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Electrostatic Potential and Capacitance · TOPIC LESSON

Electrostatic Potential and Energy

Use scalar potentials, spatial gradients and signed energy changes to simplify electrostatics.

JEE Main · JEE Advanced · NEET UG80 min study guide9 indexed sections

BY THE END, YOU CAN

  • Relate voltage, field work and signed potential-energy changes.
  • Compute potentials by superposition and recover field from gradients.
  • Interpret equipotentials, conducting spheres, dipoles and ring symmetry.
  • Calculate assembly energy and charged-particle energy gains without double counting.

BEFORE YOU START

  • Coulomb’s law, vector field direction and work–energy.
  • Differentiation, simple integration and scientific notation.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. Potential is energy per charge, not force per charge

Electric field E describes force per unit test charge, while electric potential V describes electrostatic potential energy per unit test charge in a fixed source configuration. A test charge q has U = qV and a potential-energy change ΔU = qΔV. Potential is a scalar measured in volts, with 1 V = 1 J C⁻¹; field is a vector measured in N C⁻¹ or V m⁻¹.

For a displacement from A to B in an electrostatic field, V_B − V_A = −∫_A^B E·dl. The field’s work on charge q is q(V_A − V_B). If an external agent moves the charge quasistatically with negligible kinetic-energy change and no other energy transfer, its work is q(V_B − V_A). These two works have opposite signs.

A positive charge released from rest tends to accelerate toward lower potential, reducing qV. A negative charge tends toward higher potential while also reducing qV. Saying that every charge naturally moves “downhill in voltage” ignores the sign of q. The field direction itself is defined by the force on a positive test charge.

Only potential differences determine this work. Adding a constant to V everywhere changes the chosen zero but not the field or any potential difference. For an isolated finite charge distribution, choosing V = 0 at infinity is often convenient. For ideal infinite charge distributions, that reference may not be finite, so a finite reference point is needed.

ORIGINAL WORKED EXAMPLE

Signs of field and external work

A charge q = −2.0 μC is moved quasistatically from potential 10 V to 40 V in a fixed electrostatic configuration. Find ΔU, field work and external-agent work.

  1. Potential change is +30 V.
  2. ΔU = qΔV = (−2.0 × 10⁻⁶)(30) = −6.0 × 10⁻⁵ J.
  3. Field work is −ΔU = +6.0 × 10⁻⁵ J. With negligible ΔK, external work is ΔU = −6.0 × 10⁻⁵ J.
  4. The negative external work means the agent removes energy while restraining motion favored by the field.

Answer ΔU = −60 μJ; field work +60 μJ; external work −60 μJ.

02 / UNDERSTAND & APPLY

2. Add signed source potentials as scalars

For a point charge Q with zero at infinity, V(r) = kQ/r, where k = 1/(4πε₀) in vacuum. The sign comes from Q and distance r is a positive magnitude. A positive source gives positive potential and a negative source gives negative potential under this reference; neither statement alone gives the direction of a vector field.

For several fixed point charges, V(P) = kΣQᵢ/rᵢ. Add algebraically rather than using vector components. To obtain E directly, however, each charge’s field needs its direction. A symmetry point can have zero potential and a nonzero field, or zero field and a nonzero potential. A scalar cancellation does not imply a vector cancellation.

Two equal positive charges have zero field at their midpoint but positive potential there. Equal opposite charges have zero potential at the midpoint but their fields point in the same direction there, from positive toward negative. These contrasting cases are a useful diagnostic of whether a calculation has confused the two quantities.

For a continuous distribution, replace the sum by V(P) = k∫dq/r. Choose dq = λdl, σdA or ρdV according to the charge distribution. Integrating a scalar potential can be easier than integrating vector field components, but the observation distance r must still be expressed correctly for each source element.

ORIGINAL WORKED EXAMPLE

Potential at a point between unlike charges

Charges +4.0 nC and −1.0 nC are 0.30 m apart. Find potential at a point between them 0.20 m from the positive charge, using k = 9.0 × 10⁹ N m² C⁻² and zero at infinity.

  1. The point is 0.10 m from the negative charge.
  2. V = k[(4.0 × 10⁻⁹)/0.20 + (−1.0 × 10⁻⁹)/0.10].
  3. The two contributions are +180 V and −90 V, giving +90 V. Their electric fields would require a separate directional calculation.

Answer +90 V.

03 / UNDERSTAND & APPLY

3. Field is the negative spatial gradient

In one dimension, E_x = −dV/dx. In Cartesian three-dimensional form, E = −(∂V/∂x)î − (∂V/∂y)ĵ − (∂V/∂z)k̂. Field points toward decreasing potential, independently of the charge later placed in it. A negative test charge then experiences force opposite to that field.

For a uniform field along +x, V(x) = V(0) − Ex. Potential decreases linearly with x. Between ideal parallel plates far from edges, field magnitude equals the magnitude of voltage difference divided by separation. For a nonuniform field, dividing a finite potential difference by distance gives an average component along that path, not necessarily the field at every point.

A constant nonzero potential throughout a region gives zero field there. Conversely, a potential equal to zero at just one point does not imply zero gradient. A graph crossing the zero-potential axis with nonzero slope represents a nonzero field at the crossing. Read slope rather than height when extracting field from a V–x graph.

At a surface charge layer, the potential can remain continuous while the normal field jumps, so the slope of potential changes abruptly. The ideal point-charge potential diverges at the charge position. These limiting models require care when taking derivatives exactly at their singularities; use the appropriate one-sided or distributional interpretation rather than assuming every curve is smooth everywhere.

ORIGINAL WORKED EXAMPLE

A polynomial potential

In a region, V(x,y) = ax² + by, where a = 3 V m⁻² and b = 4 V m⁻¹. Find E at x = 2 m and the force on q = −1 μC there.

  1. Partial derivatives are ∂V/∂x = 2ax and ∂V/∂y = b.
  2. At x = 2 m, E = (−12î − 4ĵ) V m⁻¹.
  3. Force qE = (−10⁻⁶)(−12î − 4ĵ) N = (12î + 4ĵ) μN. The negative charge reverses both components.

Answer E = (−12, −4) V m⁻¹; force = (12, 4) μN.

04 / UNDERSTAND & APPLY

4. Equipotential surfaces reveal direction and field strength

An equipotential surface has one constant value of V. Motion along it has dV = 0, so the electrostatic field does no work on a test charge during that tangential displacement. This does not say the field is zero; it says the field has no tangential component along the equipotential.

Where the field is nonzero and the potential is differentiable, E is normal to the equipotential surface. For a positive point charge, equipotentials are concentric spheres and field lines are outward radii. For a uniform field, equipotentials are parallel planes perpendicular to that field.

On a contour map drawn with equal voltage intervals, more closely spaced equipotentials indicate a greater field magnitude in the normal direction, approximately |ΔV|/Δn over a small spacing. Comparing unequal voltage intervals without accounting for their values can be misleading. Field-line spacing and equipotential spacing are different graphical conventions and must be interpreted as labeled.

Two distinct equipotential surfaces with different values cannot intersect, because one point cannot have two potentials. At critical points where E = 0, the usual unique normal-direction interpretation can fail. A conductor’s electrostatic equilibrium surface is an equipotential, but an arbitrary insulating surface is not automatically one.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Field lines cross equipotentials normallyThree concentric equipotential circles around a positive point charge have radii R, 2 R and 3 R. Their potentials are 3 V 0, 1.5 V 0 and V 0 when V 0=kQ/(3 R). Radial field arrows point outward, perpendicular to each circle.3V₀1.5V₀V₀+QE points outwardE ⟂ equipotentialsV = kQ/rV decreases outward
A planar cross-section of spherical equipotentials. The selected potentials are not equally spaced; the radii are. The zero is at infinity.

05 / UNDERSTAND & APPLY

5. A conductor can have nonzero constant potential inside

In electrostatic equilibrium, the electric field within the conducting material is zero. Its connected material therefore has constant potential. Excess charge resides on its surfaces under the usual ideal conductor model, and any field just outside is normal to the surface. These statements apply to static equilibrium, not arbitrary currents or rapidly varying fields.

For an isolated conducting sphere of radius R and charge Q, with no external charges and zero at infinity, outside potential is kQ/r and inside potential is the constant kQ/R. The surface joins them continuously. Field is zero inside and kQ/r² radially outside, with the direction set by Q. A jump in normal field at the surface corresponds to surface charge.

If two far-separated conducting spheres are joined by a thin wire and reach equilibrium, their potentials are equal. In the approximation that mutual influence and wire capacitance are negligible, Q₁/R₁ = Q₂/R₂. The larger sphere carries more charge, but the smaller sphere has the larger surface charge density and field for equal potential.

Grounding fixes a conductor’s potential relative to a large charge reservoir; it does not universally mean the conductor has zero charge. Nearby external charges can induce net charge exchange while the grounded potential remains fixed. Similarly, a hollow conductor shields its charge-free enclosed cavity in electrostatic equilibrium, but a charge placed inside the cavity changes the inner-surface distribution and cavity field.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

A charged conducting sphere has constant interior potentialThe normalized potential equals one from the center through the surface r/R=1, then decreases as R/r outside. Interior field vanishes although the potential is nonzero.01231V / (kQ/R)r/RInside: E = 0Outside: V ∝ 1/rPotential is continuous at r = R
Quantitative profile for an isolated positively charged conducting sphere of radius R, electrostatic equilibrium and V(infinity)=0.

ORIGINAL WORKED EXAMPLE

Charge sharing between distant spheres

Two conducting spheres of radii R and 2R are far apart. Initially the first has charge Q and the second is uncharged. They are joined by a thin wire, then disconnected after equilibrium. Neglect mutual influence and wire capacitance. Find final charges and surface-field ratio.

  1. Charge is conserved: Q₁ + Q₂ = Q.
  2. Equal potential gives Q₁/R = Q₂/(2R), so Q₂ = 2Q₁.
  3. Hence Q₁ = Q/3 and Q₂ = 2Q/3. Surface-field magnitudes are proportional to |Qᵢ|/Rᵢ², so E₁/E₂ = (1/3)/(2/12) = 2.

Answer Charges Q/3 and 2Q/3; the smaller sphere has twice the surface-field magnitude for nonzero Q.

06 / UNDERSTAND & APPLY

6. Symmetry simplifies useful extended-source potentials

An electric dipole consists of equal opposite charges separated by a displacement from negative to positive, defining dipole moment p. Its exact potential is kq(1/r_plus − 1/r_minus). Far from the separation scale, V ≈ k(p·r̂)/r² = kp cosθ/r², where θ is measured from p. This is a far-field approximation, not an exact formula next to either charge.

On the dipole’s perpendicular bisector, distances to the two charges are equal, so potential is zero under the infinity reference. The electric field there is generally nonzero and points opposite to p. Knowing that V is zero along a whole plane does not make the normal derivative zero; field can cross that plane.

For a uniformly charged ring of radius a and total charge Q, every element is the same distance √(a² + x²) from an axis point x. Therefore V(x) = kQ/√(a² + x²). Differentiating gives axial field E_x = kQx/(a² + x²)^(3/2). At the centre, potential is kQ/a while field vanishes by symmetry.

The ring result illustrates a reliable sequence: exploit equal source distances in the scalar integral, then differentiate with respect to the observation coordinate. In the far-field limit |x| much larger than a, it approaches the point-charge potential kQ/|x|. This limit checks both dimensions and the physical interpretation of total charge.

07 / UNDERSTAND & APPLY

7. A system’s energy counts each interacting pair once

To assemble several point charges quasistatically from infinity, bring them in one at a time. The first requires no mutual-interaction work. Each later charge qᵢ requires qᵢ times the potential of charges already present. The final total mutual energy is U = kΣ_(i<j)qᵢqⱼ/rᵢⱼ, counting each distinct pair once.

Equivalently, U = (1/2)ΣqᵢV_others(rᵢ), where V_others excludes the charge’s own singular self-potential. The factor one half corrects the double count in that version. For one test charge placed in an already fixed external potential, its interaction energy is qV_external without a half. These are different bookkeeping situations.

Like-charge pairs contribute positive energy and unlike-charge pairs negative energy under the infinite-separation reference. A negative total indicates lower mutual potential energy than the separated state, but does not by itself guarantee a stable free arrangement of unconstrained charges. Static equilibrium and stability need force analysis as well.

If all distances in a fixed charge configuration scale by a positive factor s, its mutual Coulomb energy scales as 1/s. The external work for a quasistatic rearrangement equals the energy difference if no kinetic-energy change or dissipative transfer occurs. If charges are released, lost potential energy can become kinetic energy instead.

ORIGINAL WORKED EXAMPLE

Three charges at an equilateral triangle

Three equal charges +q are fixed at the vertices of an equilateral triangle of side a. Find their total mutual electrostatic energy and the quasistatic external work to expand the triangle uniformly to side 2a.

  1. There are three distinct pairs, each contributing kq²/a, so U_initial = 3kq²/a.
  2. At side 2a, U_final = 3kq²/(2a).
  3. External work is ΔU = −3kq²/(2a). The field does positive work as the charges move apart; the agent removes that energy in a quasistatic expansion.

Answer Initial energy 3kq²/a; external work −3kq²/(2a).

08 / UNDERSTAND & APPLY

8. Potential differences determine kinetic-energy changes with the charge sign

If the only work on a particle is electrostatic and the sources remain fixed, K + qV is conserved. Thus K_B − K_A = q(V_A − V_B). A particle released from rest can reach a point only if the predicted kinetic energy is nonnegative, subject also to trajectory constraints. Potential energy alone determines a speed change, not necessarily the complete path or travel time.

An electron-volt is an energy unit: 1 eV = e × 1 V ≈ 1.602 × 10⁻¹⁹ J. A particle with charge magnitude ze gaining kinetic energy through an accelerating voltage magnitude ΔV gains zΔV eV when the force and motion have the appropriate sign. Electron-volts are not volts, and a doubly charged ion does not gain the same energy as a singly charged ion through the same accelerating difference.

For nonrelativistic motion from rest, speed follows v = √(2ΔK/m). If the inferred speed approaches the speed of light, this formula ceases to be valid and relativistic energy is required. For multi-particle release, momentum conservation and recoil can distribute the potential-energy decrease among several moving particles; the fixed-source test-particle formula alone is not enough.

ORIGINAL WORKED EXAMPLE

A negative particle accelerates toward higher potential

An electron starts from rest and moves from 0 V to +100 V in a fixed electrostatic field. Ignore other forces and use e = 1.60 × 10⁻¹⁹ C and m_e = 9.1 × 10⁻³¹ kg. Find kinetic-energy gain and approximate speed.

  1. q = −e, so ΔK = q(V_initial − V_final) = (−e)(−100 V) = 100 eV.
  2. In joules this is 1.60 × 10⁻¹⁷ J.
  3. v = √[2(1.60 × 10⁻¹⁷)/(9.1 × 10⁻³¹)] ≈ 5.9 × 10⁶ m s⁻¹, well below c for this approximation.

Answer 100 eV gained; approximately 5.9 × 10⁶ m s⁻¹.

ORIGINAL PRACTICE · 1

A +3 μC charge moves through ΔV = −20 V. Find ΔU.

Reveal answer & reasoning

−60 μJ.

ΔU = qΔV; positive charge loses potential energy on moving to lower potential.

ORIGINAL PRACTICE · 2

A potential is constant at 50 V throughout a region. What is its electric field there?

Reveal answer & reasoning

Zero.

Field is minus the spatial gradient, not proportional to the potential value itself.

ORIGINAL PRACTICE · 3

Can V = 0 at one point coexist with nonzero E?

Reveal answer & reasoning

Yes.

A potential can cross zero with nonzero spatial slope, as on a dipole’s perpendicular bisector.

ORIGINAL PRACTICE · 4

For V(x) = 10 − 4x in volts with x in metres, find E_x.

Reveal answer & reasoning

+4 V m⁻¹.

E_x = −dV/dx = −(−4).

ORIGINAL PRACTICE · 5

Equipotentials differ by 10 V across a small normal spacing 2 mm. Estimate field magnitude.

Reveal answer & reasoning

5.0 × 10³ V m⁻¹.

|E|≈|ΔV|/Δn = 10/0.002.

ORIGINAL PRACTICE · 6

What electrostatic work is done moving a charge along an equipotential?

Reveal answer & reasoning

Zero.

The potential difference is zero, so field work q(V_A−V_B) vanishes.

ORIGINAL PRACTICE · 7

Outside a point charge, radius doubles. How do potential and field magnitude scale?

Reveal answer & reasoning

Potential halves; field magnitude becomes one quarter.

V∝1/r and |E|∝1/r².

ORIGINAL PRACTICE · 8

At the centre of a uniformly charged ring, is potential necessarily zero?

Reveal answer & reasoning

No; it is kQ/a with zero at infinity.

All source elements are distance a away; field cancels by symmetry but potentials add.

ORIGINAL PRACTICE · 9

For an isolated charged conducting sphere, what is potential at an interior point?

Reveal answer & reasoning

The constant kQ/R.

Interior field is zero, and continuity sets the constant equal to the surface value.

ORIGINAL PRACTICE · 10

Does grounding always remove all net charge from a conductor?

Reveal answer & reasoning

No.

Grounding fixes potential; nearby sources can induce charge exchange needed to maintain it.

ORIGINAL PRACTICE · 11

How many pair terms enter mutual energy for four point charges?

Reveal answer & reasoning

Six.

Each distinct pair is counted once: 4×3/2.

ORIGINAL PRACTICE · 12

Should a factor one half multiply qV for a test charge in an already fixed external potential?

Reveal answer & reasoning

No.

The half appears when summing each mutual pair twice across all charges, not for a single test-charge interaction.

ORIGINAL PRACTICE · 13

A +2e ion gains energy through an accelerating difference of 30 V. Find the gain in eV.

Reveal answer & reasoning

60 eV.

Energy magnitude is charge magnitude times voltage, giving 2×30 electron-volts.

ORIGINAL PRACTICE · 14

Does a zero potential on the entire dipole equatorial plane prove zero field there?

Reveal answer & reasoning

No.

The tangential potential derivative vanishes, but the normal derivative can be nonzero.

09 / UNDERSTAND & APPLY

9. Separate scalar sums, gradients and energy bookkeeping

Choose the potential reference first. Add signed source potentials as scalars, obtain field from the spatial gradient when useful, and multiply by the signed test charge only when calculating energy or force. A zero value, zero slope and zero work along one direction are three different statements.

For assembly, count each pair once and exclude self-potentials. For motion, state whether the sources stay fixed and whether the particle is nonrelativistic. For conductors, specify electrostatic equilibrium and the surrounding charges. These distinctions keep a compact potential method from silently applying the wrong physical model.

  • Potential is scalar energy per charge; field is vector force per charge.
  • Field work is minus the potential-energy change.
  • Negative charges can gain kinetic energy by moving to higher potential.
  • Equipotential normals give field direction where the field is nonzero.
  • A conductor can have nonzero constant potential with zero interior field.
  • Pair-energy formulas and single-test-charge formulas have different counting factors.

KEEP THE BIG IDEAS

Quick revision

  • A potential value and its gradient contain different information.
  • The sign of charge controls whether higher potential means higher energy.
  • Electrostatic geometry can make fields cancel while potentials do not, or the reverse.
  • Energy methods require clear source, reference and pair-counting assumptions.
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