I. Electrostatic Potential Energy and Conservative Forces

Potential energy is introduced as work done by an external force in moving a body against a conservative force, such as spring force or gravitational force. This work is stored as potential energy.
When the external force is removed, the body gains kinetic energy and loses an equal amount of potential energy, indicating that the sum of kinetic and potential energies is conserved.
Conservative forces are those for which the sum of kinetic and potential energies is conserved. Examples include spring force, gravitational force, and Coulomb force between stationary charges.
The Coulomb force is a conservative force, similar to gravitational force, as both have an inverse-square dependence on distance.
Electrostatic potential energy of a charge in an electrostatic field can be defined, analogous to gravitational potential energy.
To bring a test charge q from point R to point P against a repulsive electric force, an external force (F_ext) is applied equal and opposite to the electric force (F_E = -F_ext). This ensures no net force or acceleration, meaning the charge is moved with infinitesimally slow constant speed.
In this scenario, the work done by the external force is the negative of the work done by the electric force and is fully stored as potential energy of the charge.
Potential energy difference (ΔU) between two points P and R is defined as the work done (W_RP) by the external force in moving a charge q from R to P: ΔU = U_P - U_R = W_RP.
The work done by an electrostatic field in moving a charge depends only on the initial and final points and is independent of the path taken. This path-independence is a fundamental characteristic of a conservative force and makes the concept of potential energy meaningful.
Actual potential energy values are not physically significant; only the potential energy difference is significant. An arbitrary constant can always be added to the potential energy at every point without changing the difference.
A convenient choice for defining potential energy is to set electrostatic potential energy to zero at infinity.
With this choice, the potential energy of charge q at a point is defined as the work done by the external force in bringing the charge q from infinity to that point.

II. Electrostatic Potential (Voltage)

Electrostatic potential (V) is derived by dividing the work done on a test charge q by the amount of charge q, making the quantity independent of q and characteristic of the electric field.
Potential difference (V_P - V_R) is defined as the work done by an external force in bringing a unit positive charge from point R to P: V_P - V_R = (U_P - U_R) / q.
Similar to potential energy, only the potential difference is physically significant, not the actual value of potential.
If potential is chosen to be zero at infinity, then the electrostatic potential (V) at any point is the work done in bringing a unit positive charge (without acceleration) from infinity to that point.
To accurately obtain work done per unit test charge, one should consider an infinitesimal test charge dq, find the work dW, and determine the ratio dW/dq.

III. Potential Due to Different Charge Configurations

Potential Due to a Point Charge (Q)
- For a point charge Q placed at the origin, the potential V(r) at any point P with position vector r is given by: V(r) = Q / (4πε₀r).
- If Q < 0, V < 0, meaning work done by external force per unit positive test charge is negative. This aligns with the attractive electrostatic force in such a case.
- The electrostatic potential V varies inversely with distance (1/r), while the electrostatic field E varies inversely with the square of the distance (1/r²).
Potential Due to an Electric Dipole
- An electric dipole consists of two charges q and -q separated by a small distance 2a, characterized by a dipole moment vector p = q × 2a pointing from -q to q.
- The electric field of a dipole falls off as 1/r³ at large distances, unlike the 1/r² dependence of a single charge.
- Due to the superposition principle, the potential due to a dipole is the algebraic sum of potentials due to its individual charges.
- For distances r much greater than a (r >> a), the electric potential V of a dipole is given by: V(r) = (1 / 4πε₀) * (p.r̂ / r²).
- Key contrasting features compared to a single charge:
  - The potential due to a dipole depends not only on r but also on the angle (θ) between the position vector r and the dipole moment vector p. It is axially symmetric about p.
  - The electric dipole potential falls off as 1/r² at large distances, unlike the 1/r characteristic of a single charge.
- Potential on the dipole axis (θ = 0 or θ = π) is V = ±p / (4πε₀r²).
- The potential in the equatorial plane (θ = π/2) is zero.
Potential Due to a System of Charges
- For a system of multiple point charges (q₁, q₂,..., q_n), the total potential V at any point P is the algebraic sum of the potentials due to the individual charges (superposition principle).
- V = V₁ + V₂ + ... + V_n.
- For a continuous charge distribution, the potential is found by integrating contributions from small volume elements.
- For a uniformly charged spherical shell:
  - Outside the shell (r ≥ R), the potential is as if the entire charge q is concentrated at the center: V = q / (4πε₀r).
  - Inside the shell (r < R), the electric field is zero, which implies the potential is constant and equal to its value at the surface: V = q / (4πε₀R).

IV. Equipotential Surfaces

An equipotential surface is a surface where the potential is constant at all points.
For a single point charge, equipotential surfaces are concentric spherical surfaces centered at the charge.
A crucial property: the electric field E at every point is normal (perpendicular) to the equipotential surface passing through that point. If E had a tangential component, work would be required to move a charge along the surface, contradicting the definition of an equipotential surface.
The electric field is in the direction in which the potential decreases steepest.
The magnitude of the electric field |E| is given by the change in potential per unit displacement normal to the equipotential surface: |E| = -dV/dl.
For a uniform electric field, equipotential surfaces are planes normal to the field direction.

V. Potential Energy of a System of Charges (Self-Interaction)

The potential energy stored in a system of charges is defined as the work done by an external agency in assembling the charges at their given locations from infinity.
For a system of two charges q₁ and q₂:
- The work done to bring q₁ is zero (no external field).
- The work done to bring q₂ is q₂ times the potential at r₂ due to q₁.
- The potential energy U of the system is U = q₁q₂ / (4πε₀r₁₂) where r₁₂ is the distance between them.
- If q₁q₂ > 0 (like charges), U is positive, indicating positive work is needed against repulsion to bring them together.
- If q₁q₂ < 0 (unlike charges), U is negative, indicating work is done by the attractive electrostatic force.
For a system of three charges q₁, q₂, q₃:
- The total potential energy U is the sum of the potential energies of all unique pairs: U = (1 / 4πε₀) * (q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃).
The final expression for potential energy is independent of the order in which the charges are assembled due to the conservative nature of the electrostatic force.

VI. Potential Energy in an External Field

This section addresses the potential energy of a charge (or system of charges) in an external field, where the field sources are external and not produced by the charge(s) in question.
Potential energy of a single charge q in an external potential V(r) is qV(r).
- The electron volt (eV) is a common unit of energy in atomic and nuclear physics: 1 eV = 1.6 × 10⁻¹⁹ J.
Potential energy of a system of two charges (q₁ at r₁, q₂ at r₂) in an external field:
- U = q₁V(r₁) + q₂V(r₂) + q₁q₂ / (4πε₀r₁₂). This includes energy due to interaction with the external field and mutual interaction between the charges.
Potential energy of a dipole (p) in a uniform external electric field E:
- The dipole experiences a torque τ = p × E.
- The potential energy U(θ) is given by U = -p.E. The zero potential energy is conventionally chosen when the dipole moment is perpendicular to the electric field (θ = π/2).

VII. Electrostatics of Conductors

Conductors contain mobile charge carriers (e.g., free electrons in metals, ions in electrolytes).
Key results regarding electrostatics of conductors in static situations:
1. Inside a conductor, the electrostatic field is zero. Free charges redistribute until the field becomes zero everywhere inside.
2. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point. This is because any tangential component would cause free charges to move.
3. The interior of a conductor can have no excess charge; any excess charge resides only on the surface in static situations. This is a consequence of Gauss's Law and the zero field inside.
4. Electrostatic potential is constant throughout the volume of the conductor and has the same value on its surface. No work is done in moving a test charge within or on the surface of a conductor.
5. The electric field at the surface of a charged conductor is given by E = (σ/ε₀) n̂, where σ is the surface charge density and n̂ is the outward normal unit vector.
6. Electrostatic shielding: The electric field inside a charge-free cavity within a conductor is zero, regardless of the cavity's shape or size, or external fields. This principle is used to protect sensitive instruments. Shielding works only from outside to inside, not vice-versa.

VIII. Dielectrics and Polarisation

Dielectrics are non-conducting substances with negligible charge carriers.
When a dielectric is placed in an external electric field, it does not fully cancel the field like a conductor but reduces it.
The external field induces dipole moments in the dielectric through polarisation.
Non-polar molecules have coincident centers of positive and negative charges; an external field displaces these centers, creating an induced dipole moment.
Polar molecules have inherent permanent dipole moments, which are randomly oriented in the absence of a field. An external field causes these permanent dipoles to tend to align with the field, resulting in a net dipole moment.
Polarisation (P) is the dipole moment per unit volume. For linear isotropic dielectrics, P = ε₀χ_e E, where χ_e is the electric susceptibility.
The uniform polarisation of a dielectric leads to induced surface charge densities (±σ_p) on its surfaces normal to the field. These induced charges produce a field that opposes the external field, thereby reducing the total field inside the dielectric. These are bound (not free) charges.

IX. Capacitors and Capacitance

A capacitor is a system of two conductors separated by an insulator (dielectric).
The conductors typically hold equal and opposite charges (+Q and -Q), with a potential difference V = V₁ - V₂ between them. The total charge of the capacitor is zero.
Capacitance (C) is defined as the ratio of the charge on one conductor to the potential difference between them: C = Q/V.
C is independent of Q or V; it depends only on the geometrical configuration (shape, size, separation) of the conductors and the nature of the insulator (dielectric) between them.
The SI unit of capacitance is the Farad (F), where 1 F = 1 Coulomb/Volt (C V⁻¹).
A capacitor with large capacitance can hold a large amount of charge Q at a relatively small potential difference V. This is important because high potential differences can lead to strong electric fields that ionize the surrounding air and cause charge leakage.
Dielectric strength is the maximum electric field a dielectric medium can withstand without breakdown.
The Farad (F) is a very large unit in practice; common units are microfarads (μF), nanofarads (nF), and picofarads (pF).
Parallel Plate Capacitor (with vacuum between plates):
- Consists of two large parallel conducting plates of area A separated by a small distance d.
- The electric field E between the plates is uniform and given by E = σ/ε₀ = Q/(Aε₀).
- The potential difference V is V = E_0 d = Qd / (Aε₀).
- The capacitance C is given by: C = ε₀A/d.

X. Effect of Dielectric on Capacitance

When a dielectric is inserted fully between the plates of a capacitor, the capacitance increases.
The dielectric is polarized by the field, creating induced surface charges ±σ_p.
The net electric field inside the dielectric becomes E = (σ - σ_p) / ε₀, which is less than E₀ (the field without dielectric).
The potential difference across the plates is reduced (V = E d), but the charge Q remains the same (if disconnected from source) or is increased (if connected to source).
The relationship is (σ - σ_p) = σ/K, where K is the dielectric constant of the material.
The new capacitance C with the dielectric is given by: C = K * (ε₀A/d) = KC₀.
The product ε₀K is called the permittivity of the medium (ε = ε₀K).
The dielectric constant K is a dimensionless ratio K = ε/ε₀. K > 1.
K is the factor by which the capacitance increases when the dielectric is inserted.

XI. Combination of Capacitors

Capacitors in Series
- When capacitors are connected in series (Fig. 2.26), the charge Q on each capacitor is the same.
- The total potential drop V across the combination is the sum of the potential drops across individual capacitors: V = V₁ + V₂ + ... + V_n.
- The formula for the effective capacitance C of n capacitors in series is: 1/C = 1/C₁ + 1/C₂ + ... + 1/C_n.
Capacitors in Parallel
- When capacitors are connected in parallel (Fig. 2.28), the same potential difference V is applied across all capacitors.
- The total charge Q of the equivalent capacitor is the sum of the charges on individual capacitors: Q = Q₁ + Q₂ + ... + Q_n.
- The formula for the effective capacitance C of n capacitors in parallel is: C = C₁ + C₂ + ... + C_n.

XII. Energy Stored in a Capacitor

A capacitor stores energy in its electric field.
The energy U stored in a capacitor of capacitance C with charge Q and voltage V can be expressed in several equivalent ways:
- U = (1/2)QV
- U = (1/2)CV²
- U = (1/2)Q²/C
This stored energy is the work done by an external agency in charging the capacitor. It is independent of the charging process.
The energy density u (energy stored per unit volume) in a region with an electric field E is given by: u = (1/2)ε₀E². This result is general, not just for parallel plate capacitors.
When a charged capacitor is connected to an uncharged capacitor, energy can be lost during the transient period as heat and electromagnetic radiation.

XIII. Important Clarifications / Points to Ponder

In electrostatics, when discussing forces on charges at rest, it's implied that other unspecified forces are present to oppose the net Coulomb force, keeping the charges static.
Capacitors are designed to confine electric field lines to a small region, allowing for significant field strength with a relatively small potential difference.
The electric field is discontinuous across the surface of a charged conducting shell (zero inside, σ/ε₀ n̂ outside), but the electric potential is continuous across the surface.
A dipole in an electric field experiences a torque that tends to align it with the field. Without a dissipative mechanism, it would oscillate; dissipation damps oscillations and leads to alignment.
The potential due to a charge q at its own location is not defined (it is infinite).
In the expression qV(r) for potential energy, V(r) specifically refers to the potential due to external charges, not the potential generated by the charge q itself.
Electrostatic shielding works in one direction only: it shields the inside of a conductor's cavity from outside electrical influences. However, charges placed inside the cavity are not shielded from the exterior of the conductor.

Here is a detailed conceptual breakdown of Chapter 2 – Electrostatic Potential and Capacitance from Class 12 CBSE Physics.

📘 Chapter 2: Electrostatic Potential and Capacitance

🔑 Key Concepts

Electrostatic Potential
- Work done per unit charge in bringing a test charge from infinity to that point:
  $V = \frac{W}{q}$
Potential due to:
- A point charge:
  $V = \frac{kQ}{r}$
- A system of charges: Algebraic sum of potentials
- An electric dipole at a point:
  $V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p\cos\theta}{r^2}$
Equipotential Surfaces
- Surface where potential is constant
- Always perpendicular to electric field lines
Electric Field and Potential Relation
$\vec{E} = -\nabla V \quad \text{or} \quad E = -\frac{dV}{dr}$
Potential Energy of:
- System of two charges:
  $U = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1q_2}{r}$
- Dipole in external field:
  $U = -\vec{p} \cdot \vec{E}$
Capacitor
- Device to store electrical energy
- Capacitance:
  $C = \frac{Q}{V}$
Parallel Plate Capacitor
$C = \frac{\varepsilon_0 A}{d} \quad \text{(without dielectric)} \quad , \quad C = \frac{K\varepsilon_0 A}{d} \quad \text{(with dielectric)}$
Effect of Dielectric
- Increases capacitance by factor K (dielectric constant)
Combination of Capacitors
- In Series:
  $\frac{1}{C_\text{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$
- In Parallel:
  $C_\text{eq} = C_1 + C_2 + \cdots$
Energy Stored in a Capacitor
$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$
Van de Graaff Generator
- Device to develop high voltage
- Uses electrostatic induction and action of points

🔢 Important Formulas

Concept	Formula
Electrostatic Potential	$V = \frac{W}{q}$ , $V = \frac{kQ}{r}$
Electric Field from Potential	$E = -\frac{dV}{dr}$
Capacitance	$C = \frac{Q}{V}$
Parallel Plate Capacitor	$C = \frac{\varepsilon_0 A}{d}$
Energy Stored	$U = \frac{1}{2}CV^2$
Series Combination	$\frac{1}{C} = \sum \frac{1}{C_i}$
Parallel Combination	$C = \sum C_i$

🔍 Applications

Calculating work done in electrostatics
Designing capacitors with required capacitance
Using capacitors in circuits for energy storage
High-voltage generation in physics experiments (Van de Graaff)

📘 NCERT Exercise Questions with Solutions (Class 12 Physics Chapter 2)

Q1. Two charges $5 \times 10^{-8} \, \text{C}$ and $-3 \times 10^{-8} \, \text{C}$ are located 16 cm apart. At what point on the line joining them is the electric potential zero?

Solution:
Let the point be at a distance $x$ from the 5 µC charge where potential is zero.
Apply:
$\frac{k \cdot 5 \times 10^{-8}}{x} = \frac{k \cdot 3 \times 10^{-8}}{16 - x}$
Solving:
$\frac{5}{x} = \frac{3}{16 - x} \Rightarrow 5(16 - x) = 3x \Rightarrow x = 10 \, \text{cm}$
Answer: 10 cm from the 5 μC charge.

Q2. A regular hexagon of side 10 cm has a charge 5 μC at each vertex. Find the potential at the center.

Solution:
All distances from center to each vertex are equal. So total potential:
$V = 6 \cdot \frac{k \cdot q}{r}$
Where $r = 10 \, \text{cm} = 0.1 \, \text{m}$ , $q = 5 \times 10^{-6}$
$V = 6 \cdot \frac{9 \times 10^9 \cdot 5 \times 10^{-6}}{0.1} = 2.7 \times 10^6 \, \text{V}$

Q3. Two charges 2 μC and –2 μC are placed at (0, 0) and (0, 6 cm). Find the potential at (0, 3 cm).

Solution:
The point is equidistant from both charges → net potential = 0
Answer: 0 V

Q4. A charge of 2 μC is placed at the center of a cube. What is the flux through one face?

Solution:
Total flux:
$\Phi = \frac{q}{\varepsilon_0} = \frac{2 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 2.26 \times 10^5 \, \text{Nm}^2/\text{C}$
Each face gets 1/6 of the flux:
$\Phi_{\text{face}} = \frac{1}{6} \cdot \Phi \approx 3.77 \times 10^4 \, \text{Nm}^2/\text{C}$

Q5. Derive expression for potential due to electric dipole at axial and equatorial line.

Solution:

Axial line:

$V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r^2}$

Equatorial line:

$V = 0 \quad \text{(because contributions cancel out)}$

Q6. Derive expression for capacitance of parallel plate capacitor.

Solution:
Without dielectric:
$C = \frac{\varepsilon_0 A}{d}$
With dielectric:
$C = \frac{K\varepsilon_0 A}{d}$

📋 CBSE Previous Year Questions (PYQs) with Best Answers

🔹 Q (2023)

Define electric potential. Draw equipotential surfaces for point charge.
Answer:
Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point.
Equipotential surfaces are concentric spheres around the charge. No work is done when moving along them.

🔹 Q (2022)

Derive the expression for energy stored in a capacitor.
Answer:
Work done:
$U = \int_0^Q \frac{q}{C} \, dq = \frac{1}{2} \cdot \frac{Q^2}{C}$
Also expressed as:
$U = \frac{1}{2}CV^2 = \frac{1}{2}QV$

🔹 Q (2021)

Capacitors 6 μF and 12 μF in series with 12V. Find:

Equivalent capacitance:
$\frac{1}{C} = \frac{1}{6} + \frac{1}{12} \Rightarrow C = 4 \, \mu F$

Energy:
$U = \frac{1}{2} \cdot 4 \cdot (12)^2 \times 10^{-6} = 2.88 \times 10^{-4} \, \text{J}$

🔹 Q (2020)

Explain Van de Graaff Generator.
Answer:

Works on electrostatic induction and action of points

Charge is transferred to a large conducting sphere via a moving belt

Used to generate high voltages (~10^7 V) for experiments like nuclear physics

Key parts: hollow sphere, belt, pulley, spray comb, discharge sphere

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Electrostatic Potential and Capacitance Class 12 – NCERT Concepts, Formulas & CBSE PYQs