An electric charge 2 $\times$ 10^–3 μC placed at the origin (0, 0) of x – y coordinate system. The points A and B are situated at $(\sqrt{2}, \sqrt{2})$ and (2, 0) respectively. Find the potential difference between A and B.

Solution

Given, an electric charge 10^-3 μC is placed at origin of (x,-y) coordinate system.
Point A is at $(\sqrt{2}, \sqrt{2})$
Point B is at (2,0).
Let, r_Abe the distance from point A to origin and,
r_B be the distance from point B to (0,0).

Therefore,
r_{A= $\sqrt{(\sqrt{2)^{2}} + (\sqrt{2)^{2}}} = \sqrt{2 + 2} = 2 m$}r_B= $\sqrt{(2)^{2} + (0)^{2}} = \sqrt{4} = 2 m$
and, charge is, q = $2 \times 10^{- 3} μC = 2 \times 10^{- 9} C$

Now,
Potential difference between points A and B is given by,
V_A-V_B= $\frac{1}{4 {πε}_{\circ}} (\frac{q}{r_{A}} - \frac{q}{r_{B}}) = \frac{q}{4 π ε_{\circ}} (\frac{1}{r_{A}} - \frac{1}{r_{B}}) = 2 \times 10^{- 9} \times 9 \times 10^{9} \times (\frac{1}{2} - \frac{1}{2}) = 0$

Therfore, the potential difference is 0 between points A and B.

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An electric charge 2 $\times$ 10^–3 μC placed at the origin (0, 0) of x – y coordinate system. The points A and B are situated at $(\sqrt{2}, \sqrt{2})$ and (2, 0) respectively. Find the potential difference between A and B.