Analyzing Electric Dipole Reorientation and Heat Release

Let's analyze the energy changes involved when electric dipoles realign in a changing electric field direction. We are given a substance with molecules that have a permanent electric dipole moment.

Given Information:

• Magnitude of electric dipole moment of one molecule, \(p = 10^{-31}\) Cm
• Number of molecules = 1 mole = \(N = 6 \times 10^{23}\)
• Magnitude of applied electrostatic field, \(E = 10^8\) Vm\(^{-1}\)
• Initial state: Substance is 100% polarised along the original field direction.
• The direction of the field is suddenly changed by \(60^\circ\).
• Final state: Dipoles align along the new field direction.

Understanding Dipole Potential Energy

The potential energy \(U\) of an electric dipole \(\vec{p}\) in an electric field \(\vec{E}\) is given by the formula:

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

where \(p\) is the magnitude of the dipole moment, \(E\) is the magnitude of the electric field, and \(\theta\) is the angle between the dipole moment vector \(\vec{p}\) and the electric field vector \(\vec{E}\).

For a mole of substance with \(N\) molecules, the total potential energy is the sum of the potential energies of individual dipoles. If all dipoles are oriented at the same angle \(\theta\) with respect to the field, the total potential energy is \(U_{total} = N \times (-pE \cos\theta)\).

Initial and Final States for Reorientation

Initially, the molecules are 100% polarised along the original field direction. Let's call the original field \(\vec{E}_{old}\). The dipoles \(\vec{p}\) are aligned with \(\vec{E}_{old}\), so the angle between \(\vec{p}\) and \(\vec{E}_{old}\) is \(0^\circ\).

The direction of the field suddenly changes by \(60^\circ\). Let the new field be \(\vec{E}_{new}\). The angle between \(\vec{E}_{old}\) and \(\vec{E}_{new}\) is \(60^\circ\).

The question asks for the heat released in aligning the dipoles along the new direction. This process starts when the new field \(\vec{E}_{new}\) is present, but the dipoles are still pointing in the original direction (along \(\vec{E}_{old}\)). The process ends when the dipoles are aligned with the new field (\(\vec{E}_{new}\)).

Initial Potential Energy (Before Re-alignment, After Field Change)

At the start of the re-alignment process, the dipoles are oriented along the original field direction, while the new field \(\vec{E}_{new}\) is at \(60^\circ\) to this direction. So, the angle between the dipole moment \(\vec{p}\) and the new field \(\vec{E}_{new}\) is \(\theta_1 = 60^\circ\).

The total initial potential energy for \(N\) dipoles in the new field \(\vec{E}_{new}\) is:

\(U_1 = N (-pE \cos\theta_1) = -NpE \cos(60^\circ)\)

Final Potential Energy (After Re-alignment)

After re-alignment, the dipoles are aligned along the new field direction \(\vec{E}_{new}\). The angle between the dipole moment \(\vec{p}\) and the new field \(\vec{E}_{new}\) is \(\theta_2 = 0^\circ\).

The total final potential energy for \(N\) dipoles in the new field \(\vec{E}_{new}\) is:

\(U_2 = N (-pE \cos\theta_2) = -NpE \cos(0^\circ) = -NpE\)

Calculating the Change in Potential Energy

The change in potential energy \(\Delta U\) during the re-alignment is the final potential energy minus the initial potential energy:

\(\Delta U = U_2 - U_1 = -NpE - (-NpE \cos(60^\circ))\)

\(\Delta U = -NpE + NpE \cos(60^\circ) = NpE (\cos(60^\circ) - 1)\)

We know that \(\cos(60^\circ) = \frac{1}{2}\).

\(\Delta U = NpE \left(\frac{1}{2} - 1\right) = NpE \left(-\frac{1}{2}\right) = -\frac{1}{2} NpE\)

Substituting Values

Now, substitute the given values for \(N\), \(p\), and \(E\):

\(N = 6 \times 10^{23}\)

\(p = 10^{-31}\) Cm

\(E = 10^8\) Vm\(^{-1}\)

\(\Delta U = -\frac{1}{2} \times (6 \times 10^{23}) \times (10^{-31}) \times (10^8)\)

\(\Delta U = -\frac{1}{2} \times 6 \times 10^{(23 - 31 + 8)}\)

\(\Delta U = -\frac{1}{2} \times 6 \times 10^0\)

\(\Delta U = -\frac{1}{2} \times 6 \times 1 = -3\)

So, the change in potential energy is \(\Delta U = -3\) J.

Calculating Heat Released

When a system releases energy, its potential energy decreases, and this energy is often converted into heat or work. In this case, the decrease in potential energy is released as heat to the surroundings.

Heat released \(Q = -\Delta U\)

\(Q = -(-3 \text{ J}) = +3\) J

The heat released by the substance is 3 J.

Summary of Calculations:

Revision Table: Electric Dipole Energy in Field

Additional Information: Polarization and Energy Release in Dielectrics

When a substance containing polar molecules is subjected to an external electric field, the molecules experience a torque that tends to align their permanent electric dipole moments with the field. This alignment process is known as orientation polarization. In a strong field, this alignment can be significant, leading to a net macroscopic dipole moment within the substance.

The potential energy of the dipoles is minimized when they are aligned parallel to the electric field. When the direction of the field changes, the dipoles will rotate to align with the new field direction, moving towards a lower potential energy state if they were initially at a non-zero angle relative to the new field.

The energy difference between the initial potential energy (dipoles at 60° to the new field) and the final potential energy (dipoles at 0° to the new field) is released. This released energy is typically converted into heat within the substance due to internal damping mechanisms as the dipoles rotate. This is an example of energy dissipation during a relaxation process in a dielectric material.