CLASS 12 PHYSICS · CHAPTER 2 · STUDY NOTES
Electrostatic Potential and Capacitance
A comprehensive study of electrostatic potential, potential energy, and the principles of capacitance in conductors and dielectrics, based on Class 12 Physics curriculum.
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Two paths, one energy change
Relate external work, field work and potential energy.Two paths, one energy change
A fixed positive source has kQ = 60 V·m. Move a +1 μC test charge slowly from r = 4 m to 1 m: U rises from 15 to 60 μJ. External work is +45 μJ; field work is −45 μJ on either path. No kinetic-energy change is assumed.
- Electrostatic potential-energy change
- The change in stored electric energy: ΔU = q(V_final − V_initial).
In a static electric field, work depends on endpoints, not the route. Move a test charge slowly so its kinetic energy does not change: W_external = ΔU, while W_field = −ΔU.
For a positive test charge near a positive source, moving closer raises U. An external agent supplies that energy against repulsion. Moving away reverses these work signs.
Let’s look a little closer.
Go deeper: The zero is a choice
Only changes in potential energy affect the work. An additive constant changes every U value together without changing the forces. If a charge is released and electric force alone acts, ΔK = −ΔU; the slow-motion condition used by the control no longer applies.
Take a moment. What do you think?
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Question
A +2 μC test charge moves slowly from 8 V to 26 V. Find external and field work.
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Separate potential from the test charge
Explain why V belongs to the source configuration.The same potential, a different test charge
At r = 2 m, V = 60/r = 30.0 V. A 1 μC test charge has U = 30.0 μJ. Doubling the test charge doubles U, not the source potential V. Reference: V at infinity = 0.
- Electric potential
- Potential energy per unit test charge due to external sources, measured relative to a chosen zero.
Potential difference is external work per unit positive test charge: ΔV = W_external/q for slow movement. Its unit is the volt, 1 V = 1 J/C.
For a fixed source configuration, changing the small test charge changes U = qV, but does not change V. For a negative charge, the potential-energy sign reverses. The test charge must be small enough not to disturb the sources.
Go deeper: Choose a reference that exists
For localised charges we usually choose V at infinity = 0. That convention cannot be applied to an ideal uniform field extending to infinity; instead choose a finite reference point. An electron moving from lower to higher V loses potential energy because its charge is negative. One electronvolt is an energy of approximately 1.60 × 10⁻¹⁹ J, not a voltage.
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Question
At a point where V = −12 V, compare the energy of +3 μC and −3 μC test charges.
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Read the potential from its source
Compare a point charge, a dipole and a charged shell.A source, a dipole and a spherical shell
For this source, kQ = 60 V·m. V = kQ/r = 30.0 V. Reverse Q to reverse the potential sign; the graph uses V at infinity = 0.
- Scalar superposition
- Add the signed potential of each source at the observation point.
With V at infinity = 0, a point source gives V = kQ/r. Potential is a signed scalar: for several sources, add kQ_i/r_i algebraically, not as vectors. A continuous distribution uses V = k∫ρ d(volume)/r.
For charges +q and −q, exact dipole potential is kq(1/r₊ − 1/r₋). Far from a dipole of moment p, V ≈ kp cosθ/r². On its equatorial plane V = 0, although the electric field there is generally nonzero.
For a thin uniformly charged spherical shell of radius R, V = kQ/R everywhere inside and kQ/r outside. The potential joins continuously at R. E is zero inside but has different one-sided values across an ideal charged surface.
Go deeper: Distance laws and limits
The dipole approximation requires r much greater than half the charge separation a. Unlike a point-charge potential, its leading far-field potential falls as 1/r² and depends on direction. For volume density ρ, each small element contributes kρ d(volume)/r; integrate over the source. Point-charge V diverges at that source itself, so never include its own potential in qV for interaction energy.
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Question
At P, +2 nC is 0.30 m away and −1 nC is 0.15 m away. Find V using k = 9 × 10⁹ and explain whether zero V proves zero E.
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 4, 5, 6, 7, 8, 9, 32, 34.
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Read the voltage map
Connect surface spacing, direction and field strength.Read the voltage map
Move the test point along one circular section of a spherical equipotential: ΔV = 0 and field work = 0. Radial E is perpendicular to the surface. These radii are 1, 1.5 and 3 m for V = 120, 80 and 40 V: equal 40 V steps lie closer together near the charge, where E is stronger.
- Equipotential surface
- A surface on which electric potential has one constant value.
Every point on an equipotential surface has the same V. Moving a test charge along it gives ΔU = qΔV = 0, so the electric field does no work.
Where the field is nonzero, E is perpendicular to the surface and points towards decreasing V. Equal voltage steps over a smaller normal distance mean a larger field magnitude: E ≈ |ΔV|/Δs.
Go deeper: The slope gives the field
In one dimension E_x = −dV/dx; in three dimensions E = −∇V. A point charge has spherical equipotentials; a uniform field has parallel planes. A dipole or pair of like charges has differently shaped surfaces, so concentric spheres are not a universal rule. Distinct equipotential surfaces cannot intersect at the same point because V has only one value there. At a zero-field point the field direction is undefined.
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Question
Parallel equipotential planes drop from 72 V to 48 V over 0.08 m. Find the uniform field and its direction.
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Count each interaction once
Distinguish mutual energy from external-field energy.Count interactions, then find the energy minimum
Three charges of magnitude 1 μC sit at vertices of a 0.30 m equilateral triangle. Count each pair once: 0.030 + 0.030 + 0.030 = 0.090 J. Each pair magnitude is 0.030 J. Self-energy is not included.
- Interaction energy
- External work needed to assemble distinct charges slowly from infinite separation, excluding their self-energies.
Two point charges separated by r have interaction energy U = kq₁q₂/r. Like charges give positive U; unlike charges give negative U, relative to infinite separation.
For several charges, sum over each distinct pair once. If another fixed source produces external potential V_ext, add Σq_i V_ext(r_i). Do not include a charge’s own potential.
A dipole in a uniform external field has U = −pE cosθ, choosing U = 0 at 90°. Aligned is a stable energy minimum; exactly antiparallel is an unstable energy maximum. Both have zero torque at that exact angle.
Go deeper: An energy minimum is not instant alignment
The torque p × E tends to rotate a displaced dipole towards alignment. Without damping it can oscillate about that direction; eventual rest requires energy transfer to the surroundings. For rotation from θ₁ to θ₂, slow external work is pE(cosθ₁ − cosθ₂). For two charges in an external field, include both mutual kq₁q₂/r and external q₁V₁ + q₂V₂ terms.
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Question
A dipole has pE = 8 μJ. Rotate it slowly from 60° to 120°. Find ΔU and external work.
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A conductor rearranges charge
Explain equilibrium, surface charge and cavity shielding.A conductor rearranges charge
At electrostatic equilibrium, excess charge lies on the conductor’s surface, E = 0 in the metal, and V is constant throughout. E = 0 does not imply V = 0. A charged sphere illustrates the general equilibrium rule.
- Electrostatic equilibrium
- A static distribution in which mobile charges no longer drift under an internal electric field.
| Region | Equilibrium result |
|---|---|
| Metal interior | E = 0; V constant |
| Just outside charged surface | E normal; magnitude |σ|/ε₀ |
| Charge-free cavity | Shielded from external static fields |
Mobile charges redistribute until E = 0 inside the metal at electrostatic equilibrium. The entire conductor, including its surface, has one constant potential; that value need not be zero.
Excess charge lies on the surface. Just outside a charged surface, E = (σ/ε₀)n̂: it is normal, outward for positive σ and inward for negative σ. A tangential field would drive charge motion.
A charge-free closed cavity is shielded from external electrostatic fields. That condition matters: placing charge in the cavity changes the configuration, and this statement does not promise shielding of every electromagnetic wave.
Go deeper: Why the charge stays on the surface
An arbitrarily small Gaussian surface wholly in the metal has zero electric flux because E = 0; it therefore encloses zero net excess charge. Potential differences inside are also zero because no electric work is needed between interior points. The surface relation follows from a small pillbox crossing the boundary: only its outside face contributes flux. Each separate conductor may have a different constant potential.
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Question
An isolated charged conductor is at +18 V. What are E and V inside its metal? Does E = 0 imply the conductor is uncharged?
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From molecules to a surface field
Distinguish induced displacement from permanent-dipole alignment.From molecular dipoles to a surface field
Without an applied field, non-polar charge centres coincide; polar molecules have permanent dipoles with differing orientations. Choose a molecule type, then apply the field.
- Polarisation P
- Net electric dipole moment per unit volume, in C/m².
A dielectric has bound charges rather than freely mobile metal electrons. In a non-polar molecule an applied field shifts positive and negative charge centres apart, inducing a dipole.
Polar molecules already have a permanent dipole. The field encourages alignment, while thermal agitation resists it; the alignment is preferential, not perfect. Both mechanisms produce a net dipole moment per volume called P.
A uniformly polarised slab has bound charges on its opposite faces. Their field opposes the applied field. With fixed free source charge, the net field is reduced; a battery-connected capacitor can instead maintain its voltage by exchanging free charge.
Go deeper: The molecular dipoles add up
For a linear isotropic dielectric P = ε₀χ_e E, where E is the total local macroscopic field and χ_e is electric susceptibility. Adjacent dipole ends compensate in a uniformly polarised bulk, leaving bound face charges. Non-polar O₂ and polar H₂O illustrate the distinction; polar molecules can also acquire an induced contribution. The classroom drawing simplifies molecular shape and thermal statistics.
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Question
A polar dielectric is placed in a uniform rightward field. Which slab face becomes negatively bound-charged, and which way does the bound-charge field point?
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Change the geometry; watch C, Q and V
Predict what stays fixed before inserting a dielectric.Change the geometry; watch C, Q and V
C = Kε₀A/d = 8.854 pF. The battery holds V = 6 V, so increasing C draws more charge. Here |Q| = 53.124 pC and V = 6.000 V. K fills the entire gap; Edge fringing is neglected.
- Capacitance
- For a linear capacitor, the geometry- and dielectric-dependent ratio C = Q/V, in farads (C/V).
A capacitor holds +Q and −Q on two conductors. Its quoted charge Q means the magnitude on one plate, not the net charge of the pair. Capacitance C = Q/V measures charge stored per potential difference.
For ideal parallel plates, C = ε₀A/d in vacuum. A linear dielectric filling the whole gap gives C = Kε₀A/d. Larger area or K raises C; a larger gap lowers it. Neglect fringing when d² is much smaller than A.
With a connected battery, V stays fixed and Q changes with C. After isolation, Q stays fixed and V changes inversely with C. An insulating material has a breakdown limit, so these ideal formulas do not allow unlimited voltage.
Go deeper: Partial filling needs two field regions
A slab of thickness t and relative permittivity K leaves vacuum thickness d − t. At fixed Q, E_vac = Q/(ε₀A) and E_slab = E_vac/K. Add the two voltage drops: V = E_vac(d − t) + E_vac t/K, giving C = ε₀A/(d − t + t/K). For fixed geometry and linear material, changing Q and V together does not change C. 1 μF = 10⁻⁶ F, 1 nF = 10⁻⁹ F and 1 pF = 10⁻¹² F.
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Question
An isolated 20 pF capacitor initially has V = 9 V. A dielectric K = 3 fills its gap. Find final C, Q and V.
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Follow the connections
Use shared voltage or equal series charge to choose the rule.Follow the connections, not the positions
For 3 μF and 6 μF in series across 6 V: C_eq = 2 μF; each capacitor carries |Q| = 12 μC. Voltage divides: 4 V + 2 V = 6 V. The connecting conductor starts neutral.
- Equivalent capacitance
- The single capacitance that gives the same terminal charge–voltage relation as a network.
Series capacitors carry equal charge magnitudes when the intermediate conductors start neutral. Their voltage drops add, so 1/C_eq = Σ1/C_i. The smaller capacitor takes the larger voltage drop.
Parallel capacitors share the same two nodes, hence the same V. Their branch charges add, giving C_eq = ΣC_i. A larger branch capacitance stores a larger charge at that common voltage.
Go deeper: Reduce a network by its nodes
Two elements that look side by side are not necessarily parallel; trace their electrical connections. Series means an intermediate connection with no additional charge branch under the stated neutral-node assumption. For a mixed network, reduce an unambiguous series or parallel group, then repeat. Series C_eq is smaller than the smallest member; parallel C_eq exceeds any individual positive member.
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Question
A 4 μF and 12 μF capacitor are placed in series across 8 V. Find C_eq, charge magnitude and each voltage.
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Where does the charging work go?
Relate the charging graph, stored field energy and charge sharing.Where does the charging work go?
Charging a 2 μF capacitor to 6 V gives Q = 12 μC. Energy = area under V(q) = ½QV = 36.0 μJ. It is not QV because the voltage rises during charging.
- Electric-field energy density
- Stored electric energy per unit volume; in vacuum u = ½ε₀E², in J/m³.
As charge accumulates, V grows from zero to Q/C. Each small transferred charge dq requires work V dq. The area under this straight V(q) graph gives U = ½QV = ½CV² = Q²/(2C).
In vacuum the energy density is u = ½ε₀E². Multiplying by a uniform field-filled volume gives the stored energy there. Doubling E quadruples energy density.
Joining an isolated charged capacitor to an equal uncharged one conserves charge but halves their total stored energy after equilibrium. The remaining energy transfers as heat and electromagnetic radiation during the transient; it does not disappear.
Go deeper: Which variable stays fixed matters
Increasing C raises ½CV² at fixed V but lowers Q²/(2C) at fixed Q. A connected battery can do work when the capacitor changes, so the capacitor alone is not an isolated energy system. In charge sharing, the diagram selects initial and settled states rather than claiming an instantaneous transfer. For a uniform vacuum plate field, C = ε₀A/d and V = Ed give U = ½ε₀E²Ad.
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Question
An isolated 5 μF capacitor at 4 V is connected to an equal uncharged capacitor. Find initial energy, final voltage and total final stored energy.
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