An electric dipole of dipole moment \( \mathbf{P} \) is rotated in an electric field of magnitude \( E \) from the most stable orientation to the most unstable orientation. The work done in the process is:
\( 2pE \)
Understanding Work Done on an Electric Dipole
The problem asks us to calculate the work done when an electric dipole is rotated in a uniform electric field from its most stable orientation to its most unstable orientation. To solve this, we need to understand the potential energy of an electric dipole in an electric field and the relationship between work done and potential energy change.Potential Energy of an Electric Dipole
An electric dipole with dipole moment \( \mathbf{P} \) placed in a uniform electric field \( \mathbf{E} \) has a potential energy \( U \) given by the dot product of the dipole moment and the electric field vector, with a negative sign: $$ U = -\mathbf{P} \cdot \mathbf{E} $$ If \( \theta \) is the angle between the direction of the electric dipole moment \( \mathbf{P} \) and the direction of the electric field \( \mathbf{E} \), the potential energy can be written as: $$ U = -PE \cos\theta $$ where \( P \) is the magnitude of the dipole moment and \( E \) is the magnitude of the electric field.Initial and Final Orientations
The question specifies two orientations: the most stable and the most unstable.- Most Stable Orientation: This occurs when the dipole aligns itself parallel to the electric field, i.e., the angle \( \theta \) between \( \mathbf{P} \) and \( \mathbf{E} \) is \( 0^\circ \). This corresponds to the minimum potential energy. The initial potential energy, \( U_{initial} \), is: $$ U_{initial} = -PE \cos(0^\circ) = -PE(1) = -PE $$
- Most Unstable Orientation: This occurs when the dipole is aligned antiparallel to the electric field, i.e., the angle \( \theta \) between \( \mathbf{P} \) and \( \mathbf{E} \) is \( 180^\circ \). This corresponds to the maximum potential energy. The final potential energy, \( U_{final} \), is: $$ U_{final} = -PE \cos(180^\circ) = -PE(-1) = +PE $$
Calculating the Work Done
The work done by an external agent to rotate the electric dipole from an initial orientation to a final orientation is equal to the change in potential energy of the dipole. $$ W = \Delta U = U_{final} - U_{initial} $$ Substituting the values of \( U_{initial} \) and \( U_{final} \) we found: $$ W = (+PE) - (-PE) $$ $$ W = PE + PE $$ $$ W = 2PE $$ Thus, the work done in rotating the electric dipole from the most stable to the most unstable orientation is \( 2PE \).| Orientation | Angle \( \theta \) | Potential Energy \( U = -PE \cos\theta \) |
|---|---|---|
| Most Stable | \( 0^\circ \) | \( -PE \) |
| Most Unstable | \( 180^\circ \) | \( +PE \) |
| Perpendicular | \( 90^\circ \) | \( 0 \) |
The calculated work done, \( 2PE \), matches one of the given options.
Revision Table: Electric Dipole Work and Potential Energy
| Concept | Formula/Description |
|---|---|
| Electric Dipole Moment \( \mathbf{P} \) | Vector pointing from negative to positive charge, magnitude \( q \times d \) (charge times distance) |
| Potential Energy \( U \) of Dipole in \( \mathbf{E} \) | \( U = -\mathbf{P} \cdot \mathbf{E} = -PE \cos\theta \) |
| Torque \( \mathbf{\tau} \) on Dipole in \( \mathbf{E} \) | \( \mathbf{\tau} = \mathbf{P} \times \mathbf{E} = PE \sin\theta \) (magnitude) |
| Work Done \( W \) by External Agent | \( W = \Delta U = U_{final} - U_{initial} \) |
| Most Stable Orientation | \( \theta = 0^\circ \), \( U_{min} = -PE \) |
| Most Unstable Orientation | \( \theta = 180^\circ \), \( U_{max} = +PE \) |
Additional Information: Dipoles and Fields
An electric dipole consists of two equal and opposite charges separated by a small distance. The electric dipole moment is a measure of the strength and direction of the dipole.
When an electric dipole is placed in a uniform electric field, it experiences a torque that tends to align it with the field. The magnitude of this torque is given by \( \tau = PE \sin\theta \). The torque is zero when the dipole is parallel (\( \theta = 0^\circ \)) or antiparallel (\( \theta = 180^\circ \)) to the field.
The electric field does work on the dipole as it rotates. The work done by the electric field is \( W_{field} = -\Delta U \). The work done by the external agent is the negative of the work done by the field, assuming no change in kinetic energy. So, \( W_{external} = -W_{field} = \Delta U \), which is what we used in the calculation.
Rotating the dipole from the most stable position (\( \theta=0^\circ \)) to the most unstable position (\( \theta=180^\circ \)) requires positive work done by an external agent because the system's potential energy increases.
Important Questions from Electrostatic Potential and Capacitance
A copper ball of density 8.0 g/cc and 1 cm in diameter is immersed in oil of density 0.8 g/cc. The charge on the ball if it remains just suspended in oil in an electric field of intensity 600π V/m acting in the upward direction is:
A metal wire is subjected to a constant potential difference. When the temperature of the metal wire increases, the drift velocity of the electrons in it:
A cube of side 'a' has a charge Q at each of its vertices. What is the potential due to this charge array at the centre of the cube?
A parallel plate capacitor with air between the plates has a capacitance of 6pF. What will be the capacitance if the distance between the plates is reduced to half and the space is filled with a dielectric constant 5?
What is the unit of electric flux in terms of the base units of SI?
