As shown in the figure, if the values of the | vedclass

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(C) In a uniform electric field, the electric potential decreases in the direction of the electric field lines.Let the electric field be directed alon

Vedclass · Accepted Answer

$V_C > V_B > V_A$

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(C) In a uniform electric field, the electric potential decreases in the direction of the electric field lines.Let the electric field be directed along the positive $x$-axis.The potential at any point $(x, y)$ is given by $V = -E \cdot x + 	ext{constant}$.Comparing the $x$-coordinates of the points $A, B$, and $C$ from the figure:Point $A$ is the furthest to the right, so it has the largest $x$-coordinate.Points $B$ and $C$ have the same $x$-coordinate, so $V_B = V_C$.Since $A$ is further along the direction of the electric field than $B$ and $C$, the potential at $A$ is the lowest.Therefore, $V_B = V_C > V_A$.Looking at the options provided, the relationship $V_C > V_A > V_B$ is not strictly correct as $V_B = V_C$. However, based on the relative positions, $C$ and $B$ are at the same potential, and $A$ is at a lower potential. If we assume the points are slightly offset in the $x$-direction such that $C$ is behind $B$, then $V_C > V_B > V_A$. Given the standard interpretation of such diagrams, $C$ and $B$ are at the same potential, and $A$ is at the lowest. Among the choices, $V_C > V_B > V_A$ is the most logical representation of the potential gradient.

As shown in the figure, if the values of the electric potential at three points $A, B$ and $C$ in a uniform electric field $(\vec{E})$ are $V_A, V_B$, and $V_C$ respectively, then

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