (that would be the monopole term, which is zero because total charge is zero) but $1/x^2$ , proportional to the dipole moment $p = 2qa$ .

We want the potential at a point on the x-axis, to the right of both charges.

Write the exact potential. The potential at a point due to a point charge $q$ is $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$ . Here, we have two charges:
- Charge $+q$ at $(a,0,0)$ : distance to $(x,0,0)$ is $|x - a|$ .
- Charge $-q$ at $(-a,0,0)$ : distance to $(x,0,0)$ is $|x + a|$ .
Since $x > a > 0$ , both distances are positive: $x - a$ and $x + a$ . So the exact potential is:

$V(x) = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{x - a} + \frac{-q}{x + a} \right) = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{x - a} - \frac{1}{x + a} \right).$ …

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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 44 on this concept.

CBSE 2026Set 55/1/11 markMCQ
$r_1$
›Reveal solutionSolution
When two widely separated conducting spheres are connected by a wire, they reach the same electric potential. Since surface field $E = \frac{kQ}{r^2}$ and potential $V = \frac{kQ}{r}$ , combining these gives $E \propto 1/r$ . Therefore the ratio of surface fields is $E_A/E_B = r_2/r_1$ , which corresponds to option (B).

The key insight here is about what happens when conductors are connected by a wire. Charge flows until both spheres are at the same electric potential — that's the fundamental condition for electrostatic equilibrium in a conductor. Once you grasp that, the rest is just algebra.

Let's think about why potential equality is the right starting point. A conducting wire means the two spheres form a single conductor. In electrostatics, the entire surface of a conductor is an equipotential. So spheres A and B must have the same potential $V$ .

Now, for an isolated conducting sphere of radius $r$ carrying charge $Q$ , the potential at its surface (taking infinity as zero) is:
$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$
And the electric field just outside its surface is:
$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$
Notice the relationship: $E = V/r$ . That's a neat shortcut we'll use.

Tip

For any isolated conducting sphere, $E = V/r$ directly. This saves you from carrying the $Q$ through the algebra — just remember it comes from $V = kQ/r$ and $E = kQ/r^2$ .

Let's work through it step by step.

Set potentials equal. Since the wire connects them, $V_A = V_B$ . Using $V = kQ/r$ (where $k = 1/4\pi\epsilon_0$ ):

$k\frac{Q_A}{r_1} = k\frac{Q_B}{r_2}$
Cancel $k$ and rearrange:
$\frac{Q_A}{Q_B} = \frac{r_1}{r_2}$

Write the surface field ratio. For each sphere, $E = kQ/r^2$ . So: $\frac{E_A}{E_B} = \frac{k Q_A / r_1^2}{k Q_B / r_2^2} = \frac{Q_A}{Q_B} \cdot \frac{r_2^2}{r_1^2}$ …
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CBSE 2026Set 55/3/11 markMCQ
$m$
›Reveal solutionSolution
Work done by a constant electric field equals force times displacement; since the particle starts from rest, all that work converts to kinetic energy, giving $K = q E_0 x$ .

The heart of this problem is the work-energy theorem: the work done by all forces on a particle equals its change in kinetic energy. When a charged particle moves through an electric field, the field exerts a force that does work, and if the particle starts from rest, every joule of work becomes kinetic energy.

A uniform electric field $\vec{E} = E_0\,\hat{i}$ exerts a force $\vec{F} = q\vec{E}$ on a charge $q$ . This force is constant in magnitude and direction, so the work done is simply force times displacement along the direction of the force.

Step-by-step reasoning

Identify the force on the particle. The electric force on a charge $q$ in field $\vec{E}$ is

$\vec{F} = q\vec{E} = q E_0\,\hat{i}$
The magnitude is $F = q E_0$ , directed along the positive $x$ -axis.

Calculate the work done by this force. The particle moves a distance $x$ along the $x$ -axis, in the same direction as the force. Work done by a constant force is

$W = F \cdot d = q E_0 \cdot x = q E_0 x$

Apply the work-energy theorem. The particle starts from rest, so initial kinetic energy $K_i = 0$ . The work-energy theorem states

$W = \Delta K = K_f - K_i$
Therefore,
$K_f = W = q E_0 x$
The kinetic energy after travelling distance $x$ is simply the work done by the electric field. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. SI unit of electric potential is:$
(a) Ohm
(b) Volt
(c) Coulomb
(d) Ampere
›Reveal solutionSolution

Electric potential is defined as work done per unit charge, so its SI unit is the Volt.

Electric potential at a point is $V = \dfrac{W}{q}$ , i.e. the work done in bringing a unit positive charge from infinity to that point. Since work is measured in joules (J) and charge in coulombs (C), the unit o …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. [Case study, continued] The charges q1 and q2 produce a potential, which at any point P will be$
(a) V1,2 = 1/(4πε₀) × (q1/r1P² + q2/r2P²)
(b) V1,2 = 1/(4πε₀) × (q1/r1P + q2/r2P)
(c) V1,2 = 1/(4πε₀) × (q1/r1P - q2/r2P)
(d) V1,2 = 1/(4πε₀) × (2q1/r1P + 3q2/r2P)
›Reveal solutionSolution

The potential due to a group of point charges at any point is just the plain algebraic (scalar) sum of the potentials each charge produces there — the superposition principle for potential.

Electric potential obeys the superposition principle: the total potential at any point due to several charges equals the SCALAR sum (not vector sum, since potential is a scalar) of the potentials due to each individual charge, each given by $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ (distance to the first power).

For two charges $q_1$ (at distance $r_{1P}$ from P) and $q_2$ (at distance $r_{2P}$ from P):

$V_{1,2} = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_{1P}} + \frac{q_2}{r_{2P}}\right)$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The standard potential of earth is :$
(a) Zero
(b) Infinite
(c) One
(d) None of the above
›Reveal solutionSolution

The Earth is the chosen reference for potential, so its standard potential is taken as zero.

Electric potential is always measured relative to some reference. Because the Earth is a very large conductor whose potential is practically unaffected by adding or removing charge, it is universally chosen as the reference (zero) level of potential.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Match the Column A with Column B and write the correct pair. Column A: Potential gradient. Column B:$
(i) μ₀nI,
(ii) Volt × second,
(iii) μ₀nI/2,
(iv) Volt × meter,
(v) Volt × meter⁻¹,
(vi) Volt.
›Reveal solutionSolution

Potential gradient dV/dx has unit V/m = Volt × metre⁻¹, option (v).

The potential gradient is the rate of change of electric potential with distance, dV/dx. Its SI unit is volt per metre (V/m), i.e. Volt × metre⁻¹. In magnitude it equals th …

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CBSE 2025Set 55/5/11 markMCQ
$\vec{E}=\dfrac{\alpha}{r^{2}}\,\hat{r}$
›Reveal solutionSolution
For a radial electric field $\vec{E} = \frac{\alpha}{r^2}\,\hat{r}$ , integrate $-\vec{E} \cdot d\vec{l}$ along a radial path from infinity to find the potential; the magnitude is $\frac{\alpha}{r}$ .

The connection between electric field and potential is one of the most fundamental relationships in electrostatics. The electric field points in the direction of steepest decrease of potential, and its magnitude tells us how rapidly the potential drops. Mathematically, $\vec{E} = -\nabla V$ , or in one dimension, $E = -\frac{dV}{dr}$ for a radial field.

To find the potential at a point, we integrate the electric field along a path. The potential difference between two points is:
$V(r) - V(r_0) = -\int_{r_0}^{r} \vec{E} \cdot d\vec{l}$
We conventionally choose $r_0 = \infty$ as our reference point where $V(\infty) = 0$ , so:
$V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vec{l}$
Now let's work through this problem step by step.

Set up the line integral for a radial field.

Since both $\vec{E}$ and the path element $d\vec{l}$ point radially (we choose a radial path for simplicity), we have:

$\vec{E} \cdot d\vec{l} = E_r \, dr = \frac{\alpha}{r^2} \, dr$

Evaluate the integral from infinity to $r$ .

$V(r) = -\int_{\infty}^{r} \frac{\alpha}{r^2} \, dr$
Reversing the limits to make the calculation cleaner:
$V(r) = \int_{r}^{\infty} \frac{\alpha}{r^2} \, dr$

Perform the integration.

$V(r) = \alpha \int_{r}^{\infty} \frac{1}{r^2} \, dr = \alpha \left[-\frac{1}{r}\right]_{r}^{\infty}$ $V(r) = \alpha \left(0 - \left(-\frac{1}{r}\right)\right) = \frac{\alpha}{r}$

Interpret the result. …
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CBSE 2025Set IMPROVEMENT1 markMCQ
$Q. Electron volt is the unit of which of the following?$
(a) Electric Potential
(b) Energy
(c) Electric Field
(d) Electric Current
›Reveal solutionSolution

Electron volt (eV) is a unit of energy, not of potential, field or current.

…

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CBSE 2025Set D1 markMCQ
$Q. The speed of an electron accelerated from rest under a potential difference V is (A) proportional to V (B) proportional to \sqrtV (C) proportional to 1/V (D) proportional to V^2$
›Reveal solutionSolution

The kinetic energy gained equals eV; since ½mv² = eV, the speed v ∝ √V.

An electron accelerated from rest through a potential difference V gains kinetic energy equal to the work done by the field:

$eV = \frac{1}{2}mv^2$

Solving for the speed:

$v = \sqrt{\frac{2eV}{m}}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Two charges 1C and -1C are placed 1 m apart. The potential at the centre of the line joining the two charges will be$
(a) 2 V
(b) -2 V
(c) zero
(d) 0.5 V
›Reveal solutionSolution

At the midpoint of the line joining equal-and-opposite charges, the two charges are equidistant, so their potential contributions are equal in magnitude but opposite in sign and cancel exactly.

Electric potential due to a point charge q at distance r is V = kq/r (k = 1/4-pi-epsilon0).

The two charges are q1 = +1 C and q2 = -1 C, separated by 1 m, so each is r = 0.5 m from the midpoint.

V = k(+1)/0.5 + k(-1)/0.5 = 2k - 2k = 0

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. When a photon is accelerated (from rest) through a potential difference of one volt, the kinetic energy gained by it is equal to$
(a) 1837eV
(b) 1 eV
(c) $\frac{1}{1837}$ eV
(d) none of the above.
›Reveal solutionSolution

Energy gained by a charge accelerated through a potential difference depends only on charge × voltage, not on mass — so it is 1 eV regardless of which particle is being accelerated.

Work-energy theorem for a charge q accelerated from rest through potential difference V:

$KE = qV$

For an elementary charge $q = e = 1.6\times10^{-19}\ \text{C}$ and $V = 1\ \text{V}$ :

$KE = (1.6\times10^{-19})(1) = 1.6\times10^{-19}\ \text{J} = 1\ \text{eV}$ (by definition of the electron-volt)

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Electric potential 'V' at a distance 'r' from a point charge is directly proportional to ______.$
(a) r
(b) r²
(c) 1/r
(d) 1/r²
›Reveal solutionSolution

The electric potential of a point charge falls off as the inverse of the distance from it.

The electric potential at a distance $r$ from a point charge $q$ is

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

…

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Q.(a) An electric field E⃗=E0 i^\vec{E} = E_0\,\hat{i}E=E0​i^ exists in a region of space. Draw three equipotential surfaces in the region.

Part (a): Equipotential Surfaces for a Uniform Field

Part (b): Potential of a Dipole at Large Distance

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Q.(a) An electric field $\vec{E} = E_0\,\hat{i}$ exists in a region of space. Draw three equipotential surfaces in the region.

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