Comprehensive Study Notes: Electrostatic Potential and Capacitance

• Introduction to Electrostatic Potential and Conservative Forces

The principles of potential energy, as previously established in Class XI (Chapters 6 and 8), serve as the foundation for understanding electrostatics. When an external force performs work to move a body against a restoring force, such as a spring force or a gravitational force, that work is stored as potential energy ($U$). If the external force is subsequently removed, the body translates this stored energy into kinetic energy ($K$), losing an equivalent amount of potential energy such that the total mechanical energy ($K + U$) remains conserved. Forces that possess this property are classified as conservative forces. Examples include spring forces and gravitational forces. In electrostatics, the Coulomb force between two stationary charges is also a conservative force. This shared characteristic arises because both gravitational and Coulomb forces demonstrate an inverse-square dependence on distance ($1/r^2$), with the primary difference being the proportionality constants: masses in the law of gravitation are replaced by charges in Coulomb's law.

In the context of an electrostatic field ($\mathbf{E}$) generated by a charge configuration, one can define electrostatic potential energy for a test charge ($q$). If a test charge ($q$) is moved from a point ($R$) to a point ($P$) against the repulsive force ($\mathbf{F}_E$) of a source charge ($Q$) at the origin, work must be performed by an external force ($\mathbf{F}_{\text{ext}}$). For this definition, it is assumed that the test charge ($q$) is small enough not to disturb the original configuration of ($Q$). Furthermore, the movement occurs without acceleration, meaning the external force is exactly equal and opposite to the electric force ($\mathbf{F}_{\text{ext}} = -\mathbf{F}_E$). The work done by the external force is fully stored as the potential energy of the charge at its final position. If the external force is removed at point ($P$), the electric force will propel the charge away, converting the stored potential energy into kinetic energy while maintaining energy conservation.

Mathematical Definition of Potential Energy Difference

The work done ($W_{RP}$) by an external force in moving a charge ($q$) from point ($R$) to point ($P$) is given by the line integral of the external force: $W_{RP} = \int_R^P \mathbf{F}_{\text{ext}} \cdot d\mathbf{r} = -\int_R^P \mathbf{F}_E \cdot d\mathbf{r}$. This work increases the potential energy of the charge, and the potential energy difference ($\Delta U$) between points ($P$) and ($R$) is defined as $\Delta U = U_P - U_R = W_{RP}$. It is important to note that because the displacement is in the opposite sense to the electric force, the work done by the electric field itself is negative ($-W_{RP}$). The fundamental characteristic of this relationship is that the work done depends only on the initial ($R$) and final ($P$) positions and is entirely independent of the path taken. The concept of potential energy would not be meaningful if work varied with the path taken between points.

The actual value of potential energy at a specific point is not physically significant; rather, it is the difference in potential energy that matters. An arbitrary constant ($\alpha$) can be added to the potential energy at every point without altering the difference: $(U_P + \alpha) - (U_R + \alpha) = U_P - U_R$. Traditionally, the point of zero potential energy is chosen to be at infinity ($R \rightarrow \infty$). Under this convention, the potential energy ($U_P$) of a charge ($q$) at point ($P$) is defined as the work done by an external force in bringing the charge ($q$) from infinity to that point: $U_P = W_{\infty P}$.

Electrostatic Potential

Electrostatic potential ($V$) is introduced to provide a characteristic of the electric field that is independent of the test charge ($q$). Since the work done in moving a charge is proportional to its magnitude ($W \propto q$), dividing the work by the charge yields a constant value relevant to the field itself. The potential difference ($V_P - V_R$) between points ($P$) and ($R$) is the work done per unit positive charge in moving it from ($R$) to ($P$): $V_P - V_R = \frac{1}{q} (U_P - U_R)$. Like potential energy, it is the potential difference that is physically meaningful, and a zero point must be established. By choosing potential to be zero at infinity, the electrostatic potential at any point ($P$) is defined as the work done by an external force in bringing a unit positive charge from infinity to that point without acceleration.

Historically, the unit of potential is the Volt, named after Count Alessandro Volta (1745–1827), an Italian physicist who established that electricity could be generated by wet bodies sandwiched between dissimilar metals, leading to the development of the first battery (the voltaic pile). To precisely determine potential, one should consider an infinitesimal test charge ($\delta q$), calculate the corresponding work ($\delta W$), and take the ratio $\delta W / \delta q$. In all paths, the external force must be equal and opposite to the electrostatic force at every point.

Potential due to a Point Charge and System of Charges

For a point charge ($Q$) located at the origin, the potential at a point ($P$) with position vector ($\mathbf{r}$) is derived by calculating work along a radial path from infinity. At an intermediate point ($P'$) at distance ($r'$), the force on a unit positive charge is given by $\mathbf{F} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r'^2} \mathbf{\hat{r}}'$. The total work done by integrating from $r' = \infty$ to $r' = r$ is $W = \int_\infty^r -\frac{Q}{4\pi\varepsilon_0 r'^2} dr' = \frac{Q}{4\pi\varepsilon_0 r}$. Thus, the potential due to a point charge is $V(r) = \frac{Q}{4\pi\varepsilon_0 r}$. This formula is valid for any sign of ($Q$). If (Q < 0), the potential ($V$) is negative, indicating that the electrostatic force does positive work (attraction), while the external force does negative work. The potential ($V$) varies as ($1/r$), while the electric field ($E$) varies as ($1/r^2$).

For a system of multiple charges ($q_1, q_2, \dots, q_n$), the total potential at a point is the algebraic sum of individual potentials due to each charge, according to the superposition principle: $V = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^n \frac{q_i}{r_{iP}}$. For a continuous charge distribution with density ($\rho$), the potential is found by integrating the contributions from small volume elements ($\rho \Delta v$). In the specific case of a uniformly charged spherical shell of radius ($R$): outside the shell ($r \geq R$), the potential is $V = \frac{q}{4\pi\varepsilon_0 r}$, acting as if all charge is at the center. Inside the shell (r < R), the electric field is zero, meaning the potential is constant and equal to its value at the surface: $V = \frac{q}{4\pi\varepsilon_0 R}$.

Potential due to an Electric Dipole

An electric dipole consists of two charges ($q$) and ($-q$) separated by a distance ($2a$). The potential at a distance ($r$) from the center of the dipole is the sum of potentials from both charges ($V = \frac{1}{4\pi\varepsilon_0} (\frac{q}{r_1} - \frac{q}{r_2})$). For distances large compared to the dipole size ($r \gg a$), geometry and binomial expansion lead to the approximate potential: $V = \frac{p \cos(\theta)}{4\pi\varepsilon_0 r^2} = \frac{1}{4\pi\varepsilon_0} \frac{\mathbf{p} \cdot \mathbf{\hat{r}}}{r^2}$. Here, ($p = 2qa$) is the dipole moment and ($\theta$) is the angle between the position vector ($\mathbf{r}$) and the dipole moment ($\mathbf{p}$).

Contrasting point charges and dipoles:

1. Dipole potential depends on both ($r$) and angle ($\theta$), whereas point charge potential is spherically symmetric (depends only on $r$). However, dipole potential is axially symmetric about ($\mathbf{p}$).

2. Dipole potential falls off as ($1/r^2$), faster than the ($1/r$) fall-off of a point charge potential.

3. Potential in the equatorial plane ($\theta = \pi/2$) is zero for a dipole, while on the axis ($\theta = 0, \pi$) it is maximal ($V = \pm \frac{p}{4\pi\varepsilon_0 r^2}$).

Equipotential Surfaces

An equipotential surface is defined as a surface where the potential remains constant at every point. For a single point charge, equipotential surfaces are a series of concentric spheres centered on the charge. For a uniform electric field, these surfaces are planes oriented normal to the direction of the field. A critical property of equipotential surfaces is that the electric field ($\mathbf{E}$) is always normal to the surface at every point. If ($\mathbf{E}$) was not normal, it would have a tangential component along the surface, requiring work to move a charge between points on the surface, which contradicts the definition of an equipotential surface where potential difference is zero.

The relationship between electric field and potential is expressed by the gradient: $|E| = -\frac{\delta V}{\delta l}$. This implies two conclusions: first, the electric field points in the direction where the potential decreases most steeply. Second, the magnitude of the electric field is equal to the change in potential per unit displacement normal to the equipotential surface.

Potential Energy of a System of Charges

The potential energy ($U$) of a system of charges is the work done to assemble the configuration from infinity. For two charges ($q_1$) and ($q_2$) separated by distance ($r_{12}$): work to bring ($q_1$) is zero; work to bring ($q_2$) in the field of ($q_1$) is $q_2 V_1(r_2)$, giving $U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}}$. If (q_1 q_2 > 0), work is positive (repulsive force), and if (q_1 q_2 < 0), work is negative (attractive force).

For a system of three charges ($q_1, q_2, q_3$), the total potential energy is the sum of work for each step: $U = \frac{1}{4\pi\varepsilon_0} (\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}})$. This logic extends to any number of charges. This potential energy is a characteristic of the current configuration and is independent of the assembly sequence.

Potential Energy in an External Field

When a charge ($q$) is placed in an external field ($\mathbf{E}$) with potential ($V(\mathbf{r})$), its potential energy is $U = qV(\mathbf{r})$. Energy in such contexts often uses the electron volt ($1\,eV = 1.6 \times 10^{-19}\,J$). For two charges ($q_1$) and ($q_2$) in an external field, the total potential energy includes the interaction with the external field and their mutual interaction: $U = q_1 V(\mathbf{r}_1) + q_2 V(\mathbf{r}_2) + \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}}$.

For a dipole with moment ($\mathbf{p}$) in a uniform external field ($\mathbf{E}$), the dipole experiences a torque ($\tau = \mathbf{p} \times \mathbf{E}$). The work done to rotate the dipole from ($\theta_0$) to ($\theta_1$) is $W = \int pE \sin(\theta) d\theta = pE(\cos \theta_0 - \cos \theta_1)$. Taking ($\theta_0 = \pi/2$) as the reference zero, the potential energy is $U(\theta) = -pE \cos(\theta) = -\mathbf{p} \cdot \mathbf{E}$. This formula accounts for the work done against the external field while ignoring the constant mutual energy of the internal dipole charges.

Electrostatics of Conductors

Metallic conductors consist of free electrons that act as charge carriers within a lattice of fixed positive ions. Key properties include:

1. Inside a conductor, the electrostatic field is zero: Charge carriers drift until the internal field is cancelled in the static state.

2. At the surface, the field must be normal to the surface: Tangential components would cause charges to move along the surface, which is not possible in a static situation.

3. Inside a conductor, there is no excess charge: By Gauss's law, since ($E=0$) inside, flux through any internal surface is zero, meaning net charge resides only on the exterior.

4. Potential is constant throughout the volume: Since ($E=0$) inside and has no tangential component on the surface, no work is required to move charges within or on the surface.

5. Surface electric field magnitude: $E = \frac{\sigma}{\varepsilon_0}$, where ($\sigma$) is surface charge density. The vector form is $\mathbf{E} = \frac{\sigma}{\varepsilon_0} \mathbf{\hat{n}}$.

6. Electrostatic Shielding: The field inside any cavity in a conductor is zero, regardless of the size/shape of the cavity or external fields. This shields the interior from external electrical influences.

Dielectrics and Polarisation

Dielectrics are insulators with no free charge carriers. In an external field ($\mathbf{E}_0$), a dielectric becomes polarised. Non-polar molecules (e.g., $O_2, H_2$) develop induced dipole moments as positive and negative centers shift. Polar molecules (e.g., $HCl, H_2O$) have permanent moments that align with the field despite thermal agitation. Polarisation ($\mathbf{P}$) is the dipole moment per unit volume. For linear isotropic dielectrics, $\mathbf{P} = \chi_e \mathbf{E}$, where ($\chi_e$) is electrical susceptibility.

The induced surface charges on the dielectric create an internal field that opposes and reduces ($\mathbf{E}_0$). The total field is reduced but not zeroed as in conductors. We define electric displacement ($\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}$). In a linear medium, $\mathbf{D} = \varepsilon_0 K \mathbf{E}$, where ($K$) is the dielectric constant and ($\varepsilon = \varepsilon_0 K$) is the permittivity of the medium. The dielectric constant is related to susceptibility by $K = 1 + \frac{\chi_e}{\varepsilon_0}$.

Capacitors and Capacitance

A capacitor consists of two conductors separated by an insulator. It stores charge ($Q$) at a potential difference ($V$), where the ratio ($C = Q/V$) is the capacitance. ($C$) depends on geometry and the dielectric between the conductors. The SI unit is the Farad ($1\,F = 1\,C/V$), though common sub-multiples like ($\mu F, nF, pF$) are used. Dielectric strength is the maximum field a material can withstand before breakdown; for air it is ($3 \times 10^6\,V\,m^{-1}$).

For a parallel plate capacitor in vacuum with plate area ($A$) and separation ($d$), the field is $E = \sigma / \varepsilon_0 = Q / (A \varepsilon_0)$. The potential is $V = Ed$, and capacitance is $C_0 = \frac{\varepsilon_0 A}{d}$. When a dielectric is inserted, the potential decreases to ($V_0/K$) and capacitance increases by the factor ($K$): $C = K C_0 = \frac{\varepsilon A}{d}$.

Combinations and Energy Stored

Capacitors can be combined in series or parallel:

• Series: The charge ($Q$) is the same for all. The inverse of the equivalent capacitance ($C_s$) is the sum of inverses: $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n}$.

• Parallel: The potential ($V$) is the same for all. Equivalent capacitance ($C_p$) is the sum of individual capacitances: $C_p = C_1 + C_2 + \dots + C_n$.

The energy ($U$) stored in a capacitor is the work done to charge it. For an intermediate charge ($Q'$) and potential ($V' = Q'/C$), the work to add ($\delta Q'$) is $\delta W = V' \delta Q'$. Integrating from 0 to ($Q$) yields $U = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} CV^2 = \frac{1}{2} QV$. This energy is stored in the electric field between the plates.

Questions & Discussion

Question (Example 2.2): Two charges ($3 \times 10^{-8}\,C$) and ($-2 \times 10^{-8}\,C$) are 15 cm apart. Where is the potential zero? Response: Let ($x$) be the distance from the positive charge. Setting total potential to zero: $\frac{1}{4\pi\varepsilon_0} (\frac{3 \times 10^{-8}}{x} + \frac{-2 \times 10^{-8}}{15-x}) = 0$ gives $x = 9\,cm$. On the extended line beyond the negative charge, $\frac{3}{x} = \frac{2}{x-15}$ gives $x = 45\,cm$.

Question (Example 2.3): Comparison of potential and work for points near positive vs negative charges. Reponse: For a positive charge, potential (V_P > V_Q) for points ($P, Q$) where ($P$) is closer. Potential energy of a small negative charge is higher at ($Q$). Work done by the field in moving a positive charge ($Q \rightarrow P$) is negative as it is against the repulsive force. For a negative charge, potential (V_B > V_A) if ($B$) is further away (less negative). A negative charge's kinetic energy decreases moving from ($B \rightarrow A$) due to repulsion from the central negative charge.

Question (Example 2.7): Conceptual applications of electrostatics. Response: A comb attracts paper because it is charged by friction and polarizes molecules in paper. Wet hair reduces friction, thus preventing charging. Aircraft tires are made conducting to discharge static electricity to ground. Birds don't get shocked on wires because there is no potential difference between their feet; a grounded man creates a potential difference circuit.