electric field $$\vec{E}$$ is given by:

$$ W = pE \cos(\theta) - pE \cos(0) $$

For a rotation by $$180^{\circ}$$, $$\theta = 180^{\circ}$$, 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:

$$ 5 \times 10^{-25} \text{ J} $$

And the electric field strength is:

$$ E = 2.5 \times 10^4 \text{ N/C} $$

We can find the dipole moment, $$p$$:

$$ 5 \times 10^{-25} = -2p(2.5 \times 10^4) $$

Solving for $$p$$:

$$ 5 \times 10^{-25} = -5 \times 10^4 p $$

$$ p = \frac{5 \times 10^{-25}}{-5 \times 10^4} $$

$$ p = -1 \times 10^{-29} \text{ C m} $$

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):

$$ q = 1.6 \times 10^{-19} \text{ C} $$

Therefore, we can solve for the separation $$d$$:

$$ d = \frac{p}{q} $$

$$ d = \frac{1 \times 10^{-29}}{1.6 \times 10^{-19}} $$

$$ d = \frac{1}{1.6} \times 10^{-10} $$

$$ d \approx 0.625 \times 10^{-10} \text{ m} $$

So, the approximate separation of the centers of charges is:

$$ 0.625 \times 10^{-10} \text{ m} $$

Hence, the correct option is:

Option B: $$0.625 \times 10^{-10} \mathrm{~m}$$

Electrostatics

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