Solution:

The potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by:

\[
V = \frac{kq}{r}
\]

1. Distance from the charge at the origin:
- Point \( A(\sqrt{2}, \sqrt{2}) \): Distance \( r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{4} = 2 \).
- Point \( B(2, 0) \): Distance \( r_B = \sqrt{(2)^2 + 0^2} = 2 \).

2. Potentials at \( A \) and \( B \):
\[
V_A = \frac{kq}{r_A} = \frac{kq}{2}, \quad V_B = \frac{kq}{r_B} = \frac{kq}{2}.
\]

3. Potential difference:
\[
V_A - V_B = \frac{kq}{2} - \frac{kq}{2} = 0.
\]

Thus, \( V_A - V_B = 0 \).