Question

Obtain the formula for work done by an electric dipole in rotating \(\theta\) from equilibrium in a uniform electric field.

Hint:

remember that work done equals the change in potential energy (\(W = \Delta U\)). Since \(U = -pE\cos\theta\), the work done from \(0\) to \(\theta\) is \(U(\theta) - U(0) = -pE\cos\theta - (-pE\cos 0) = pE(1-\cos\theta)\).

Answer 1

Step 1: Understanding the Concept:
When an electric dipole with dipole moment \(\vec{p}\) is placed in a uniform electric field \(\vec{E}\), it experiences a torque that tries to align it with the field. To rotate the dipole from its equilibrium position, external work must be done against this restoring torque. This work is stored as potential energy in the dipole.

Step 2: Key Formula or Approach:
The torque (\(\tau\)) experienced by the dipole is given by \(\tau = pE\sin\theta\), where \(\theta\) is the angle between \(\vec{p}\) and \(\vec{E}\).
The work done (\(dW\)) in rotating the dipole through a small angle \(d\theta\) is \(dW = \tau d\theta\).
To find the total work done, we must integrate this expression from the initial position to the final position.

Step 3: Detailed Explanation (Derivation):
The equilibrium position for a dipole is when it is aligned with the electric field, so the initial angle is \(\theta_1 = 0^\circ\). We want to find the work done in rotating it to a final angle \(\theta_2 = \theta\). The total work done \(W\) is: \[ W = \int_{\theta_1}^{\theta_2} dW = \int_{0}^{\theta} \tau d\theta' \] Substitute the expression for torque: \[ W = \int_{0}^{\theta} pE\sin\theta' d\theta' \] Since \(p\) and \(E\) are constant: \[ W = pE \int_{0}^{\theta} \sin\theta' d\theta' \] Performing the integration: \[ W = pE [-\cos\theta']_{0}^{\theta} = pE [ (-\cos\theta) - (-\cos 0) ] \] Since \(\cos 0 = 1\): \[ W = pE (-\cos\theta + 1) \] \[ W = pE(1 - \cos\theta) \]

Step 4: Final Answer:
The formula for the work done in rotating an electric dipole by an angle \(\theta\) from its equilibrium position is \(W = pE(1 - \cos\theta)\).