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Chapter: Electric Field

The Electric Field

The electric field strength is defined as the force experienced per unit positive charge.

\[ E = \frac{F}{q} \]

Where:

  • \(E\) = Electric field strength
  • \(F\) = Force experienced by the test charge
  • \(q\) = Magnitude of the test charge

The electric field is a vector quantity — it has both magnitude and direction. It is directed away from positive charges and toward negative charges.

Representation:

Electric fields are often represented using field lines around a charge.
(Diagram shown: A positive charge with circular field lines labeled "E Field")

Gravitational Analogy (for better understanding)

Just like an object of mass \(m\) experiences a gravitational force when near Earth:

\[ F = mg \]

Where:

  • \(g\) = Gravitational field (acceleration due to gravity)
  • \(m\) = Mass of the object
  • \(F\) = Gravitational force

Similarly:

\[ g = \frac{F}{m} \]

In the same way, electric field strength is:

\[ E = \frac{F}{q} \]

This analogy helps understand how fields work — whether electric or gravitational — as force per unit "something" (mass or charge).

Electric Field — Lines of Force

For a small test charge \(q_0 \to 0\), the electric field is defined as:

\[ \vec{E} = \lim_{{q_0 \to 0}} \frac{\vec{F}}{q_0} \]

The Lines of Force

The relationship between the imaginary lines of force and the electric field strength \(\vec{E}\):

① Tangent Property:

The tangent to a line of force at any point gives the direction of \(\vec{E}\) at that point.
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(Illustration shows field lines going from a positive to a negative charge with curved paths — this represents a dipole field.)

② Density and Magnitude:

The lines of force are drawn such that the number of lines per unit cross-sectional area is proportional to the magnitude of \(\vec{E}\).

  • Where the lines are close together, \(\vec{E}\) is large.
  • Where the lines are far apart, \(\vec{E}\) is small.
Calculation of \(\vec{E}\) (Electric Field Strength)

Let a test charge \(q_0\) be placed at a distance \(r\) from a point charge \(q\). From Coulomb’s Law, the force between the two charges is given by:

\[ \vec{F} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q q_0}{r^2} \]
➤ Electric Field Strength

The electric field strength due to charge \(q\) at a distance \(r\) is:
$$
\vec{E} = \frac{\vec{F}}{q_0}
= \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2}
$$
This gives the magnitude and direction of the electric field produced by a single point charge.

➤ For a Group of Charges

If multiple point charges \(q_1, q_2, q_3, \ldots\) are present, the net electric field at a point is the vector sum of the individual electric fields:

\[ \vec{E} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \cdots + \vec{E}_n \]
Electric Dipole – Field on the Perpendicular Bisector

Problem:
A positive and a negative charge of equal magnitude \(q\) are placed at a distance \(2a\) apart, forming an electric dipole. Find the electric field \(\vec{E}\) at point \(P\), located a distance \(r\) from the center of the dipole along the perpendicular bisector.
Assume \(r \gg a\).

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Solution:
The total electric field at point \(P\) is the vector sum of the fields due to \(+q\) and \(-q\):

\[ \vec{E} = \vec{E}_1 + \vec{E}_2 \]

Due to symmetry, vertical components cancel and horizontal components add.
Let dipole moment \(p = 2aq\). Then, the net field at point \(P\) is:

\[ \vec{E}_\perp = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{(r^2 + a^2)^{3/2}} \approx \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r^3} \quad \text{(for } r \gg a\text{)} \]

Let a positive charge \(+q\) and a negative charge \(-q\) be separated by a distance \(2a\), forming a dipole. Point \(P\) lies along the axis of the dipole, at a distance \(r\) from its center.

Each charge contributes an electric field at \(P\), both having magnitude:

\[ E_1 = E_2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{(r^2 + a^2)} \]

From the geometry, only the horizontal components add, while vertical components cancel.

Using the angle \(\theta\), where:

\[ \cos\theta = \frac{r}{\sqrt{r^2 + a^2}} \]

The horizontal component of each field is:

\[ E_x = E_1 \cos\theta = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r^2 + a^2} \cdot \frac{r}{\sqrt{r^2 + a^2}} \]

So, total electric field:

\[ E = 2E_1 \cos\theta = \frac{2q r}{4\pi\varepsilon_0 (r^2 + a^2)^{3/2}} \]

For \(r \gg a\), using the dipole moment \(p = 2aq\), it simplifies to:

\[ E_{\text{axial}} \approx \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r^3} \]

Is the direction of the electric field from positive to negative? I'm also confused about the difference in direction between the electric field and the dipole moment. Can you explain it?

Answer:
Yes, the electric field (\(\vec{E}\)) is directed from the positive charge to the negative charge. It represents the force experienced by a positive test charge placed in the field.

In contrast, the electric dipole moment (\(\vec{p}\)) is a vector quantity that points from the negative charge to the positive charge. It is defined as \(\vec{p} = q \cdot \vec{d}\), where \(\vec{d}\) is the vector from −q to +q.

Thus,
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* Electric field direction: \(+ \rightarrow -\)
* Dipole moment direction: \(- \rightarrow +\)

A Dipole in a Uniform Electric Field

🔷 Question (Enhanced for University Level):

Explain the behavior of an electric dipole placed in a uniform electric field. Derive the expression for the torque acting on the dipole and discuss the conditions under which equilibrium occurs. Include proper diagrams and assumptions in your explanation.

🔷 Answer:

1. Introduction:

An electric dipole consists of two equal and opposite point charges \(+q\) and \(-q\), separated by a fixed distance \(2a\). The dipole moment \(\vec{p}\) is a vector quantity defined as:
$$
\vec{p} = q \cdot 2a \cdot \hat{d}
$$
where \(\hat{d}\) is a unit vector pointing from the negative charge to the positive charge.

2. Dipole in a Uniform Electric Field:

When an electric dipole is placed in a uniform electric field \(\vec{E}\), the positive and negative charges experience forces of equal magnitude but in opposite directions.

\[ \vec{F}_{+} = +q\vec{E}, \quad \vec{F}_{-} = -q\vec{E} \]

These forces are equal in magnitude and opposite in direction, so the net force on the dipole is zero:

\[ \vec{F}_{\text{net}} = \vec{F}_{+} + \vec{F}_{-} = 0 \]

However, these forces form a couple, resulting in a torque on the dipole, which tends to rotate it.

3. Torque on the Dipole:

If the dipole makes an angle \(\theta\) with the direction of the electric field \(\vec{E}\), then the torque \(\vec{\tau}\) experienced by the dipole is given by:

\[ \vec{\tau} = \vec{p} \times \vec{E} \]

In magnitude,

\[ \tau = pE \sin\theta \]

This torque tries to align the dipole with the direction of the electric field.

4. Equilibrium Conditions:

  • Stable Equilibrium: When the dipole is aligned with the electric field (\(\theta = 0^\circ\)), the torque is zero and the dipole remains at rest.

  • Unstable Equilibrium: When the dipole is anti-aligned (\(\theta = 180^\circ\)), torque is again zero, but any slight disturbance will cause it to rotate.

5. Energy of a Dipole in an Electric Field:

The potential energy of a dipole in a uniform electric field is:

\[ U = -\vec{p} \cdot \vec{E} = -pE \cos\theta \]
  • Minimum Energy at \(\theta = 0^\circ\): Stable equilibrium
  • Maximum Energy at \(\theta = 180^\circ\): Unstable equilibrium

🔷 Diagram:

(You can include the neat diagram from your notes showing the dipole, charges, and angle with the field direction, but redrawn properly with labels for \(\vec{p}\), \(\vec{E}\), and torque arrows.)

🔷 Conclusion:

Although the net force on an electric dipole in a uniform electric field is zero, it experiences a torque that tends to align it with the field. The torque depends on the dipole moment, electric field strength, and the angle between them. Understanding this interaction is crucial in fields such as molecular physics, electromechanical systems, and dielectrics in external fields.

The magnitude of the electric dipole moment is given by the product of the charge and the distance between the charges.
i.e., Electric dipole moment = q × 2a = 2aq = p (say)

Unit of p = (Coulomb × meter) = Cm

If the dipole is placed in an electric field E, then
Torque τ = p × E = pE sinθ (Vector quantity)

Let us consider a dipole placed in a uniform electric field, such that the angle between the dipole moment and the direction of the field is θ.

A torque τ = pE sinθ acts on the dipole, which tends to rotate the dipole and align it along the direction of the field.

So, the dipole experiences a torque but no net force, as the two forces on the positive and negative charges cancel each other out.

Potential Energy:

Let the dipole be rotated in a uniform electric field from an angle θ₁ to θ₂. The work done in this process is stored as the potential energy of the dipole.

Let the dipole moment = p
E = strength of electric field
θ = angle between p and E

Then, the potential energy of the dipole is given by
U = -pE cosθ
(Minimum when p and E are parallel, maximum when anti-parallel)

Electric Dipole (তড়িৎ দ্বিমেরু)

When two equal and opposite electric charges are placed very close to each other, an electric dipole is formed.
Since the electric potential at any point on the perpendicular bisector of an electric dipole is zero, the amount of work required to move a positive charge along this line is also zero.

Two point charges that are equal in magnitude but opposite in nature and are placed very close to each other constitute an electric dipole.

Dipole Moment (তড়িৎ দ্বিমেরু ভ্রামক)

The dipole moment of an electric dipole is defined as the product of the magnitude of either charge and the distance between them.
Let the magnitude of each charge in the dipole be q
and the distance between the charges be 2l.

\[ \therefore \text{Dipole moment, } p = q \times 2l \]

The vector form of dipole moment is:

\[ \vec{p} = 2q \vec{l} \]

Its direction is from the negative charge to the positive charge.

Why is no work done when a positive charge is moved along the perpendicular bisector of an electric dipole?

Since the electric potential is zero at any point on the perpendicular bisector of an electric dipole,
the work required to move a positive charge along this line is also zero.
In other words, no work is done.

Why is no work done when a positive charge is moved along the perpendicular bisector of an electric dipole?

If a positive charge is moved along the perpendicular bisector of an electric dipole, then the electric potential at each point along that path is the same.
This path lies along the equipotential surface, and since there's no potential difference, no work is done in moving the charge.
Also, the electric field is always perpendicular to the direction of motion in this case.

2.4.2 Torque on a dipole in a uniform electric field

Suppose we have an electric dipole consisting of charges +q and –q, separated by distance AB = 2l. Let the dipole be placed in a uniform electric field E such that it makes an angle θ with the field.

Force on charge +q = qE (along the field)
Force on charge –q = –qE (opposite to the field)

As both forces are equal and opposite but act on different points, they form a couple. This couple creates a torque that tries to rotate the dipole to align with the field.

Hence, they exert a torque on the dipole.

The magnitude of the torque is:
τ = Magnitude of one force × Perpendicular distance between the two forces

\[ \tau = qE \cdot 2l \cdot \sin \theta = pE \sin \theta \quad [p = 2ql] \tag{2.36} \]

Here, p is the dipole moment.
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In vector form, equation (2.36) is written as:

\[ \vec{\tau} = \vec{p} \times \vec{E} \tag{2.37} \]

Equation (2.37) shows the relation between the torque acting on a dipole and the angle between its dipole moment p and the external electric field E.

  • When θ = 90°, torque becomes τ = pE sin 90° = pE → Maximum torque,
    i.e., τₘₐₓ = pE

  • When θ = 0°, torque becomes τ = pE sin 0° = 0 → Minimum torque,
    i.e., τₘᵢₙ = 0

Work done to deflect a dipole in an electric field

In absence of external forces, an electric dipole in a uniform electric field aligns itself parallel to the field direction.

To displace or rotate it from that equilibrium position, work must be done, which gets stored as potential energy.

Let at any instant the dipole makes an angle θ with the field, then torque is:

\[ \tau = pE \sin \theta \]

To rotate it by a small angle dθ, the infinitesimal work done is:

\[ dW = \tau d\theta = pE \sin \theta \, d\theta \]

So, total work done in rotating the dipole from 0 to α:

\[ W = \int_0^{\alpha} pE \sin \theta \, d\theta = pE (1 - \cos \alpha) \]

Hence, potential energy of the dipole in that position:

\[ U = pE (1 - \cos \alpha) \]

Special cases:

  1. When α = 90°,
\[ W = pE(1 - \cos 90°) = pE \]
  1. When α = 180°,
\[ W = pE(1 - \cos 180°) = pE (1 + 1) = 2pE \tag{2.38} \]

Here is a cleaned-up, easy-to-understand version of your handwritten notes, written in simple language but structured like a textbook — non-bookish yet formal, perfect for a non-native student trying to master the concept of Electric Potential.

Electric Potential – Explained Simply

💡 What is Electric Potential?

Electric potential is the energy (or work done) required to bring a unit positive charge from one point to another in an electric field.

🔁 Potential Difference Between Two Points

If we take two points A and B in an electric field and move a small test charge \(q_0\) from A to B, the potential difference is given by:

\[ V_B - V_A = \frac{W_{AB}}{q_0} \]

Where:

  • \(V_A, V_B\) are the electric potentials at A and B.
  • \(W_{AB}\) is the work done to move the charge.
  • SI unit of electric potential is volt (V) = joule per coulomb (J/C).

🔄 Work and Electric Field

We know that work done by force \(\vec{F}\) over displacement \(\vec{d}\) is:

\[ W = \vec{F} \cdot \vec{d} \]

In an electric field, \(\vec{F} = q_0 \vec{E}\). So,

\[ W_{AB} = q_0 \vec{E} \cdot \vec{d} \]

Therefore, potential difference becomes:

\[ V_B - V_A = \frac{W_{AB}}{q_0} = \vec{E} \cdot \vec{d} \]

🧠 Simple Analogy

Think of electric potential like height. Water naturally flows from high to low ground. Similarly, charge flows from high to low potential. The difference in height is like the potential difference — it tells you how much energy change happens while moving something.

🔄 Electric Potential Using Integration

If the electric field varies from point to point, total work is found by adding up small amounts:

\[ W_{AB} = \int_A^B \vec{F} \cdot d\vec{l} = \int_A^B q_0 \vec{E} \cdot d\vec{l} \]
\[ \Rightarrow V_B - V_A = \frac{1}{q_0} \int_A^B q_0 \vec{E} \cdot d\vec{l} = \int_A^B \vec{E} \cdot d\vec{l} \]

Or in reverse:

\[ V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l} \]

If point A is at infinity and \(V_A = 0\), then:

\[ V = -\int_\infty^B \vec{E} \cdot d\vec{l} \]

Electric Potential Due to a Point Charge

Consider two points \(A\) and \(B\) near an isolated point charge \(q\). Let us find the potential difference between points \(A\) and \(B\).

Assume a test charge \(q_0\) is moved from point \(A\) to \(B\).

In the figure, electric field \(\vec{E}\) points to the right and displacement \(\vec{dl}\) is to the left.

\[ \vec{E} \cdot \vec{dl} = E \cdot dl \cdot \cos(180^\circ) = -E \cdot dl \]

If \(\vec{dl} = -dr\), then:

\[ \vec{E} \cdot \vec{dl} = E \cdot dr \tag{2} \]

Now, the electric potential difference is:

\[ V_B - V_A = - \int_{A}^{B} \vec{E} \cdot \vec{dl} \]

Given that:

\[ E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \]

Substituting into the integral:

\[ V_B - V_A = - \int_{A}^{B} \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} dr \]
\[ V_B - V_A = \frac{q}{4 \pi \varepsilon_0} \left( \frac{1}{r_A} - \frac{1}{r_B} \right) \tag{4} \]

If we take the reference point at infinity where \(V = 0\), then the potential at a distance \(r\) becomes:

\[ V = \frac{q}{4 \pi \varepsilon_0 r} \]

This is the electric potential due to a point charge.

🔹 For a Group of Point Charges:

The total potential at a point is the sum of potentials due to each charge:

\[ V = \sum \frac{q_n}{4 \pi \varepsilon_0 r_n} \]

Where:

  • \(q_n\): value of the nth charge
  • \(r_n\): distance from the point to the nth charge

🔹 For Continuous Charge Distribution:

If the charge is spread continuously, the potential is:

\[ V = \frac{1}{4 \pi \varepsilon_0} \int \frac{dq}{r} \]

Where:

  • \(dq\): small element of charge
  • \(r\): distance from this element to the point of observation

🤝 Easy Explanation + Analogy:

💡 What Is Electric Potential?

Imagine you're climbing a hill. The higher you go, the more energy it takes. That energy is like potential — it's not moving, but it has the capacity to do work.

In electricity, when you move a test charge near another charge (like a +q), you need to do some work. That "work per unit charge" is called electric potential.

🔌 Analogy:

Think of the electric field like gravity and the test charge like a ball.

  • Moving a ball uphill takes effort → similar to moving a charge against an electric field.
  • If you drop the ball from a height, it gains speed → just like a charge would start moving if there’s a potential difference.

📌 Summary (Simple Terms):

  • Electric Potential tells us how much work is needed to move a charge in an electric field.
  • The closer you go to a positive charge, the higher the potential.
  • Formula:

$$
V = \frac{q}{4 \pi \varepsilon_0 r}
$$
* It helps in calculating how charges interact and how much energy is involved.

Let me know if you'd like a diagram or if you want me to explain vector directions or integration steps one by one in Bangla.

✅ Final Summary (Easy Words)

  • Electric potential tells us how much energy a charge would need (or release) to move in an electric field.
  • Potential difference is like the "height gap" that lets charge flow, just like water flows downhill.
  • It is closely related to electric field, but while field pushes the charge, potential tells you how much energy is stored or used.
  • Equations help you calculate this potential in different setups: one charge, many charges, or continuous charge spread.

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