Torque on a dipole in a uniform electric field, electric flux, Gauss law and applications

Torque on an Electric Dipole in a Uniform Electric Field

An electric dipole consists of two equal and opposite charges, $+q$ and $-q$, separated by a small distance $2a$. When this dipole is placed in a uniform external electric field $\vec{E}$, it experiences a net torque that tends to align it with the field.

Consider a dipole with charges $+q$ and $-q$ placed in a uniform electric field $\vec{E}$ such that the dipole moment vector $\vec{p}$ makes an angle $\theta$ with the direction of $\vec{E}$. The electric field exerts a force on each charge. The force on $+q$ is $\vec{F}_1 = q\vec{E}$ and the force on $-q$ is $\vec{F}_2 = -q\vec{E}$. These forces are equal in magnitude and opposite in direction, so the net force on the dipole is zero ($\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = q\vec{E} - q\vec{E} = 0$). However, these forces act at different points, creating a couple that produces a torque.

The magnitude of the force on each charge is $F = qE$. The perpendicular distance between the line of action of these forces is $d = (2a)\sin\theta$. The torque $\tau$ is the product of the magnitude of one of the forces and the perpendicular distance between the forces.

$\tau = F \times d = (qE) \times (2a\sin\theta)$

Since the magnitude of the electric dipole moment is $p = q(2a)$, we can write the torque as:

$\tau = pE\sin\theta$

This torque tends to rotate the dipole clockwise, aligning it with the electric field. The torque is a vector quantity. The direction of the torque vector is perpendicular to the plane containing $\vec{p}$ and $\vec{E}$, given by the right-hand rule. The torque can be expressed in vector form as:

$\vec{\tau} = \vec{p} \times \vec{E}$

Special Cases:

• When $\theta = 0^\circ$ (dipole aligned with the field), $\tau = pE\sin(0^\circ) = 0$. This is a position of stable equilibrium.
• When $\theta = 90^\circ$ (dipole perpendicular to the field), $\tau = pE\sin(90^\circ) = pE$. This is the maximum torque.
• When $\theta = 180^\circ$ (dipole anti-aligned with the field), $\tau = pE\sin(180^\circ) = 0$. This is a position of unstable equilibrium.

Work Done in Rotating a Dipole: The work done by the external agent to rotate the dipole from an angle $\theta_1$ to $\theta_2$ against the torque of the electric field is given by:

$W = \int_{\theta_1}^{\theta_2} \tau d\theta = \int_{\theta_1}^{\theta_2} pE\sin\theta d\theta = pE [-\cos\theta]_{\theta_1}^{\theta_2} = pE (\cos\theta_1 - \cos\theta_2)$

If the dipole is initially at rest at an angle $\theta$ with the field and is rotated from $\theta = 0^\circ$ to $\theta$, the work done is:

$W = pE(\cos 0^\circ - \cos\theta) = pE(1 - \cos\theta)$

This work done is stored as potential energy $U$ of the dipole in the electric field. The potential energy is given by:

$U = -pE\cos\theta = -\vec{p} \cdot \vec{E}$

The potential energy is minimum ($U = -pE$) when $\theta = 0^\circ$ (stable equilibrium) and maximum ($U = +pE$) when $\theta = 180^\circ$ (unstable equilibrium). When $\theta = 90^\circ$, $U = 0$.

Electric Flux

Electric flux ($\Phi_E$) is a measure of the total electric field passing through a given area. It quantifies the number of electric field lines that pierce through a surface.

Imagine an electric field $\vec{E}$ and a small, flat surface element of area $dA$. Let $\hat{n}$ be the unit vector normal to the surface element. The electric flux $d\Phi_E$ through this small area is given by the dot product of the electric field vector and the area vector ($d\vec{A} = dA \hat{n}$):

$d\Phi_E = \vec{E} \cdot d\vec{A} = E dA \cos\phi$

where $\phi$ is the angle between $\vec{E}$ and the normal vector $\hat{n}$.

For a larger, arbitrary surface, the total electric flux is the integral of the flux through all the infinitesimal area elements that make up the surface:

$\Phi_E = \oint_S \vec{E} \cdot d\vec{A}$

The integral is taken over the entire closed surface $S$.

Units of Electric Flux: The SI unit of electric flux is Newton-meter squared per Coulomb ($Nm^2/C$). It can also be expressed as Volt-meter ($Vm$).

Interpretation:

• If the electric field lines are directed outwards from the surface, the flux is positive.
• If the electric field lines are directed inwards into the surface, the flux is negative.
• If the electric field lines are parallel to the surface (tangential), the flux through that surface is zero.
• If the electric field is uniform and the surface is a flat plane, the flux is simply $\Phi_E = EA\cos\phi$, where $A$ is the area of the plane and $\phi$ is the angle between $\vec{E}$ and the normal to the plane.

Electric flux is a scalar quantity. It depends on the strength of the electric field, the area of the surface, and the orientation of the surface with respect to the electric field.

Gauss's Law

Gauss's law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the net electric charge enclosed within that surface. It is one of Maxwell's equations.

Gauss's law states that the total electric flux through any closed surface (also called a Gaussian surface) is proportional to the total electric charge enclosed within that surface. Mathematically, it is expressed as:

$\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$

where:

• $\Phi_E$ is the total electric flux through the closed surface.
• $\vec{E}$ is the electric field vector.
• $d\vec{A}$ is the infinitesimal area vector, with magnitude equal to the area element and direction normal to the surface.
• $S$ is the closed surface (Gaussian surface).
• $Q_{enc}$ is the net electric charge enclosed within the surface $S$.
• $\epsilon_0$ is the permittivity of free space, a fundamental constant approximately equal to $8.854 \times 10^{-12} C^2/(Nm^2)$.

Key Aspects of Gauss's Law:

• Enclosed Charge: Only the charge inside the Gaussian surface contributes to the flux. Charges outside the surface do not affect the net flux, although they do contribute to the electric field at the surface.
• Symmetry: Gauss's law is most useful for calculating the electric field in situations with high symmetry (spherical, cylindrical, or planar symmetry).
• Gaussian Surface: The Gaussian surface is an imaginary closed surface chosen for convenience in applying Gauss's law. It does not have to coincide with a physical surface.
• Vector Nature: The law relates a flux (a scalar) to enclosed charge (a scalar). The dot product $\vec{E} \cdot d\vec{A}$ correctly accounts for the direction of the electric field relative to the surface.

Gauss's law is equivalent to Coulomb's law but is more general. It can be used to derive Coulomb's law and is particularly powerful for calculating electric fields in symmetric charge distributions.

Applications of Gauss's Law

Gauss's law is a powerful tool for calculating the electric field produced by various charge distributions, especially those with symmetry.

1. Electric Field due to an Infinitely Long Straight Charged Wire

Consider a thin, straight wire with a uniform linear charge density $\lambda$ (charge per unit length). To find the electric field $\vec{E}$ at a distance $r$ from the wire, we choose a cylindrical Gaussian surface of radius $r$ and length $L$, coaxial with the wire.

The electric field lines due to the wire are radial, pointing outwards if $\lambda$ is positive.

The Gaussian surface has three parts: two flat circular end caps and the curved cylindrical surface.

• Flux through the end caps: The electric field is radial, and the area vectors of the end caps are axial. Thus, $\vec{E}$ is perpendicular to $d\vec{A}$ on the end caps, so $\vec{E} \cdot d\vec{A} = 0$. The flux through the end caps is zero.
• Flux through the curved surface: On the curved surface, the electric field $\vec{E}$ is radial and parallel to the outward normal vector $d\vec{A}$. The magnitude of $\vec{E}$ is constant at a distance $r$ from the wire. So, $\vec{E} \cdot d\vec{A} = E dA$.

The total flux through the Gaussian surface is $\Phi_E = \oint \vec{E} \cdot d\vec{A} = \int_{curved} E dA + \int_{end1} E dA + \int_{end2} E dA = EA + 0 + 0 = EA$.

The area of the curved surface is $A = 2\pi rL$. So, $\Phi_E = E(2\pi rL)$.

The charge enclosed by the Gaussian surface is $Q_{enc} = \lambda L$.

Applying Gauss's law:

$\Phi_E = \frac{Q_{enc}}{\epsilon_0}$

$E(2\pi rL) = \frac{\lambda L}{\epsilon_0}$

Solving for $E$:

$E = \frac{\lambda}{2\pi\epsilon_0 r}$

The direction of $\vec{E}$ is radially outward from the wire if $\lambda > 0$ and radially inward if $\lambda < 0$. The electric field strength is inversely proportional to the distance from the wire.

2. Electric Field due to an Infinite Plane Sheet of Charge

Consider an infinite plane sheet of charge with uniform surface charge density $\sigma$ (charge per unit area). To find the electric field $\vec{E}$ at a point near the sheet, we choose a cylindrical Gaussian surface with cross-sectional area $A$ and length $2r$, with its axis perpendicular to the plane and passing through the point where we want to find $\vec{E}$. The plane sheet passes through the center of the cylinder.

The electric field lines are perpendicular to the plane, pointing outwards if $\sigma$ is positive.

The Gaussian surface has three parts: two flat circular end caps and the curved cylindrical surface.

• Flux through the end caps: On both end caps, the electric field $\vec{E}$ is parallel to the outward normal vector $d\vec{A}$. The magnitude of $\vec{E}$ is constant at a distance $r$ from the plane. So, $\vec{E} \cdot d\vec{A} = E dA$. The flux through each end cap is $EA$.
• Flux through the curved surface: The electric field is perpendicular to the plane of the sheet, and thus parallel to the curved surface of the cylinder. The area vectors of the curved surface are radial and perpendicular to $\vec{E}$. So, $\vec{E} \cdot d\vec{A} = 0$. The flux through the curved surface is zero.

The total flux through the Gaussian surface is $\Phi_E = \int_{end1} E dA + \int_{end2} E dA + \int_{curved} E dA = EA + EA + 0 = 2EA$.

The charge enclosed by the Gaussian surface is $Q_{enc} = \sigma A$.

Applying Gauss's law:

$\Phi_E = \frac{Q_{enc}}{\epsilon_0}$

$2EA = \frac{\sigma A}{\epsilon_0}$

Solving for $E$:

$E = \frac{\sigma}{2\epsilon_0}$

The direction of $\vec{E}$ is perpendicular to the plane, away from the sheet if $\sigma > 0$ and towards the sheet if $\sigma < 0$. Importantly, the electric field strength is constant and independent of the distance $r$ from the plane, as long as the plane is infinite.

3. Electric Field due to a Uniformly Charged Spherical Shell

Consider a thin spherical shell of radius $R$ with total charge $Q$ distributed uniformly over its surface. We want to find the electric field at a distance $r$ from the center. We use a spherical Gaussian surface of radius $r$ concentric with the shell.

Due to spherical symmetry, the electric field $\vec{E}$ must be radial, and its magnitude $E$ must be constant on the Gaussian surface.

Case 1: Outside the Shell ($r > R$)

We choose a spherical Gaussian surface of radius $r > R$. The electric field $\vec{E}$ is radial and parallel to the outward normal vector $d\vec{A}$ everywhere on the surface.

The total flux is $\Phi_E = \oint \vec{E} \cdot d\vec{A} = EA$, where $A = 4\pi r^2$ is the surface area of the Gaussian sphere.

The charge enclosed by the Gaussian surface is the total charge of the shell, $Q_{enc} = Q$.

Applying Gauss's law:

$EA = \frac{Q_{enc}}{\epsilon_0}$

$E(4\pi r^2) = \frac{Q}{\epsilon_0}$

$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$

This is the same as the electric field due to a point charge $Q$ located at the center of the shell. The field is directed radially outward if $Q > 0$ and inward if $Q < 0$.

Case 2: Inside the Shell ($r < R$)

We choose a spherical Gaussian surface of radius $r < R$. Since the charge is uniformly distributed only on the surface of the shell (radius $R$), there is no charge enclosed within the Gaussian surface of radius $r$.

$Q_{enc} = 0$.

Applying Gauss's law:

$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} = \frac{0}{\epsilon_0} = 0$

Since the flux is zero, and the field must be radial and have constant magnitude on the Gaussian surface, the electric field inside the shell must be zero:

$E = 0$