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Step-by-Step Solution
Step 1: Recall the expression for electric potential due to a point charge
The electric potential $V$ at a distance $r$ from a point charge $q$ is given by
$ V = \dfrac{kq}{r}, $
where $k$ is the electrostatic constant (often written as $ \dfrac{1}{4\pi\epsilon_0} $).
Step 2: Identify the given quantities
Point charge: $ q = 10^{-3}\,\mu C = 10^{-3} \times 10^{-6}\,C = 10^{-9}\,C.$
Position of the charge: Origin $(0,0).$
Point $A$: $(\sqrt{2}, \sqrt{2}).$
Point $B$: $(2,0).$
Step 3: Calculate distance of each point from the origin
For point $A$ at $(\sqrt{2}, \sqrt{2})$, the distance $r_A$ from the origin is
$ r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2. $
For point $B$ at $(2,0)$, the distance $r_B$ from the origin is
$ r_B = \sqrt{(2)^2 + 0^2} = \sqrt{4} = 2. $
Step 4: Compute the individual potentials at $A$ and $B$
Since both $r_A$ and $r_B$ are equal to $2$, the potential at $A$ due to the charge is
$ V_A = \dfrac{kq}{r_A} = \dfrac{k\, (10^{-9}\,C)}{2}, $
and the potential at $B$ is
$ V_B = \dfrac{kq}{r_B} = \dfrac{k\, (10^{-9}\,C)}{2}. $
Step 5: Find the potential difference
The potential difference $V_{AB}$ is
$ V_{AB} = V_A - V_B. $
Since
$ V_A = V_B = \dfrac{k \, (10^{-9}\,C)}{2}, $
we have
$ V_{AB} = \dfrac{k\, (10^{-9}\,C)}{2} - \dfrac{k\, (10^{-9}\,C)}{2} = 0. $
Step 6: Conclude the result
Because the distances of both points $A$ and $B$ from the origin are equal, they experience the same electric potential. Hence, the potential difference between $A$ and $B$ is zero.
