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(D) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$,where $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.F

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(D) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$,where $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.For point $A(0, a)$,the distance from the origin is $r_A = \sqrt{0^2 + a^2} = a$.Thus,the potential at $A$ is $V_A = \frac{kq}{a}$.For point $B(a, 0)$,the distance from the origin is $r_B = \sqrt{a^2 + 0^2} = a$.Thus,the potential at $B$ is $V_B = \frac{kq}{a}$.The potential difference between points $A$ and $B$ is $\Delta V = V_A - V_B = \frac{kq}{a} - \frac{kq}{a} = 0$.

$A$ charge of $10 \,\mu C$ is placed at the origin of an $x-y$ coordinate system. The potential difference between two points $(0, a)$ and $(a, 0)$ in volts will be

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