Question

Two electric charges \(+3.2\times10^{-19}\,\text{C}\) and \(-3.2\times10^{-19}\,\text{C}\) are placed \(2.4\,\text{\AA}\) apart to form an electric dipole. It is placed in a uniform electric field of intensity \(4\times10^5\,\text{V m}^{-1}\). The work done to rotate the electric dipole from equilibrium position by \(180^\circ\) is:

• A. \(3\times10^{-25}\,\text{J}\)
• B. \(6\times10^{-23}\,\text{J}\)
• C. \(12\times10^{-23}\,\text{J}\)
• D. Zero

Hint:

Potential energy of dipole: \[ U=-pE\cos\theta \] Hence rotating dipole from stable equilibrium \((0^\circ)\) to unstable equilibrium \((180^\circ)\): \[ W=2pE \]

Answer 1

The Correct Option is B

Step 1: Find the dipole moment.
A dipole moment is charge times the gap between the charges. \[ p = q d \] Here $q = 3.2\times10^{-19}$ C and $d = 2.4\times10^{-10}$ m.

Step 2: Multiply to get p.
\[ p = (3.2\times10^{-19})(2.4\times10^{-10}) = 7.68\times10^{-29}\,\text{C m} \]
Step 3: Recall the rotation work formula.
To turn a dipole from lined up ($0^\circ$) to fully reversed ($180^\circ$), the work needed is \[ W = pE(\cos0^\circ - \cos180^\circ) = 2pE \]
Step 4: Put in the field.
The field is $E = 4\times10^{5}$ V/m. \[ W = 2 (7.68\times10^{-29})(4\times10^{5}) \]
Step 5: Multiply the numbers.
\[ W = 2 \times 7.68 \times 4 \times 10^{-29+5} = 61.44 \times 10^{-24} \]
Step 6: Write in standard form.
\[ W = 6.14\times10^{-23}\,\text{J} \approx 6\times10^{-23}\,\text{J} \] \[ \boxed{6\times10^{-23}\,\text{J}} \]