Question

An electric dipole consists of two point charges $+1\ \mu\text{C}$ and $-1\ \mu\text{C}$, held $10\text{ cm}$ apart. It is subjected to a uniform electric field of $100\text{ N/C}$. Calculate the amount of work done in turning the dipole from its position of stable equilibrium to the position of unstable equilibrium, in the field.

Hint:

Remember key rotation work formulas:
$0^\circ \to 90^\circ \implies W = pE$
$0^\circ \to 180^\circ \implies W = 2pE$ (Stable to Unstable equilibrium)

Answer 1

Concept:
• Electric dipole moment is $p = q \times (2a)$.

• Potential energy of a dipole in a uniform electric field $\vec{E}$ at angle $\theta$ is $U(\theta) = -p E \cos \theta$.

• Work done in rotating a dipole from angle $\theta_1$ to $\theta_2$ is $W = U(\theta_2) - U(\theta_1) = -p E (\cos \theta_2 - \cos \theta_1)$.

• Stable equilibrium corresponds to $\theta_1 = 0^\circ$ ($\vec{p}$ parallel to $\vec{E}$).

• Unstable equilibrium corresponds to $\theta_2 = 180^\circ$ ($\vec{p}$ antiparallel to $\vec{E}$).

Step 1: Calculate electric dipole moment
Given charge magnitude $q = 1\ \mu\text{C} = 1 \times 10^{-6}\text{ C}$.
Separation distance $2a = 10\text{ cm} = 0.1\text{ m}$.
Electric field intensity $E = 100\text{ N/C}$.
Dipole moment $p$:
\[ p = q \times (2a) = (1 \times 10^{-6}\text{ C}) \times (0.1\text{ m}) = 10^{-7}\text{ C}\cdot\text{m} \]

Step 2: Formulate work done equation
For stable equilibrium position: $\theta_1 = 0^\circ$.
For unstable equilibrium position: $\theta_2 = 180^\circ$.
Work done $W$:
\[ W = -p E (\cos \theta_2 - \cos \theta_1) \]
\[ W = -p E (\cos 180^\circ - \cos 0^\circ) \]
Substitute $\cos 180^\circ = -1$ and $\cos 0^\circ = 1$:
\[ W = -p E (-1 - 1) = 2 p E \]

Step 3: Substitute values and compute result
\[ W = 2 \times (10^{-7}\text{ C}\cdot\text{m}) \times (100\text{ N/C}) \]
\[ W = 2 \times 10^{-5}\text{ J} \]

Step 4: Conclusion
The work done in turning the dipole from stable to unstable equilibrium is $2 \times 10^{-5}\text{ J}$.