L-16: Electric Potential and Capacitors
Physics — Class 12 · NIOS Code 312 · Module 5 · Source: 312_Physics_Eng_Lesson16.pdf
Electric Potential and Capacitors
Like water flows from higher to lower level and heat from hotter to colder body, positive charge moves from higher to lower electric potential. This lesson connects electric field E and potential V, introduces capacitors for storing charge and energy, and explains dielectric polarization.
NIOS objectives: electric potential and potential difference; potential of point charge and dipole; capacitors and applications; parallel plate capacitance; series/parallel grouping; energy stored; dielectric polarization.
16.1 Electric Potential and Potential Difference
Work done against the electric force in moving a unit positive charge from outside the field to a point defines the electric potential at that point — a scalar quantity. Positive potential means work done against the field; negative means work done by the field.
For points A and B, work done moving test charge q₀ from A to B:
q₀ = test charge (C)
VA, VB = potentials at A and B (V)
Work is path independent — electrostatic field is conservative
1 volt = 1 joule per coulomb
- Potential at a point depends on choice of zero (usually infinity).
- Potential difference between two points is unique.
- Positive charge tends to move from higher to lower potential.
16.1.1 Potential due to a Point Charge
For charge +q at O, potential at P (OP = r):
q = source charge (C)
r = distance from charge (m)
∝ 1/r; sign follows sign of q
Superposition: V = Σ qᵢ/(4πε₀rᵢ)
16.1.2 Potential due to an Electric Dipole
Dipole: −q at A, +q at B, separation 2l, centre O. At P with polar coordinates (r, θ):
θ = angle between dipole axis and OP
Unlike point charge (∝ 1/r), dipole potential ∝ 1/r²
- Axial (θ = 0°): V = p/(4πε₀r²)
- Axial opposite side (θ = 180°): V = −p/(4πε₀r²)
- Equatorial (θ = 90°): V = 0 at every point on perpendicular bisector
An equipotential surface has the same potential at every point. E is perpendicular to it. No work is done moving a charge along an equipotential surface.
16.1.3 Potential Energy of a System of Point Charges
Energy possessed by charges by virtue of their positions. At infinite separation, U = 0. For two charges q₁ and q₂ separated by r₁₂:
Same sign → U increases when brought closer (work against repulsion)
Opposite sign → U decreases when brought closer
For a dipole in uniform field E:
Stable equilibrium when p aligned with E (θ = 0)
16.2 Relation Between E and Potential
In uniform field, moving unit + charge from A to B over distance Δr against the field:
Negative sign: work done against field increases potential
Potential gradient (rate of change of V) equals E in magnitude
Potential is scalar; potential gradient is vector (= E)
16.2.1 Behaviour of Conductors in an Electric Field
Free electrons redistribute on the surface until internal E cancels the external field — equilibrium in ~10⁻¹⁶ s.
- No electric field inside a conductor in electrostatic equilibrium.
- External E is perpendicular to the conductor surface.
- All excess charge resides on the surface.
- Field inside a cavity in a conductor is zero.
- Electrostatic shielding: hollow conductor blocks external fields — why cars/buses are safer in lightning.
16.3 Capacitance
Two conductors with equal and opposite charges ±Q and potential difference V form a capacitor. Capacitance measures ability to store charge:
1 F = 1 C/V — very large; practical units: μF (10⁻⁶ F), pF (10⁻¹² F)
Symbol in circuits: two parallel lines
16.3.1 Capacitance of a Spherical Conductor
Capacitance ∝ radius
Example: r = 0.18 m → C = 20 pF
16.3.2 Parallel Plate Capacitor
Two parallel plates of area A, separation d, charge ±q, uniform field between plates (d ≪ plate size):
∝ area A; ∝ 1/separation d
E between plates: E = σ/ε₀ = q/(ε₀A); V = Ed
Dielectric increases capacitance K-fold
Bringing earthed plate near isolated plate also increases C (induction)
16.3.3 Dielectric Constant
K for mica ≈ 6, paper ≈ 3.6, metals K → ∞
Force between charges in medium reduced by factor K
16.4 Grouping of Capacitors
16.4.1 Parallel Grouping
Same potential V across each capacitor; charges add:
Used for charge accumulation
Equivalent capacitance greater than any individual C
16.4.2 Series Grouping
Same charge q on each capacitor; potentials add:
Cs less than smallest individual capacitance
Capacitor with smallest C has largest V across it
Commercial types: paper (rolled foil + mylar), metal-plate (oil dielectric, variable air capacitors for tuning), electrolytic (thin oxide dielectric — observe polarity).
16.4.3 Energy Stored in a Capacitor
Charging transfers charge from battery; work stored as electrostatic energy in the field between plates:
Derived using average potential V/2 during charging
Discharge through resistor releases heat
High-voltage charged capacitors can be dangerous
16.5 Dielectrics and Dielectric Polarization
Dielectrics are insulators that transmit electric effects without conducting.
- Non-polar: N₂, O₂, CO₂, CH₄ — zero dipole moment normally; external E separates charge centres → induced dipole.
- Polar: H₂O, NH₃, HCl — permanent dipole moment.
Induced dipole moment in non-polar molecule under field E
Reduces V between plates → increases C = q/V
This is why dielectric constant K > 1
Applications of Electrostatics
- Capacitors in electronic circuits and power transmission.
- Gold-leaf electroscope for detecting charge.
- Lightning conductors (Benjamin Franklin) protect buildings.
- Photocopiers operate on electrostatic principles.
Quick Revision
- Potential: V = W/q₀; 1 V = 1 J/C; path independent work.
- Point charge: V = q/(4πε₀r); dipole: V = p cos θ/(4πε₀r²); equatorial V = 0.
- PE: U = q₁q₂/(4πε₀r); dipole U = −p·E.
- Field-potential: E = −dV/dr; E = 0 inside conductor; electrostatic shielding.
- Capacitance: C = Q/V; C₀ = ε₀A/d; C = KC₀; spherical C = 4πε₀r.
- Grouping: parallel Cₚ = ΣCᵢ; series 1/Cₛ = Σ(1/Cᵢ).
- Energy: U = ½CV² = q²/(2C); dielectric reduces E, increases C.
Q1. The SI unit of electric potential is:
Q2. Electric potential due to a point charge varies with distance r as:
Q3. Electric potential on the equatorial line of a dipole is:
Q4. The relation between electric field E and potential V is:
Q5. Inside a conductor in electrostatic equilibrium, the electric field is:
Q6. Capacitance of a parallel plate air capacitor is given by:
Q7. When capacitors are connected in parallel, equivalent capacitance is:
Q8. When capacitors are connected in series, equivalent capacitance satisfies:
Q9. Energy stored in a charged capacitor is:
Q10. Inserting a dielectric (K > 1) between parallel plates while charge Q is held constant:
PYQ — Previous Year Questions
Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L16 — Electric Potential and Capacitors only. Use Model Answer for marking points; Explanation for concept clarity.
Chapter 16 — Electric Potential and Capacitors (L16)
9 questions · Section A (MCQ & objective) + Section B (short/long) · Sources: 312/TUS/104A, 312/MAY/204A–B, 68/ESS/1-312-A, Marking Scheme
Section A — Multiple Choice (1 mark)
PYQ1. 10 capacitors, each of capacitance 5 μF, are connected first in parallel and then in series. The ratio of maximum to minimum capacitance obtained is — (A) 100 : 1 (B) 50 : 1 (C) 10 : 1 (D) 5 : 1
Model Answer
Answer: (A) 100 : 1
Cp = 10 × 5 = 50 μF
1/Cs = 10 × (1/5) → Cs = 0.5 μF
Ratio = 50/0.5 = 100 : 1
Explanation
Parallel adds capacitances; series adds reciprocals. Max is parallel combination; min is series (L16 §16.4).
PYQ2. A very thin metal foil is introduced at the centre between the plates of a parallel-plate capacitor of capacitance C. Its new capacitance will be — (A) zero (B) 2C (C) C (D) C/2
Model Answer
Answer: (C) C
Thin conducting foil at centre is an equipotential surface; plate separation and geometry unchanged → capacitance remains C.
Explanation
Foil acquires induced charges but does not alter effective distance between outer plates. C depends on A, d, and dielectric — not on inserting a thin conductor at midpoint.
PYQ3. A parallel-plate capacitor is charged by a battery. The battery is disconnected and plate separation is increased. The potential difference between the plates will now — (A) increase (B) decrease (C) remain unchanged (D) first increase then decrease
Model Answer
Answer: (A) increase
Battery disconnected → Q constant. C = ε₀A/d decreases as d increases. V = Q/C → V increases.
Explanation
Classic board trap: with battery connected, V stays fixed and Q changes; disconnected, charge is trapped on plates (L16 §16.3.2).
PYQ4. Which of the following does not affect the capacitance of a capacitor? — (A) plate separation (B) plate area (C) dielectric constant of medium (D) charge on the plate
Model Answer
Answer: (D) charge on the plate
C = ε₀KA/d — depends on geometry and dielectric only, not on Q.
Explanation
More charge raises V proportionally (Q = CV); capacitance is a property of the capacitor structure (L16 §16.3).
PYQ5. Work done in displacing a charge q on an equipotential surface through a distance r will be — (A) q/(4πε₀r) (B) 4πε₀rq (C) q/(2πε₀r) (D) zero
Model Answer
Answer: (D) zero
On equipotential surface ΔV = 0 → W = qΔV = 0 for any displacement along the surface.
Explanation
E is perpendicular to equipotential; force has no component along the path. Fundamental property of conservative electrostatic field (L16 §16.1).
PYQ6. A particle of charge −1 μC is placed where electric potential is 100 V. Its electric potential energy is — (A) 10⁻⁸ J (B) 10⁻⁴ J (C) 10⁴ J (D) −10⁻⁴ J
Model Answer
Answer: (D) −10⁻⁴ J
U = qV = (−1×10⁻⁶)(100) = −10⁻⁴ J
Explanation
PE of a charge in an external potential: U = qV. Negative charge at positive potential has negative PE (L16 §16.1.3).
PYQ7. Derive the expression for capacitance of a parallel plate capacitor with plate area A and separation d.
Model Answer
Uniform field: E = σ/ε₀ = q/(ε₀A)
Potential difference: V = Ed = (q/ε₀A)·d
C = q/V → C = ε₀A/d
With dielectric: C = Kε₀A/d
Explanation
Marking scheme: σ/ε₀ for E, V = Ed, then C = q/V. Valid when d ≪ plate dimensions (uniform field).
PYQ8. Derive the expression for electric potential at a point on the axial line of an electric dipole (r ≫ separation).
Model Answer
At axial point P: V = V₊ + V₋ = (1/4πε₀)[q/(r−a) − q/(r+a)]
For r ≫ a: V ≈ (1/4πε₀)(2qa/r²) = p/(4πε₀r²) where p = 2qa
General form: V = p cos θ/(4πε₀r²)
Explanation
Superpose scalar potentials from ±q. On equatorial line (θ = 90°) V = 0 by symmetry.
PYQ9. Derive the expression for potential energy of an electric dipole of moment p placed in a uniform electric field E. When is the dipole in stable equilibrium?
Model Answer
Torque τ = pE sin θ. Work to rotate from θ₀ to θ:
W = ∫ pE sin θ dθ = pE(cos θ₀ − cos θ)
Taking θ₀ = 90° (U = 0 reference): U = −pE cos θ = −p·E
Stable equilibrium: θ = 0° (p parallel to E) — U minimum.
Explanation
Marking scheme integrates τ = pE sin θ. Unstable at θ = 180°. Links to L15 torque τ = pE sin θ.
Problem Solving — L16 Electric Potential and Capacitors
Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.
Potential at 0.50 m from +2 μC point charge (k=9e9)?
Solution — step by step with formulas
- V = 9e9×2e-6/0.5 = 3.6×10⁴ V.
Final answer: V = 3.6×10⁴ V
Formulas used in this problem
Textbook formal language
Potential is PE per unit positive charge; for a point charge V = kQ/r (V(∞)=0).
Working formula set for this problem: V = kQ/r; ΔV = −∫E·dl. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
How much energy per coulomb you’d gain bringing +1 C from infinity.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Electric potential
Equipotential surfaces are perpendicular to field lines.
Link to chapter notes (L16 — Electric potential): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: V = kQ/r; ΔV = −∫E·dl. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write V = kQ/r; ΔV = −∫E·dl before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Find PE of system of +1 μC and −1 μC separated by 0.10 m.
Solution — step by step with formulas
- U = 9e9×(1e-6)(−1e-6)/0.1 = −0.090 J.
Final answer: U = −0.090 J
Formulas used in this problem
Textbook formal language
Electrostatic PE of a pair is kq₁q₂/r with zero at infinite separation.
Working formula set for this problem: U = k q₁q₂/r. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Opposite charges have negative PE—bound, you’d need energy to separate them.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Potential energy of charges
Total PE for many charges is sum over unique pairs.
Link to chapter notes (L16 — Potential energy of charges): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: U = k q₁q₂/r. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write U = k q₁q₂/r before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Parallel plate capacitor: A = 0.02 m², d = 1.0 mm. Find C (ε₀=8.85×10⁻¹²).
Solution — step by step with formulas
- C = ε₀A/d = 8.85e-12×0.02/0.001 = 1.77×10⁻¹⁰ F.
Final answer: C ≈ 1.77×10⁻¹⁰ F
Formulas used in this problem
Textbook formal language
Capacitance is charge stored per unit potential difference.
Working formula set for this problem: C = Q/V; C = ε₀A/d (parallel plate). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Bigger plates or smaller gap store more charge per volt.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Capacitance
Dielectric multiplies C by κ (K).
Link to chapter notes (L16 — Capacitance): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: C = Q/V; C = ε₀A/d (parallel plate). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write C = Q/V; C = ε₀A/d (parallel plate) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Capacitor C = 2 μF charged to 100 V. Energy stored?
Solution — step by step with formulas
- U = ½×2e-6×10000 = 0.10 J.
Final answer: U = 0.10 J
Formulas used in this problem
Textbook formal language
Energy resides in the electric field between plates; U = ½CV².
Working formula set for this problem: U = ½ CV² = Q²/(2C). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Half C V squared—here 0.1 joule parked in the capacitor.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Energy in capacitor
Charging from a battery dissipates equal energy in resistance for series RC ideals.
Link to chapter notes (L16 — Energy in capacitor): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: U = ½ CV² = Q²/(2C). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write U = ½ CV² = Q²/(2C) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Two 2 μF capacitors: find equivalent in series and in parallel.
Solution — step by step with formulas
- Series: 1 μF.
- Parallel: 4 μF.
Final answer: Series 1 μF; parallel 4 μF
Formulas used in this problem
Textbook formal language
Series: same charge, voltages add; parallel: same V, charges add.
Working formula set for this problem: 1/C_s = Σ1/C_i; C_p = Σ C_i. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Series shrinks capacitance; parallel adds capacitances.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Capacitor combinations
Use reductions stepwise in mixed networks.
Link to chapter notes (L16 — Capacitor combinations): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: 1/C_s = Σ1/C_i; C_p = Σ C_i. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write 1/C_s = Σ1/C_i; C_p = Σ C_i before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
How does inserting a dielectric with κ = 2 (battery disconnected) change C, Q, V, U?
Solution — step by step with formulas
- C doubles; Q fixed; V halves; U halves.
Final answer: C↑, Q same, V↓, U↓ (disconnected)
Formulas used in this problem
Textbook formal language
With charge fixed, polarisation reduces field and potential for given Q; energy decreases.
Working formula set for this problem: C′ = κC. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Battery off means charge stuck. Dielectric makes voltage drop, energy drop, capacitance up.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Dielectric
If battery remains connected, V fixed, Q and U increase.
Link to chapter notes (L16 — Dielectric): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: C′ = κC. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write C′ = κC before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).