A point charge of $$10^{-8}$$ is placed at origin. The work done in moving a point charge $$2 \mu C$$ from point A(4, 4, 2) m to B(2, 2, 1) m is _____J $$\left(\frac{1}{4\pi\epsilon _{0}}=9\times10^{9} \text{in SI units}\right)$$

The electrostatic potential at a distance $$r$$ from a point charge $$q$$ (taking zero potential at infinity) is
$$V = k\,\frac{q}{r}$$ where $$k = \frac{1}{4\pi\epsilon_0} = 9\times10^{9}\ \text{N m}^2\text{C}^{-2}$$.

The source charge at the origin is $$q = 10^{-8}\ \text{C}$$.
The test charge to be moved is $$q_0 = 2\ \mu\text{C} = 2\times10^{-6}\ \text{C}$$.

Coordinates of the two points:
A $$\left(4,\;4,\;2\right)\ \text{m}$$     B $$\left(2,\;2,\;1\right)\ \text{m}$$.

Distance of A from the origin:
$$r_A = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16+16+4} = \sqrt{36} = 6\ \text{m}$$.

Distance of B from the origin:
$$r_B = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3\ \text{m}$$.

Potentials at A and B:
$$V_A = k\,\frac{q}{r_A} = 9\times10^{9}\,\frac{10^{-8}}{6}= \frac{90}{6}=15\ \text{V}$$,
$$V_B = k\,\frac{q}{r_B} = 9\times10^{9}\,\frac{10^{-8}}{3}= \frac{90}{3}=30\ \text{V}$$.

Potential difference experienced by the test charge when moving from A to B:
$$\Delta V = V_B - V_A = 30 - 15 = 15\ \text{V}$$.

Work done $$W$$ in moving the charge is
$$W = q_0\,\Delta V = 2\times10^{-6}\,\times\,15 = 30\times10^{-6}\ \text{J}$$.

Therefore, the work done is $$30\times10^{-6}\ \text{J}$$.

Option B is correct.