A charge of 20 , mu C is placed at the origin | vedclass

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(C) The electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{kQ}{r}$.For point $A(5a, 0)$,the distance from the o

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(C) The electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{kQ}{r}$.For point $A(5a, 0)$,the distance from the origin $(0, 0)$ is $r_A = \sqrt{(5a)^2 + 0^2} = 5a$.Thus,the potential at $A$ is $V_A = \frac{kQ}{5a}$.For point $B(-3a, 4a)$,the distance from the origin $(0, 0)$ is $r_B = \sqrt{(-3a)^2 + (4a)^2} = \sqrt{9a^2 + 16a^2} = \sqrt{25a^2} = 5a$.Thus,the potential at $B$ is $V_B = \frac{kQ}{5a}$.The potential difference between points $A$ and $B$ is $V_A - V_B = \frac{kQ}{5a} - \frac{kQ}{5a} = 0$.

$A$ charge of $20 \, \mu C$ is placed at the origin. What is the potential difference between the points $(5a, 0)$ and $(-3a, 4a)$?

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