Electric Flux & Electric Dipoles

Area Vector

In our previous geometry classes, we treated area as a scalar. However, in electrostatics, the orientation of a surface relative to the electric field is crucial. Therefore, we treat area as a vector.

• Definition: The area vector $\vec{S}$ of a planar surface has a magnitude equal to the area and a direction perpendicular (normal) to the surface.
• Convention for Closed Surfaces: For a closed surface (like a sphere or cube), the area vector is always taken as the outward normal.

Electric Flux ($\Phi_E$)

Electric flux is a measure of the 'flow' of the electric field through a given area. It is proportional to the number of electric field lines crossing that area.

Mathematical Definition:

For a uniform electric field $\vec{E}$ crossing a planar area $\vec{S}$, the flux $\Phi_E$ is the dot product of the field and the area vector: $$\Phi_E = \vec{E} \cdot \vec{S} = ES \cos \theta$$ Where $\theta$ is the angle between $\vec{E}$ and the normal to the area (the area vector).

• Case 1: If the field is parallel to the area vector (perpendicular to the surface), $\theta = 0^\circ$ and $\Phi_E = ES$ (Maximum).
• Case 2: If the field is perpendicular to the area vector (parallel to the surface), $\theta = 90^\circ$ and $\Phi_E = 0$.
• SI Unit: $\text{N m}^2 \text{C}^{-1}$ or $\text{V m}$.
• Nature: Scalar quantity.

Electric Dipole

An electric dipole is a pair of equal and opposite charges, $q$ and $-q$, separated by a very small distance $2a$.

Electric Dipole Moment ($\vec{p}$):

This vector quantity measures the strength of the dipole. $$\vec{p} = q \times (2\vec{a})$$

• Direction: From negative charge ($-q$) to positive charge ($+q$).
• SI Unit: Coulomb-meter (C m).

Electric Field of a Dipole

This is a high-yield topic for both Board derivations and JEE/NEET numericals. We calculate the field at two specific locations:

A. At an Axial Point (Point on the line joining charges)

For a point at distance $r$ from the center of the dipole: $$\vec{E}_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}r}{(r^2 - a^2)^2}$$ For a short dipole ($r \gg a$): $$\vec{E}_{axial} \approx \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}}{r^3}$$ The direction of $\vec{E}_{axial}$ is the same as $\vec{p}$.

B. At an Equatorial Point (Point on the perpendicular bisector)

For a point at distance $r$ from the center: $$\vec{E}_{equatorial} = \frac{1}{4\pi\epsilon_0} \frac{-\vec{p}}{(r^2 + a^2)^{3/2}}$$ For a short dipole ($r \gg a$): $$\vec{E}_{equatorial} \approx -\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3}$$ The direction of $\vec{E}_{equatorial}$ is opposite to $\vec{p}$.

Important Comparison: For a short dipole at the same distance $r$, $|E_{axial}| = 2 |E_{equatorial}|$.

Dipole in a Uniform External Field

When a dipole is placed in a uniform electric field $\vec{E}$, the net force on it is zero (since $F_+ = qE$ and $F_- = -qE$). However, because these forces act at different points, they create a Torque ($\tau$).

Torque ($\tau$):

$$\vec{\tau} = \vec{p} \times \vec{E} \implies \tau = pE \sin \theta$$ Where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$.

• Stable Equilibrium: When $\theta = 0^\circ$, $\tau = 0$. The dipole is aligned with the field.
• Unstable Equilibrium: When $\theta = 180^\circ$, $\tau = 0$. The dipole is opposite to the field.
• Maximum Torque: When $\theta = 90^\circ$, $\tau = pE$.

🧠 Memory Capsule

• Flux: $\Phi = E A \cos \theta$. (Watch out! $\theta$ is with the normal).
• Dipole Moment: Direction is always Negative $\to$ Positive.
• Field Drop-off: For a point charge, $E \propto 1/r^2$. For a dipole, $E \propto 1/r^3$.
• Axial vs Equi: Axial field is twice as strong as equatorial field at the same large distance.
• Uniform Field: Net force on dipole is Zero, but torque is $pE \sin \theta$.
• Non-Uniform Field: In a non-uniform field, a dipole experiences both a net force and a torque.

Example 1: Basic Flux Calculation

A uniform electric field of $5 \times 10^3$ N/C passes through a circular surface of radius 10 cm. Calculate the flux if the surface is perpendicular to the field lines.

Solution:

1. Identify values: $E = 5 \times 10^3$ N/C, $r = 0.1$ m.
2. Calculate Area: $A = \pi r^2 = 3.14 \times (0.1)^2 = 0.0314 \text{ m}^2$.
3. Find angle: If surface is perpendicular to field, the normal (Area vector) is parallel to the field. So $\theta = 0^\circ$.
4. Formula: $\Phi = EA \cos 0^\circ = (5 \times 10^3) \times 0.0314 = 157 \text{ N m}^2/\text{C}$.

Example 2: Flux through a Tilted Surface

In the previous example, calculate the flux if the normal to the surface makes an angle of $60^\circ$ with the field.

Solution:

1. Given: $\theta = 60^\circ$.
2. Formula: $\Phi = EA \cos 60^\circ$.
3. Calculate: $\Phi = 157 \times 0.5 = 78.5 \text{ N m}^2/\text{C}$.

Example 3: Simple Dipole Moment

Two charges $+5 \mu C$ and $-5 \mu C$ are separated by a distance of 4 mm. Calculate the dipole moment.

Solution:

1. Given: $q = 5 \times 10^{-6}$ C, $2a = 4 \times 10^{-3}$ m.
2. Formula: $p = q \times 2a$.
3. Calculate: $p = (5 \times 10^{-6}) \times (4 \times 10^{-3}) = 20 \times 10^{-9} = 2 \times 10^{-8} \text{ C m}$.
4. Direction: From $-5 \mu C$ to $+5 \mu C$.

Example 4: Axial Field Strength

A short dipole has a dipole moment of $4 \times 10^{-9}$ C m. Find the electric field at a point on the axis 30 cm away from the center.

Solution:

1. Given: $p = 4 \times 10^{-9}$ C m, $r = 0.3$ m.
2. Condition: Short dipole ($r \gg a$).
3. Formula: $E_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} = (9 \times 10^9) \frac{2 \times 4 \times 10^{-9}}{(0.3)^3}$.
4. Calculate: $E = (9 \times 8) / 0.027 = 72 / 0.027 = 2666.67 \text{ N/C}$.

Example 5: Equatorial Field Strength

For the dipole in Example 4, find the field at the same distance (30 cm) on the equatorial line.

Solution:

1. Short Dipole Relation: $|E_{equi}| = \frac{1}{2} |E_{axial}|$.
2. Calculate: $E = 2666.67 / 2 = 1333.33 \text{ N/C}$.
3. Direction: Opposite to the direction of the dipole moment.

Example 6: Torque on a Dipole

An electric dipole with moment $5 \times 10^{-8}$ C m is aligned at $30^\circ$ with a uniform electric field of $4 \times 10^4$ N/C. Calculate the torque.

Solution:

1. Given: $p = 5 \times 10^{-8}$ C m, $E = 4 \times 10^4$ N/C, $\theta = 30^\circ$.
2. Formula: $\tau = pE \sin \theta$.
3. Calculate: $\tau = (5 \times 10^{-8}) \times (4 \times 10^4) \times \sin 30^\circ$.
4. Simplify: $\tau = (20 \times 10^{-4}) \times 0.5 = 10 \times 10^{-4} = 10^{-3} \text{ N m}$.

Example 7: Maximum Torque

What is the maximum torque the dipole in Example 6 can experience in the same field?

Solution:

1. Condition: Maximum torque occurs at $\theta = 90^\circ$.
2. Calculate: $\tau_{max} = pE = (5 \times 10^{-8}) \times (4 \times 10^4) = 20 \times 10^{-4} = 2 \times 10^{-3} \text{ N m}$.

Example 8: Work Done to Rotate (JEE Prep)

How much work is required to rotate a dipole from stable equilibrium to unstable equilibrium in a field $E$?

Solution:

1. Positions: Stable ($\theta_1 = 0^\circ$), Unstable ($\theta_2 = 180^\circ$).
2. Formula: $W = pE(\cos \theta_1 - \cos \theta_2)$.
3. Substitute: $W = pE(\cos 0^\circ - \cos 180^\circ) = pE(1 - (-1)) = 2pE$.
4. Result: Work required is $2pE$.

Example 9: System of Charges

A system has two charges $q_A = 2.5 \times 10^{-7}$ C and $q_B = -2.5 \times 10^{-7}$ C located at points A: (0, 0, -15 cm) and B: (0, 0, +15 cm). What are the total charge and electric dipole moment?

Solution:

1. Total Charge: $Q = q_A + q_B = 2.5 \times 10^{-7} - 2.5 \times 10^{-7} = 0$.
2. Distance: Separation $2a = 15 - (-15) = 30$ cm $= 0.3$ m.
3. Dipole Moment: $p = q \times 2a = (2.5 \times 10^{-7}) \times 0.3 = 7.5 \times 10^{-8} \text{ C m}$.
4. Direction: Along the negative z-axis (from positive B to negative A coordinates, following the $-q$ to $+q$ rule).

Example 10: Ratio of Fields

If the distance of a point on the axis of a short dipole is doubled, by what factor does the electric field change?

Solution:

1. Relation: $E \propto 1/r^3$.
2. Ratio: $E_2 / E_1 = (r_1 / r_2)^3$.
3. Substitute: Since $r_2 = 2r_1$, the factor is $(1/2)^3 = 1/8$.
4. Result: The field becomes $1/8$th of its original value.