Question 73: Electrostats & Capacitance

Question

A neutral water molecule is placed in an electric field $$E=2.5 	imes 10^4 \mathrm{NC}^{-1}$$. The work done to rotate it by $$180^{\circ}$$ is $$5 	imes 10^{-25} \mathrm{~J}$$. Find the approximate separation of centre of charges.

Accepted Answer

To find the approximate separation of the centers of charges in the water molecule, we need to consider the work done in rotating the dipole in an electric field. The work done $$W$$ to rotate a dipole of moment $$\vec{p}$$ by an angle $$	heta$$ in an electric field $$\vec{E}$$ is given by:
$$ W = pE \cos(	heta) - pE \cos(0) $$
For a rotation by $$180^{\circ}$$, $$	heta = 180^{\circ}$$, and $$\cos(180^{\circ}) = -1$$:
$$ W = pE \cos(180^{\circ}) - pE \cos(0) $$
Since $$\cos(0) = 1$$, the above equation simplifies to:
$$ W = pE(-1) - pE(1) $$
$$ W = -pE - pE $$
$$ W = -2pE $$
Given that the work done is:
$$ 5 	imes 10^{-25} 	ext{ J} $$
And the electric field strength is:
$$ E = 2.5 	imes 10^4 	ext{ N/C} $$
We can find the dipole moment, $$p$$:
$$ 5 	imes 10^{-25} = -2p(2.5 	imes 10^4) $$
Solving for $$p$$:
$$ 5 	imes 10^{-25} = -5 	imes 10^4 p $$
$$ p = \frac{5 	imes 10^{-25}}{-5 	imes 10^4} $$
$$ p = -1 	imes 10^{-29} 	ext{ C m} $$
Since the dipole moment $$p$$ is given by the product of the charge $$q$$ on the water molecule and the separation $$d$$ of the charges, we have:
$$ p = qd $$
For a water molecule, the charge of each center (hydrogen or oxygen) can be approximated to the elementary charge $$e$$ (since a water molecule is neutral overall, but the centers have polarity):
$$ q = 1.6 	imes 10^{-19} 	ext{ C} $$
Therefore, we can solve for the separation $$d$$:
$$ d = \frac{p}{q} $$
$$ d = \frac{1 	imes 10^{-29}}{1.6 	imes 10^{-19}} $$
$$ d = \frac{1}{1.6} 	imes 10^{-10} $$
$$ d \approx 0.625 	imes 10^{-10} 	ext{ m} $$
So, the approximate separation of the centers of charges is:
$$ 0.625 	imes 10^{-10} 	ext{ m} $$
Hence, the correct option is:
Option B: $$0.625 	imes 10^{-10} \mathrm{~m}$$

Answer

$$1.25 	imes 10^{-10} \mathrm{~m}$$

Answer

$$0.625 	imes 10^{-10} \mathrm{~m}$$

Answer

$$0.625 	imes 10^{-9} \mathrm{~m}$$

Answer

$$0.998 	imes 10^{-10} \mathrm{~m}$$

Electrostats & Capacitance

A neutral water molecule is placed in an electric field $E=2.5 \times 10^4 \mathrm{NC}^{-1}. The work done to rotate it by 180^{\circ} is 5 \times 10^{-25} \mathrm{~J}$. Find the approximate separation of centre of charges.

Choose the correct answer: