Q153 / 180 · COMEDK 2024 (Evening Shift) · Physics
A neutral water molecule, modeled as an electric dipole with an effective charge equal to one elementary charge ($e = 1.6 \times 10^{-19} \text{ C}$), is placed in an electric field $E = 2.5 \times 10^4 \text{ NC}^{-1}$. The work done to rotate it by $180^\circ$ from its stable equilibrium position is $5 \times 10^{-25} \text{ J}$. Find the approximate separation of the centre of charges.
- A$0.625 \times 10^{-9} \mathrm{~m}$
- Correct answerB$0.625 \times 10^{-10} \mathrm{~m}$
- C$1.25 \times 10^{-10} \mathrm{~m}$
- D$0.998 \times 10^{-10} \mathrm{~m}$
Solution
Work done to rotate dipole by $180°$ from stable equilibrium:
$W = pE(\cos 0° - \cos 180°) = 2pE$
$p = \dfrac{W}{2E} = \dfrac{5 \times 10^{-25}}{2 \times 2.5 \times 10^4} = \dfrac{5 \times 10^{-25}}{5 \times 10^4} = 10^{-29}$ C m
$p = qd \Rightarrow d = \dfrac{p}{q} = \dfrac{10^{-29}}{1.6 \times 10^{-19}} = 0.625 \times 10^{-10}$ m
$W = pE(\cos 0° - \cos 180°) = 2pE$
$p = \dfrac{W}{2E} = \dfrac{5 \times 10^{-25}}{2 \times 2.5 \times 10^4} = \dfrac{5 \times 10^{-25}}{5 \times 10^4} = 10^{-29}$ C m
$p = qd \Rightarrow d = \dfrac{p}{q} = \dfrac{10^{-29}}{1.6 \times 10^{-19}} = 0.625 \times 10^{-10}$ m
Hence, the correct answer is option: 'B'.