Introduction

The concept of potential energy is familiar from mechanics (e.g., gravitational or spring potential energy). Work done by an external force against a conservative force (like gravity) is stored as potential energy. Electrostatic force, like gravitational force, is an inverse-square force and is **conservative**. Therefore, the concept of electrostatic potential energy can be defined for a charge in an electrostatic field.

Consider a static electrostatic field $\vec{E}$ created by some charge configuration (e.g., a charge $Q$ at the origin). If we move a test charge $q$ from a point R to a point P against the electric force $\vec{F}_E = q\vec{E}$, we apply an external force $\vec{F}_{ext} = -\vec{F}_E$. If this is done slowly (without acceleration), the work done by the external force $W_{RP}$ is stored as the change in potential energy ($\Delta U$) of the charge $q$.

$\mathbf{W_{RP} = \int_R^P \vec{F}_{ext} \cdot d\vec{l} = - \int_R^P \vec{F}_E \cdot d\vec{l}}$

This work done by the external force increases the potential energy of charge $q$. The potential energy difference between points P and R is $\mathbf{\Delta U = U_P - U_R = W_{RP}}$.

Key points about electrostatic potential energy:

• The work done by the electrostatic field (or the negative of work done by external force) depends only on the initial and final points (R and P) and is **independent of the path** taken. This is a characteristic of a conservative force.
• Potential energy is defined up to an additive constant. Only the **potential energy difference** is physically significant.
• A convenient choice is to set the electrostatic potential energy to **zero at infinity**. With this convention, the potential energy of a charge $q$ at a point P is the work done by the external force in bringing the charge $q$ from infinity to point P. $\mathbf{U_P = W_{\infty P}}$.

Electrostatic Potential

Since the work done in moving a test charge $q$ in an electrostatic field is proportional to $q$, it is convenient to define a quantity that is independent of the test charge. This leads to the definition of **electrostatic potential** ($V$). Electrostatic potential is the work done per unit test charge.

The **electrostatic potential difference** between two points P and R, $V_P - V_R$, is defined as the work done by the external force in moving a unit positive test charge from point R to point P (without acceleration):

$\mathbf{V_P - V_R = \frac{U_P - U_R}{q} = \frac{W_{RP}}{q}}$

Like potential energy, only the potential difference is physically significant. Choosing the electrostatic potential to be **zero at infinity** ($V_\infty = 0$), the electrostatic potential $V$ at any point P is the work done by the external force in bringing a unit positive test charge from infinity to that point:

$\mathbf{V_P = \frac{W_{\infty P}}{q}}$ (for unit positive charge, $q=1$)

In simpler terms, the electrostatic potential $V$ at a point is the potential energy per unit positive charge at that point relative to the potential energy at infinity (taken as zero).

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Potential Due To A Point Charge

Consider a point charge $Q$ placed at the origin. To find the electrostatic potential $V(r)$ at a point P at a distance $r$ from the origin, we calculate the work done in bringing a unit positive test charge from infinity to P.

Choosing the radial path from infinity to P, the force on a unit positive charge at an intermediate point $P'$ at distance $r'$ is $\frac{1}{4\pi\epsilon_0} \frac{Q}{r'^2} \hat{r}'$. The work done by the external force against this electrostatic force for an infinitesimal displacement $d\vec{r}'$ is $dW = -\vec{F}_E \cdot d\vec{r}' = -(\frac{1}{4\pi\epsilon_0} \frac{Q}{r'^2} \hat{r}') \cdot (dr' \hat{r}') = -\frac{1}{4\pi\epsilon_0} \frac{Q}{r'^2} dr'$.

The total work done in bringing the unit charge from infinity ($r'=\infty$) to $r'$ is:

$W = \int_\infty^r -\frac{1}{4\pi\epsilon_0} \frac{Q}{r'^2} dr' = -\frac{Q}{4\pi\epsilon_0} \int_\infty^r r'^{-2} dr' = -\frac{Q}{4\pi\epsilon_0} \left[ \frac{r'^{-1}}{-1} \right]_\infty^r = \frac{Q}{4\pi\epsilon_0} \left[ \frac{1}{r'} \right]_\infty^r = \frac{Q}{4\pi\epsilon_0} \left(\frac{1}{r} - \frac{1}{\infty}\right) = \frac{Q}{4\pi\epsilon_0 r}$.

By definition, this work done is the electrostatic potential at distance $r$:

$\mathbf{V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}}$

This formula is valid for positive or negative $Q$. Potential $V$ is positive for $Q>0$ and negative for $Q<0$. It depends only on the distance $r$ from the point charge and is independent of direction. The potential due to a point charge falls off as **1/r** with distance.

Comparing with electric field $E(r) = \frac{1}{4\pi\epsilon_0} \frac{|Q|}{r^2}$, potential falls off slower than the magnitude of the electric field.

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Potential Due To An Electric Dipole

An electric dipole consists of two equal and opposite charges, $+q$ and $-q$, separated by a distance $2a$. The dipole moment is $\vec{p} = q(2a) \hat{p}$, directed from $-q$ to $+q$. The total charge is zero, but the electric field and potential are not.

Since electric potential obeys the **superposition principle**, the potential $V$ at any point P due to a dipole is the algebraic sum of the potentials due to $+q$ and $-q$ separately.

Let P be at a distance $r_1$ from $+q$ and $r_2$ from $-q$. The potential at P is:

$\mathbf{V = \frac{1}{4\pi\epsilon_0} \frac{q}{r_1} + \frac{1}{4\pi\epsilon_0} \frac{-q}{r_2} = \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2}\right)}$

Using the geometry (Fig. 2.5), for a point P with position vector $\vec{r}$ from the dipole center, making an angle $\theta$ with the dipole axis $\vec{p}$, the distances $r_1$ and $r_2$ can be approximated for points far from the dipole ($r \gg a$).

Using approximations $r_1 \approx r - a\cos\theta$ and $r_2 \approx r + a\cos\theta$ for $r \gg a$, the potential simplifies to:

$\mathbf{V \approx \frac{1}{4\pi\epsilon_0} \frac{q(2a\cos\theta)}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{p\cos\theta}{r^2}}$ (for $r \gg a$)

Since $p\cos\theta = \vec{p} \cdot \hat{r}$ (where $\hat{r}$ is the unit vector along $\vec{r}$), the potential is:

$\mathbf{V(\vec{r}) \approx \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2}}$ (for $r \gg a$)

For a point dipole at the origin, this formula is exact.

Contrasting features of dipole potential compared to a single charge potential:

• The potential depends on the angle $\theta$ between $\vec{r}$ and $\vec{p}$, not just the distance $r$. It is axially symmetric around the dipole axis.
• The potential falls off as **1/r$^2$** at large distances, faster than the 1/r fall-off of a single charge potential.

On the dipole axis ($\theta = 0^\circ$ or $180^\circ$, $\cos\theta = \pm 1$): $V = \pm \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$ (positive towards $+q$, negative towards $-q$).

On the equatorial plane ($\theta = 90^\circ$, $\cos\theta = 0$): $V = 0$.

Potential Due To A System Of Charges

The electrostatic potential at a point due to a system of point charges is the **algebraic sum** of the potentials due to each individual charge at that point.

For a system of $n$ charges $q_1, q_2, ..., q_n$ located at positions $\vec{r}_1, \vec{r}_2, ..., \vec{r}_n$, the potential $V(\vec{r})$ at a point P with position vector $\vec{r}$ is:

$\mathbf{V(\vec{r}) = V_1(\vec{r}) + V_2(\vec{r}) + ... + V_n(\vec{r})}$

where $V_i(\vec{r})$ is the potential at P due to charge $q_i$ alone, given by $V_i(\vec{r}) = \frac{1}{4\pi\epsilon_0} \frac{q_i}{r_{iP}}$. Here $r_{iP}$ is the distance from charge $q_i$ to point P.

So, the total potential is:

$\mathbf{V(\vec{r}) = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_{iP}}}$

For a **continuous charge distribution** with volume charge density $\rho(\vec{r}')$, the potential at point P with position vector $\vec{R}$ is obtained by integrating the potential contribution from each infinitesimal volume element $dV'$ at $\vec{r}'$: $dV = \frac{1}{4\pi\epsilon_0} \frac{dq}{|\vec{R} - \vec{r}'|} = \frac{1}{4\pi\epsilon_0} \frac{\rho(\vec{r}') dV'}{|\vec{R} - \vec{r}'|}$.

$\mathbf{V(\vec{R}) = \int \frac{1}{4\pi\epsilon_0} \frac{\rho(\vec{r}')}{|\vec{R} - \vec{r}'|} dV'}$

Similar integrals are used for surface or line charge distributions.

For a uniformly charged spherical shell of radius R and total charge q:

• Outside the shell ($r \ge R$): The potential is the same as that of a point charge $q$ at the center. $\mathbf{V(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r}}$.
• Inside the shell ($r < R$): The electric field is zero. Since $E = -dV/dr$, zero field implies constant potential. The potential inside is constant and equal to its value on the surface. $\mathbf{V(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{R}}$.

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Equipotential Surfaces

An **equipotential surface** is a surface where the electric potential $V$ is the same at every point. No work is done by the electric field when a test charge is moved from one point to another on an equipotential surface, and the potential difference between any two points on the surface is zero.

For a single point charge $q$, the potential is $V(r) = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$. For $V$ to be constant, $r$ must be constant. Thus, the equipotential surfaces for a single point charge are **concentric spheres** centered at the location of the charge.

The electric field lines are always **perpendicular** (normal) to the equipotential surfaces at every point. This is because if there were a component of the electric field tangential to the equipotential surface, work would be done moving a charge along the surface, which contradicts the definition of an equipotential surface (zero potential difference).

For a uniform electric field $\vec{E}$ (e.g., along the x-axis), the potential is constant on planes perpendicular to the field direction ($V = -\vec{E} \cdot \vec{r} + C = -E x + C$). Thus, the equipotential surfaces for a uniform field are **planes perpendicular to the direction of the field**.

Equipotential surfaces are a useful way to visualise the electric field configuration. They are always perpendicular to electric field lines.

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Relation Between Field And Potential

There is a direct mathematical relationship between the electric field $\vec{E}$ and the electrostatic potential $V$. We know that the potential difference between two points P and R is related to the work done by the field: $V_P - V_R = - \int_R^P \vec{E} \cdot d\vec{l}$.

Consider two closely spaced equipotential surfaces with potentials $V$ and $V+dV$. Let $dl$ be the perpendicular distance between them, in the direction of the electric field (where potential decreases). The work done by the field in moving a unit positive charge from the surface at $V+dV$ to the surface at $V$ along this perpendicular is $E \times dl$. This work is equal to the potential difference $(V+dV) - V = dV$. Wait, the work done by the field is $W = -\Delta U = -(V_{final} - V_{initial}) = V_{initial} - V_{final}$. So work done by the field moving from $V+dV$ to $V$ is $(V+dV) - V = dV$. The force by the field is $\vec{E}$. Displacement is $d\vec{l}$. Work $dW = \vec{E} \cdot d\vec{l}$.

If $d\vec{l}$ is a small displacement, $dV = V(\vec{r} + d\vec{l}) - V(\vec{r}) = -\vec{E} \cdot d\vec{l}$.

If we choose the displacement $d\vec{l}$ to be perpendicular to the equipotential surface (along the direction of $\vec{E}$), then $\vec{E}$ and $d\vec{l}$ are parallel, so $\vec{E} \cdot d\vec{l} = E dl$. And the potential change is $dV = V(\vec{r} + d\vec{l}) - V(\vec{r})$. Since $d\vec{l}$ is in the direction of $\vec{E}$, potential decreases, so $dV$ is negative. The potential difference between two closely spaced surfaces separated by $dl$ perpendicular to the field is $dV = -\vec{E} \cdot d\vec{l} = -E dl$.

So, $E = -\frac{dV}{dl}$, where $dl$ is the displacement in the direction of the field. This means the electric field component in any direction is the negative of the rate of change of potential with distance in that direction.

The magnitude of the electric field is given by the magnitude of the steepest decrease in potential per unit distance. This occurs in the direction normal to the equipotential surface. $\mathbf{|\vec{E}| = - \frac{dV}{dn}}$, where $dn$ is the displacement normal to the equipotential surface in the direction of decreasing potential (direction of $\vec{E}$).

Two key conclusions:

• Electric field points in the direction where the potential decreases most rapidly (steepest).
• The magnitude of the electric field at a point is the rate of decrease of potential with distance in the direction normal to the equipotential surface at that point.

In 3D Cartesian coordinates, if $V(x,y,z)$ is the potential, the electric field components are $E_x = -\frac{\partial V}{\partial x}$, $E_y = -\frac{\partial V}{\partial y}$, $E_z = -\frac{\partial V}{\partial z}$. This can be written compactly as $\mathbf{\vec{E} = -\nabla V}$, where $\nabla V$ is the gradient of $V$.

Potential Energy Of A System Of Charges

The potential energy of a system of charges is the work done by an external agency to assemble the charges at their given locations, starting from infinite separation (where potential energy is zero). Since electrostatic force is conservative, this work is independent of the path or the order in which the charges are brought.

Case 1: Potential energy of two charges $q_1$ and $q_2$.

Bring $q_1$ from infinity to its location $\vec{r}_1$. No work is done as there's no external field initially ($W_1 = 0$). Then, bring $q_2$ from infinity to its location $\vec{r}_2$. Work is done against the field created by $q_1$. The potential at $\vec{r}_2$ due to $q_1$ is $V_1(\vec{r}_2) = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_{12}}$, where $r_{12}$ is the distance between $q_1$ and $q_2$. The work done in bringing $q_2$ is $W_2 = q_2 V_1(\vec{r}_2) = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$.

The total work done, which is the potential energy $U$ of the system, is $U = W_1 + W_2 = 0 + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$.

$\mathbf{U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}}$

If $q_1 q_2 > 0$ (like charges), $U > 0$, as work is done against repulsion. If $q_1 q_2 < 0$ (unlike charges), $U < 0$, as work is done by attraction (or negative work by external force).

Case 2: Potential energy of a system of three charges $q_1, q_2, q_3$.

Bring $q_1$ to $\vec{r}_1$ ($W_1 = 0$). Bring $q_2$ to $\vec{r}_2$ ($W_2 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$). Bring $q_3$ to $\vec{r}_3$ against the fields of both $q_1$ and $q_2$. The potential at $\vec{r}_3$ due to $q_1$ and $q_2$ is $V_{12}(\vec{r}_3) = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_{13}} + \frac{1}{4\pi\epsilon_0} \frac{q_2}{r_{23}}$. The work done in bringing $q_3$ is $W_3 = q_3 V_{12}(\vec{r}_3) = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_3}{r_{13}} + \frac{1}{4\pi\epsilon_0} \frac{q_2 q_3}{r_{23}}$.

The total potential energy $U = W_1 + W_2 + W_3 = 0 + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}} + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_3}{r_{13}} + \frac{1}{4\pi\epsilon_0} \frac{q_2 q_3}{r_{23}}$.

$\mathbf{U = \frac{1}{4\pi\epsilon_0} \left(\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}}\right)}$

This generalises to any number of point charges. For $n$ charges, sum the potential energy for all unique pairs of charges.

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Potential Energy In An External Field

In contrast to the potential energy of a system of charges which considers only the interactions among those charges, the potential energy of a charge (or system of charges) in an external field considers the interaction of the charge(s) with an electric field produced by other, external sources. The external field $\vec{E}$ (or potential $V$) is given; we are not concerned with the sources creating it.

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Potential Energy Of A Single Charge

The potential $V(\vec{r})$ at a point $\vec{r}$ in an external field is the work done by the external force in bringing a unit positive charge from infinity to $\vec{r}$. The work done in bringing a charge $q$ from infinity to $\vec{r}$ is $q$ times this value. This work is stored as the potential energy $U$ of the charge $q$ at point $\vec{r}$ in the external field.

$\mathbf{U(\vec{r}) = qV(\vec{r})}$

where $V(\vec{r})$ is the external potential at $\vec{r}$. The potential at infinity is taken as zero.

The electron volt (eV) is a unit of energy commonly used in atomic and nuclear physics. $1 \text{ eV}$ is the energy gained by an electron (or any charge $e$) when it is accelerated through a potential difference of 1 volt. $1 \text{ eV} = (1.602 \times 10^{-19}\text{ C})(1 \text{ V}) = 1.602 \times 10^{-19}\text{ J}$. Multiples like keV ($10^3$ eV), MeV ($10^6$ eV), GeV ($10^9$ eV), TeV ($10^{12}$ eV) are also used.

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Potential Energy Of A System Of Two Charges In An External Field

For a system of two charges $q_1$ and $q_2$ located at $\vec{r}_1$ and $\vec{r}_2$ respectively, in an external field $V(\vec{r})$, the total potential energy is the sum of three contributions:

• The potential energy of $q_1$ in the external field: $U_1 = q_1 V(\vec{r}_1)$.
• The potential energy of $q_2$ in the external field: $U_2 = q_2 V(\vec{r}_2)$.
• The potential energy of interaction between $q_1$ and $q_2$: $U_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$, where $r_{12} = |\vec{r}_1 - \vec{r}_2|$.

The total potential energy $U$ of the system is the sum of these contributions:

$\mathbf{U = q_1 V(\vec{r}_1) + q_2 V(\vec{r}_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}}$

This is the work done to bring $q_1$ from infinity to $\vec{r}_1$ (work is $q_1V(\vec{r}_1)$ against the external field) and then bring $q_2$ from infinity to $\vec{r}_2$ (work is $q_2V(\vec{r}_2)$ against the external field PLUS $\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$ against the field of $q_1$).

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Potential Energy Of A Dipole In An External Field

Consider an electric dipole with dipole moment $\vec{p}$ placed in a uniform external electric field $\vec{E}$. We know the net force is zero, but there is a torque $\vec{\tau} = \vec{p} \times \vec{E}$ that tends to align $\vec{p}$ with $\vec{E}$.

The potential energy $U$ of the dipole depends on its orientation relative to the field, specifically the angle $\theta$ between $\vec{p}$ and $\vec{E}$. To find the potential energy at an angle $\theta$, we calculate the work done by an external torque in rotating the dipole from a reference orientation (e.g., $\theta_0$) to $\theta$. The external torque must be equal and opposite to the electric torque ($\tau_{ext} = pE\sin\theta$).

The work done by the external torque to rotate the dipole from $\theta_0$ to $\theta$ is $W = \int_{\theta_0}^\theta \tau_{ext} d\theta = \int_{\theta_0}^\theta pE\sin\theta d\theta = pE [-\cos\theta]_{\theta_0}^\theta = pE (\cos\theta_0 - \cos\theta)$.

This work is stored as potential energy. Setting the potential energy to zero when the dipole moment is perpendicular to the field ($\theta_0 = 90^\circ$, $\cos 90^\circ = 0$), the potential energy at angle $\theta$ is:

$\mathbf{U(\theta) = pE (\cos 90^\circ - \cos\theta) = -pE\cos\theta}$

In vector form:

$\mathbf{U = -\vec{p} \cdot \vec{E}}$

The potential energy is minimum ($U = -pE$) when $\vec{p}$ is parallel to $\vec{E}$ ($\theta = 0^\circ$, stable equilibrium) and maximum ($U = +pE$) when $\vec{p}$ is antiparallel to $\vec{E}$ ($\theta = 180^\circ$, unstable equilibrium).

For a non-uniform electric field, the force on the dipole is non-zero, and the calculation of potential energy is more complex, involving the rate of change of the electric field. The force is related to the spatial derivative of the potential energy.

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Electrostatics Of Conductors

Conductors (like metals) have free charge carriers (electrons) that can move throughout the material. When a conductor is in an electrostatic situation (charges are at rest, no current flow), there are specific properties of the electric field and potential associated with it:

1. **Inside a conductor, the electrostatic field is zero ($E = 0$).** Free charges will redistribute themselves instantly until the net field inside is zero, otherwise, they would move. This is the defining property of a conductor in a static field.

2. **At the surface of a charged conductor, the electrostatic field is normal to the surface at every point.** If there were a tangential component of the field, free charges on the surface would move, which contradicts the static condition.

3. **In the static situation, the interior of a conductor has no excess charge; any excess charge resides only on the surface.** This is a direct consequence of Gauss's law and $E=0$ inside. Any closed Gaussian surface drawn entirely within the conductor encloses zero net charge. Since this is true for any small volume element, there is no net charge density inside.

4. **Electrostatic potential is constant throughout the volume of the conductor and is equal to the potential on its surface.** Since $E=0$ inside and the tangential component of $E$ is zero on the surface, no work is done in moving a charge within or on the surface of the conductor. Thus, there is no potential difference between any points within or on the surface of a single conductor.

5. **The electric field at the surface of a charged conductor is proportional to the local surface charge density $\sigma$.** Using a small pillbox Gaussian surface partly inside and partly outside the conductor surface, and applying Gauss's law along with $E=0$ inside and $E$ being normal outside, we get:

$\mathbf{\vec{E} = \frac{\sigma}{\epsilon_0} \hat{n}}$

where $\sigma$ is the surface charge density at that point, and $\hat{n}$ is the unit vector normal to the surface in the outward direction. For $\sigma > 0$, $\vec{E}$ is outward; for $\sigma < 0$, $\vec{E}$ is inward.

6. **Electrostatic shielding:** The electric field inside a cavity within a conductor is zero, provided there are no charges placed inside the cavity. This is true regardless of the size and shape of the cavity, the charge on the conductor, or any external electric fields. This effect, known as electrostatic shielding, is used to protect sensitive equipment from external electric fields by enclosing them within a conducting shell (Faraday cage).

Summary of conductor properties in static fields:

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Dielectrics And Polarisation

Dielectrics are **non-conducting substances** (insulators) that have no or very few free charge carriers. When a dielectric is placed in an external electric field, unlike conductors where free charges move to cancel the field inside, charges in a dielectric are bound within molecules and cannot move freely.

Instead, the external field causes molecular dipoles to form or align. This phenomenon is called **polarisation**.

• Non-polar molecules: Have centers of positive and negative charges coinciding in the absence of a field (zero permanent dipole moment, e.g., $\textsf{O}_2$, $\textsf{H}_2$). In an external field, the charges are displaced in opposite directions, creating an **induced dipole moment** aligned with the field.
• Polar molecules: Have permanent dipole moments even in the absence of a field (e.g., $\textsf{HCl}$, $\textsf{H}_2\text{O}$). In the absence of a field, these dipoles are randomly oriented due to thermal agitation, so the net dipole moment of the material is zero. In an external field, the permanent dipoles tend to **align** with the field, resulting in a non-zero net dipole moment for the material.

In both cases, a dielectric develops a **net dipole moment per unit volume**, called **polarisation ($\vec{P}$)**, in the presence of an external electric field. For linear isotropic dielectrics, $\vec{P}$ is proportional to the external field $\vec{E}$: $\mathbf{\vec{P} = \epsilon_0 \chi_e \vec{E}}$, where $\chi_e$ is the electric susceptibility of the dielectric medium.

This polarisation effectively creates **induced surface charges** on the surfaces of the dielectric exposed to the field (where the ends of the molecular dipoles are unneutralised). These induced surface charges produce an electric field inside the dielectric that **opposes** the external field. The net electric field inside the dielectric is therefore reduced compared to the external field:

$\mathbf{\vec{E}_{\text{net}} = \vec{E}_{\text{ext}} - \vec{E}_{\text{induced}}}$

Unlike a conductor, the induced field in a dielectric does not completely cancel the external field; it only reduces its magnitude. The extent of reduction depends on the dielectric material.

Capacitors And Capacitance

A **capacitor** is a system consisting of two conductors separated by an insulator (dielectric). In typical use, the two conductors carry equal and opposite charges, $+Q$ and $-Q$. Let $V_1$ and $V_2$ be their potentials, and $V = V_1 - V_2$ be the potential difference between them (taking $V_1$ as the higher potential). $Q$ is referred to as the charge of the capacitor, even though the total charge is zero; $Q$ specifically refers to the magnitude of charge on the positive conductor.

The electric field in the region between the conductors is proportional to the charge $Q$. Consequently, the potential difference $V$ between the conductors is also proportional to $Q$. The ratio of the charge $Q$ to the potential difference $V$ is a constant value for a given capacitor called its **capacitance** ($C$).

$\mathbf{C = \frac{Q}{V}}$

The capacitance $C$ is a geometrical property of the capacitor; it depends only on the shape, size, and relative position of the two conductors, and the nature of the insulator (dielectric) between them. It does not depend on the charge $Q$ or potential difference $V$ across the capacitor.

The SI unit of capacitance is the **farad (F)**. One farad is equal to one coulomb per volt: $1 \text{ F} = 1 \text{ C V}^{-1}$.

A capacitance of 1 farad is a very large value in practice. Typical capacitances are in the microfarad ($\mu$F = $10^{-6}$ F), nanofarad (nF = $10^{-9}$ F), or picofarad (pF = $10^{-12}$ F) range.

A capacitor with large capacitance can store a large amount of charge $Q$ for a relatively small potential difference $V$. This is important because high potential differences can lead to strong electric fields that can cause the insulating medium to break down (lose its insulating properties, e.g., air ionization), leading to charge leakage. The maximum electric field a dielectric can withstand without breakdown is called its **dielectric strength**.

Symbolically, a capacitor is represented as || or for variable capacitance.

The Parallel Plate Capacitor

The simplest type of capacitor is a **parallel plate capacitor**, consisting of two large, flat parallel conducting plates separated by a small distance $d$. Let $A$ be the area of each plate and $d$ be the distance between them, with $d \ll \sqrt{A}$ (distance much smaller than plate dimensions) so that edge effects can be ignored.

When the plates are charged with $+Q$ and $-Q$, the surface charge densities are $\sigma = +Q/A$ and $-\sigma = -Q/A$. For $d \ll \sqrt{A}$, the electric field between the plates can be approximated as uniform, similar to the field between two infinite charged planes.

Outside the plates, the fields due to the two planes cancel out, $E_{\text{outer}} = 0$.

Between the plates, the fields due to the two planes add up. The field from the positive plate is $\sigma/(2\epsilon_0)$, pointing away from it. The field from the negative plate is $\sigma/(2\epsilon_0)$, pointing towards it. Both fields point from the positive plate towards the negative plate.

The net electric field between the plates is $E = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$.

Substituting $\sigma = Q/A$, the field between the plates is $\mathbf{E = \frac{Q}{\epsilon_0 A}}$. The direction is from the positive to the negative plate.

Since the field is uniform, the potential difference $V$ between the plates is the product of the field strength and the distance $d$:

$V = E d = \frac{Q}{\epsilon_0 A} d = \frac{Qd}{\epsilon_0 A}$.

By the definition of capacitance $C = Q/V$, the capacitance of a parallel plate capacitor with vacuum between the plates is:

$\mathbf{C_0 = \frac{Q}{V} = \frac{Q}{Qd/(\epsilon_0 A)} = \frac{\epsilon_0 A}{d}}$

This formula shows that the capacitance is directly proportional to the plate area $A$ and inversely proportional to the separation $d$, and depends on the permittivity of the medium ($\epsilon_0$ for vacuum).

Effect Of Dielectric On Capacitance

When a dielectric material is inserted between the plates of a capacitor, it becomes polarised by the electric field created by the charges on the plates. This polarisation creates an internal electric field within the dielectric that opposes the original field.

Let $E_0$ be the electric field between the plates with vacuum. $E_0 = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$. The potential difference is $V_0 = E_0 d = \frac{Qd}{\epsilon_0 A}$. The capacitance is $C_0 = \frac{\epsilon_0 A}{d}$.

When a dielectric is inserted, the net electric field $E$ inside the dielectric is reduced compared to $E_0$. The reduction factor is the **dielectric constant** $K$ (also called relative permittivity $\epsilon_r$).

$\mathbf{E = \frac{E_0}{K} = \frac{\sigma}{K\epsilon_0} = \frac{Q}{K\epsilon_0 A}}$

The potential difference $V$ between the plates with the dielectric is now $V = E d = \frac{E_0}{K} d = \frac{V_0}{K}$.

$V = \frac{Qd}{K\epsilon_0 A}$.

The capacitance $C$ with the dielectric is $C = \frac{Q}{V} = \frac{Q}{Qd/(K\epsilon_0 A)} = \frac{K\epsilon_0 A}{d}$.

$\mathbf{C = K C_0}$

where $C_0$ is the capacitance with vacuum. The product $K\epsilon_0$ is the permittivity of the dielectric medium, $\epsilon = K\epsilon_0$.

The dielectric constant $K$ is a dimensionless quantity ($K \ge 1$) that tells us by what factor the capacitance increases when the space between the plates is completely filled with the dielectric compared to vacuum.

Dielectric materials increase the capacitance of a capacitor. This means for the same charge $Q$, the potential difference $V$ is reduced (by a factor $K$), or for the same potential difference $V$, a larger charge $Q$ can be stored ($Q=CV=KC_0V$).

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Combination Of Capacitors

Multiple capacitors can be combined to obtain an equivalent capacitance. Two common ways of combining capacitors are in series and in parallel.

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Capacitors In Series

When capacitors are connected in **series**, they are connected end-to-end in a single path, as shown in Fig. 2.26 and 2.27. In a series combination connected to a battery, the charges on each capacitor are the **same** magnitude ($+Q$ and $-Q$). The total potential difference ($V$) across the combination is the sum of the potential differences across each individual capacitor ($V_1, V_2, ..., V_n$).

$V = V_1 + V_2 + ... + V_n$

Since $V_i = Q_i/C_i$ and $Q_i = Q$ for all $i$ in series:

$V = \frac{Q}{C_1} + \frac{Q}{C_2} + ... + \frac{Q}{C_n} = Q \left(\frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}\right)$

The equivalent capacitance $C_{eq}$ of the series combination is defined by $V = Q/C_{eq}$.

So, $\frac{Q}{C_{eq}} = Q \left(\frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}\right)$

$\mathbf{\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}}$

For a series combination, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. The equivalent capacitance in series is always less than the smallest individual capacitance.

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Capacitors In Parallel

When capacitors are connected in **parallel**, they are connected across the same two points, as shown in Fig. 2.28. In a parallel combination connected to a battery, the **potential difference ($V$) across each capacitor is the same** and equal to the battery voltage. The total charge ($Q$) supplied by the battery is the sum of the charges on each individual capacitor ($Q_1, Q_2, ..., Q_n$).

$Q = Q_1 + Q_2 + ... + Q_n$

Since $Q_i = C_i V_i$ and $V_i = V$ for all $i$ in parallel:

$Q = C_1 V + C_2 V + ... + C_n V = V (C_1 + C_2 + ... + C_n)$

The equivalent capacitance $C_{eq}$ of the parallel combination is defined by $Q = C_{eq} V$.

So, $C_{eq} V = V (C_1 + C_2 + ... + C_n)$

$\mathbf{C_{eq} = C_1 + C_2 + ... + C_n}$

For a parallel combination, the equivalent capacitance is the direct sum of the individual capacitances. The equivalent capacitance in parallel is always greater than the largest individual capacitance.

Answer:

Energy Stored In A Capacitor

A capacitor stores electrical potential energy in the electric field between its conductors. To charge a capacitor with $+Q$ and $-Q$, work must be done by an external agency to transfer charge from the negatively charged conductor to the positively charged conductor against the electric field.

Consider charging a capacitor by transferring small charge increments $dQ'$ from conductor 2 (negative) to conductor 1 (positive). At an intermediate stage when the charges are $+Q'$ and $-Q'$, the potential difference is $V' = Q'/C$. The work done $dW$ to transfer an additional small charge $dQ'$ is $dW = V' dQ' = (Q'/C) dQ'$.

The total work done $W$ to charge the capacitor from $Q'=0$ to $Q'=Q$ is the integral of $dW$:

$W = \int_0^Q \frac{Q'}{C} dQ' = \frac{1}{C} \int_0^Q Q' dQ' = \frac{1}{C} \left[\frac{Q'^2}{2}\right]_0^Q = \frac{1}{C} \left(\frac{Q^2}{2} - 0\right) = \frac{Q^2}{2C}$.

This work done is stored as the electrostatic potential energy $U$ of the capacitor:

$\mathbf{U = \frac{Q^2}{2C}}$

Using the relationship $Q=CV$, this energy can also be expressed in terms of $C$ and $V$, or $Q$ and $V$:

$\mathbf{U = \frac{(CV)^2}{2C} = \frac{1}{2} CV^2}$

$\mathbf{U = \frac{Q^2}{2(Q/V)} = \frac{1}{2} QV}$

So, the energy stored in a capacitor can be calculated using $\frac{1}{2} \frac{Q^2}{C}$, $\frac{1}{2} CV^2$, or $\frac{1}{2} QV$.

This stored energy can be thought of as residing in the electric field between the capacitor plates. For a parallel plate capacitor with vacuum, $C = \epsilon_0 A/d$ and $V = Ed$, so $Q = CV = \epsilon_0 A E$.

$U = \frac{1}{2} CV^2 = \frac{1}{2} \left(\frac{\epsilon_0 A}{d}\right) (Ed)^2 = \frac{1}{2} \epsilon_0 A d E^2$.

The volume between the plates is $Ad$. Dividing the energy by this volume gives the **energy density** ($u$) of the electric field:

$\mathbf{u = \frac{U}{\text{Volume}} = \frac{\frac{1}{2} \epsilon_0 A d E^2}{Ad} = \frac{1}{2} \epsilon_0 E^2}$

This result for the energy density of an electric field is general and applies to any region of space where an electric field exists.

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Summary

This chapter focuses on electrostatic potential, potential energy, conductors, dielectrics, and capacitors.

• Electrostatic force is conservative, allowing definition of **potential energy**. Potential energy difference $\Delta U = q(V_P - V_R) = W_{RP}$ (work by external force). Choosing $U_\infty = 0$, $U_P = W_{\infty P}$.
• **Electrostatic potential** $V$ is potential energy per unit charge ($V = U/q$). Choosing $V_\infty = 0$, $V_P = W_{\infty P}/q$ (work by external force on unit charge). Potential difference is physically significant.
• Potential due to a point charge $Q$ at distance $r$: $\mathbf{V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}}$.
• Potential due to a dipole $\vec{p}$ at distance $r \gg a$: $\mathbf{V(\vec{r}) \approx \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2}}$. Falls off as $1/r^2$.
• Potential due to system of charges: Sum of individual potentials.
• **Equipotential surfaces** have constant potential. Electric field is perpendicular to equipotential surfaces and points in direction of steepest potential decrease ($|\vec{E}| = -dV/dn$).
• Potential energy of system of charges: Sum of pairwise interaction energies ($U = \sum_{pairs} \frac{1}{4\pi\epsilon_0} \frac{q_i q_j}{r_{ij}}$).
• Potential energy of charge $q$ in external potential $V(\vec{r})$ is $qV(\vec{r})$. For system of charges $q_1, q_2$ in external field: $U = q_1 V(\vec{r}_1) + q_2 V(\vec{r}_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}$.
• Potential energy of dipole $\vec{p}$ in uniform field $\vec{E}$: $U = -\vec{p} \cdot \vec{E}$. Torque is $\vec{\tau} = \vec{p} \times \vec{E}$.
• **Conductors** in static fields: $E=0$ inside, $E$ normal to surface ($\frac{\sigma}{\epsilon_0}\hat{n}$), excess charge on surface, constant potential. Electrostatic shielding: $E=0$ inside a cavity (with no charges).
• **Dielectrics** are insulators. External field polarises dielectrics (induces/aligns dipoles), reducing net field inside.
• **Capacitor:** Two conductors separated by insulator, storing charge ($\pm Q$) with potential difference $V$. Capacitance $\mathbf{C = Q/V}$. Unit Farad (F).
• Parallel plate capacitor: $C_0 = \frac{\epsilon_0 A}{d}$ (vacuum). With dielectric $K$: $C = KC_0 = \frac{K\epsilon_0 A}{d}$. $K$ is dielectric constant.
• Capacitor combinations: Series: $\frac{1}{C_{eq}} = \sum \frac{1}{C_i}$. Parallel: $C_{eq} = \sum C_i$.
• **Energy stored** in capacitor: $\mathbf{U = \frac{1}{2} CV^2 = \frac{1}{2} QV = \frac{Q^2}{2C}}$. Energy density of electric field $u = \frac{1}{2}\epsilon_0 E^2$.

Exercises

Questions covering calculations of potential, potential energy, electric field, capacitance for various charge configurations and capacitor setups. Includes problems involving properties of conductors and dielectrics, combinations of capacitors, energy storage, and related concepts.

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