1. Chapter at a Glance
Electric potential tells us how much work is needed to bring a unit positive charge to a point. It is a scalar, so it is often easier to use than the electric field. This chapter covers potential due to charges and dipoles, equipotential surfaces, potential energy of a system of charges, conductors, capacitors, dielectrics, and the energy stored in a capacitor. NEET almost always has at least one question on capacitor combinations or dielectrics.
- Potential of point charges and the work done to move a charge.
- Relation between E and V (E = −dV/dr).
- Potential energy of a system of charges.
- Series and parallel capacitors; charge sharing and energy loss.
- Effect of a dielectric with the battery connected or disconnected.
2. Electric Potential
- Potential is a scalar; for many charges, simply add the potentials with their signs.
- Work done to move charge q from A to B: W = q(VB − VA). It does not depend on the path, because the electrostatic force is conservative.
- Potential of a dipole: V = kp cosθ / r². It is zero everywhere on the equatorial line.
- Field from potential: E = −dV/dr. The field points in the direction in which the potential falls fastest.
- 1 electron-volt (eV) = 1.6 × 10⁻¹⁹ J is the energy gained by an electron through 1 V.
At the midpoint of two equal positive charges, E = 0 but V = 2kq/r ≠ 0. At the midpoint of a dipole, V = 0 but E ≠ 0. Always check both separately.
3. Equipotential Surfaces
- An equipotential surface has the same potential at every point, so no work is done in moving a charge along it.
- Field lines are always perpendicular to equipotential surfaces.
- For a point charge, the equipotentials are concentric spheres; for a uniform field, they are parallel planes perpendicular to E.
- Equipotentials are closer together where the field is stronger.
- Two equipotential surfaces never intersect.
4. Potential Energy of a System of Charges
- U is positive for like charges (work must be done to push them together) and negative for unlike charges.
- For three charges, U = k(q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃).
- Potential energy of a dipole in a field: U = −pE cosθ.
- A charge q in an external potential V has potential energy qV.
5. Conductors in Electrostatics
- Inside a conductor, E = 0 and the potential is the same everywhere (the whole conductor is an equipotential).
- Excess charge resides on the outer surface; the field just outside is σ/ε₀, normal to the surface.
- Charge density is highest at sharp points (this is why lightning conductors are pointed).
- Electrostatic shielding: the field inside a closed conducting cavity is zero; sensitive instruments are kept in metal boxes.
- Dielectrics are insulators that get polarised in a field. The induced charges reduce the field inside by a factor K (dielectric constant).
6. Capacitors and Combinations
- The unit of capacitance is the farad (F); practical values are in μF and pF.
- C depends only on the geometry and the medium, not on Q or V.
- A slab of thickness t and dielectric constant K between plates: C = ε₀A/(d − t + t/K). A metal slab (K → ∞) gives C = ε₀A/(d − t).
| Property | Series | Parallel |
|---|---|---|
| Equivalent capacitance | 1/C = 1/C₁ + 1/C₂ + … | C = C₁ + C₂ + … |
| Same for every capacitor | Charge Q | Voltage V |
| Divided among capacitors | Voltage (V ∝ 1/C) | Charge (Q ∝ C) |
| NEET Note | Ceq is less than the smallest C | Ceq is more than the largest C |
7. Energy Stored and Dielectric Effects
| Dielectric K inserted | Battery connected (V fixed) | Battery removed (Q fixed) |
|---|---|---|
| Capacitance | K times | K times |
| Charge | K times | Unchanged |
| Voltage | Unchanged | 1/K times |
| Field between plates | Unchanged | 1/K times |
| Energy | K times | 1/K times |
When two charged capacitors are joined (positive to positive), the common potential is V = (C₁V₁ + C₂V₂)/(C₁ + C₂). Some energy is always lost as heat: ΔU = C₁C₂(V₁ − V₂)²/[2(C₁ + C₂)].
8. Formula Master Table
| Quantity | Formula |
|---|---|
| Potential of a point charge | V = kq/r |
| Work done | W = q(VB − VA) |
| Dipole potential | V = kp cosθ/r² |
| Field–potential relation | E = −dV/dr |
| Potential energy of two charges | U = kq₁q₂/r |
| Capacitance | C = Q/V |
| Parallel-plate capacitor | C = Kε₀A/d |
| Partial dielectric slab | C = ε₀A/(d − t + t/K) |
| Series / parallel | 1/C = Σ1/Cᵢ ; C = ΣCᵢ |
| Energy stored | U = ½CV² = Q²/2C |
| Energy density | u = ½ε₀E² |
| Common potential | V = (C₁V₁ + C₂V₂)/(C₁ + C₂) |
9. Master Summary Table
| Concept | Key idea | NEET Note |
|---|---|---|
| Potential | Scalar, kq/r | Add with sign |
| Equipotential | No work along it | ⊥ to field lines |
| E–V relation | E = −dV/dr | V falls along E |
| Conductor | Whole body is one equipotential | E = 0 inside |
| Dielectric | C becomes KC | Check whether the battery is connected |
| Series capacitors | Same Q | Smaller C takes larger V |
| Parallel capacitors | Same V | Larger C takes larger Q |
| Charge sharing | Energy always lost | Lost as heat and radiation |
10. Mind Map
- V = kq/r
- Dipole: kp cosθ/r²
- E = −dV/dr
- U = kq₁q₂/r
- Pairs add
- Dipole: −pE cosθ
- C = ε₀A/d
- Series / parallel
- U = ½CV²
- Polarisation
- C → KC
- Battery on / off
11. Tips and Tricks
- Two capacitors in series: C = C₁C₂/(C₁ + C₂) (product over sum).
- n identical capacitors: series C/n, parallel nC.
- In a dielectric question, first write down what stays constant: V (battery on) or Q (battery off).
- For work done by an external agent in slowly moving a charge, use W = qΔV; the kinetic energy does not change.
- Adding capacitors in series like resistors in series (it is the opposite).
- Forgetting that the energy stored decreases when a dielectric is inserted with the battery removed.
- Treating potential as a vector.
- Assuming that E = 0 means V = 0.
12. Important One-liners
- The potential on the equatorial line of a dipole is zero.
- No work is done in moving a charge on an equipotential surface.
- The surface of a charged conductor is an equipotential surface.
- Capacitance of an isolated sphere is 4πε₀R.
- Capacitance increases K times when the space is filled with a dielectric.
- Energy of a capacitor is stored in the electric field between its plates.
- 1 eV = 1.6 × 10⁻¹⁹ J.
- Pointed conductors leak charge quickly because the surface charge density is highest there.
13. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take k = 9 × 10⁹ N m² C⁻² and ε₀ = 8.854 × 10⁻¹² F m⁻¹. Click to reveal each answer and working.
Average level
Find the electric potential 30 cm from a point charge of 5 nC.
Show answer & solution
Concept: V = kq/r.
- V = 9 × 10⁹ × 5 × 10⁻⁹ / 0.3.
- V = 45/0.3 = 150 V.
Answer: 150 V
How much work is needed to carry a 2 μC charge from a point at 100 V to a point at 400 V?
Show answer & solution
Concept: W = q(VB − VA).
- ΔV = 400 − 100 = 300 V.
- W = 2 × 10⁻⁶ × 300 = 6 × 10⁻⁴ J.
Answer: 6 × 10⁻⁴ J
A parallel-plate air capacitor has plates of area 0.05 m² separated by 2 mm. Find its capacitance.
Show answer & solution
Concept: C = ε₀A/d.
- C = 8.854 × 10⁻¹² × 0.05 / 2 × 10⁻³.
- C = 2.21 × 10⁻¹⁰ F ≈ 221 pF.
Answer: about 221 pF
Capacitors of 4 μF and 12 μF are joined in series across 60 V. Find the equivalent capacitance, the charge on each and the voltage across each.
Show answer & solution
Concept: in series, both carry the same charge.
- C = 4 × 12/(4 + 12) = 3 μF.
- Q = CV = 3 × 60 = 180 μC on each.
- V₁ = 180/4 = 45 V; V₂ = 180/12 = 15 V (total 60 V ✓).
Answer: 3 μF; 180 μC each; 45 V and 15 V
A 5 μF capacitor is charged to 400 V. Find the energy stored in it.
Show answer & solution
Concept: U = ½CV².
- U = ½ × 5 × 10⁻⁶ × (400)².
- = ½ × 5 × 10⁻⁶ × 1.6 × 10⁵ = 0.4 J.
Answer: 0.4 J
Advanced level
Three charges of 2 μC each sit at the corners of an equilateral triangle of side 20 cm. Find the potential energy of the system.
Show answer & solution
Concept: add kq₁q₂/r for each of the three pairs.
- Each pair: kq²/a = 9 × 10⁹ × (2 × 10⁻⁶)² / 0.2 = 0.036/0.2 = 0.18 J.
- Three pairs: U = 3 × 0.18 = 0.54 J.
- This is also the work needed to assemble the charges from infinity.
Answer: 0.54 J
A 5 μF air capacitor is charged by a 20 V battery. A dielectric of K = 4 then fills the gap. Find the new charge, voltage and energy if (a) the battery stays connected, (b) the battery was disconnected first.
Show answer & solution
Given: initially Q = 100 μC, U = ½ × 5 × 10⁻⁶ × 400 = 1 mJ. New C = 4 × 5 = 20 μF.
- (a) V stays 20 V: Q = 20 × 20 = 400 μC; U = ½ × 20 × 10⁻⁶ × 400 = 4 mJ (K times).
- (b) Q stays 100 μC: V = 100/20 = 5 V; U = Q²/2C = (10⁻⁴)²/(2 × 20 × 10⁻⁶) = 0.25 mJ (1/K times).
Answer: (a) 400 μC, 20 V, 4 mJ; (b) 100 μC, 5 V, 0.25 mJ
A 3 μF capacitor charged to 100 V and a 6 μF capacitor charged to 40 V are joined positive plate to positive plate. Find the common potential and the energy lost.
Show answer & solution
Concept: charge is conserved; energy is not.
- Total charge = 3 × 100 + 6 × 40 = 540 μC; total C = 9 μF.
- V = 540/9 = 60 V.
- ΔU = C₁C₂(V₁ − V₂)²/[2(C₁ + C₂)] = (3 × 6 × 10⁻¹²)(60)²/(2 × 9 × 10⁻⁶).
- ΔU = 10⁻⁶ × 3600 = 3.6 × 10⁻³ J. (Check: 19.8 mJ before, 16.2 mJ after.)
Answer: 60 V; 3.6 mJ lost
The potential in a region is V = 4x² + 3y volts (x, y in metres). Find the electric field (components and magnitude) at the point (1 m, 2 m).
Show answer & solution
Concept: Ex = −∂V/∂x, Ey = −∂V/∂y.
- Ex = −8x = −8 V m⁻¹ at x = 1.
- Ey = −3 V m⁻¹ (constant).
- |E| = √(64 + 9) = √73 ≈ 8.54 V m⁻¹.
Answer: E = (−8, −3) V m⁻¹; magnitude about 8.54 V m⁻¹
A 2 μF and a 4 μF capacitor are in parallel, and this pair is in series with a 3 μF capacitor. The network is connected to 90 V. Find the equivalent capacitance and the charge on each capacitor.
Show answer & solution
Concept: reduce parallel first, then series.
- Parallel pair = 2 + 4 = 6 μF; with 3 μF in series: C = 6 × 3/9 = 2 μF.
- Total charge Q = 2 × 90 = 180 μC; this is the charge on the 3 μF capacitor.
- V across 3 μF = 180/3 = 60 V; across the pair = 30 V.
- Charge on 2 μF = 2 × 30 = 60 μC; on 4 μF = 4 × 30 = 120 μC.
Answer: 2 μF; 60 μC, 120 μC and 180 μC
14. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
The work done in moving a charge of 3 C between two points on an equipotential surface at 10 V is:
- 30 J
- 3.33 J
- Zero
- 10 J
Show solution
W = qΔV and ΔV = 0 on an equipotential surface, so W = 0. (A) wrongly uses the potential itself instead of the potential difference.
Answer: (C)
If the separation between the plates of a parallel-plate capacitor is doubled, its capacitance becomes:
- Double
- Half
- Four times
- Unchanged
Show solution
C = ε₀A/d ∝ 1/d, so doubling d halves C.
Answer: (B)
A capacitor is charged by a battery and then disconnected. A dielectric slab is now inserted. Which quantity increases?
- Charge
- Voltage
- Energy stored
- Capacitance
Show solution
With the battery disconnected, Q stays fixed. C increases to KC, so V = Q/C falls and U = Q²/2C falls. Only the capacitance increases.
Answer: (D)
At the midpoint between charges +q and −q:
- E = 0, V = 0
- E ≠ 0, V = 0
- E = 0, V ≠ 0
- E ≠ 0, V ≠ 0
Show solution
The potentials kq/r and −kq/r cancel, so V = 0. The fields of the two charges both point towards −q, so they add and E ≠ 0.
Answer: (B)
Three capacitors of 6 μF each are connected in series. The equivalent capacitance is:
- 18 μF
- 2 μF
- 3 μF
- 6 μF
Show solution
n identical capacitors in series give C/n = 6/3 = 2 μF. 18 μF is the parallel value.
Answer: (B)
Electric field lines and equipotential surfaces are:
- Parallel
- Perpendicular
- At 45°
- At any angle
Show solution
If E had a component along an equipotential surface, work would be done in moving along it. That contradicts ΔV = 0, so E must be perpendicular to the surface.
Answer: (B)
A capacitor C charged to V is joined to an identical uncharged capacitor. The fraction of the original energy lost is:
- 1/4
- 1/2
- 3/4
- Zero
Show solution
The common potential is V/2. Energy after = 2 × ½C(V/2)² = CV²/4, which is half of the initial CV²/2. So half the energy is lost as heat. Charge is conserved, but energy is not, so (D) is wrong.
Answer: (B)
A metal slab of thickness d/2 is inserted between the plates of a parallel-plate capacitor of plate separation d and capacitance C₀. The new capacitance is:
- C₀/2
- C₀
- 2C₀
- Infinite
Show solution
For a metal slab, C = ε₀A/(d − t) = ε₀A/(d/2) = 2C₀. The capacitance is infinite only if the slab fills the whole gap, which would short the plates.
Answer: (C)
15. Quick Revision Summary
- V = kq/r (scalar); W = qΔV; E = −dV/dr.
- Equipotentials ⊥ field lines; conductors are equipotential.
- U = kq₁q₂/r for each pair.
- C = Kε₀A/d; series adds reciprocals, parallel adds directly.
- U = ½CV²; battery on → V fixed, battery off → Q fixed.