Four charges of 1 mu C, 2 mu C, 3 mu C, and - | vedclass

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(A) The electric potential $V$ at any point $(x, y, z)$ due to a system of point charges $q_i$ at positions $(x_i, y_i, 0)$ is given by $V = \sum \fra

Vedclass · Accepted Answer

The electric potential is zero at the origin.

Vedclass · Answer

(A) The electric potential $V$ at any point $(x, y, z)$ due to a system of point charges $q_i$ at positions $(x_i, y_i, 0)$ is given by $V = \sum \frac{k q_i}{r_i}$,where $r_i = \sqrt{(x-x_i)^2 + (y-y_i)^2 + z^2}$.At the origin $(0, 0, 0)$,the distance $r$ from each corner of a square of side $a=1\ m$ to the center is $r = \frac{a}{\sqrt{2}} = \frac{1}{\sqrt{2}}\ m$.The potential at the origin is $V = \frac{k}{r} (q_1 + q_2 + q_3 + q_4) = \frac{k}{r} (1 + 2 + 3 - 6) \mu C = \frac{k}{r} (0) = 0$.Thus,the electric potential is zero at the origin.For any point on the $z$-axis,the potential is $V(z) = \sum \frac{k q_i}{\sqrt{r_i^2 + z^2}}$. Since the sum of the charges is zero,the potential at large distances $z$ approaches zero,but it is not zero everywhere along the $z$-axis. Therefore,only option $A$ is correct.

Four charges of $1\ \mu C, 2\ \mu C, 3\ \mu C,$ and $-6\ \mu C$ are placed at the corners of a square of side $1\ m$. The square lies in the $x-y$ plane with its center at the origin.

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