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(B) The electric potential decreases in the direction of the electric field. Since the electric field $\vec{E}$ is in the positive $x$-direction,the p

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$V_A > V_B$

Vedclass · Answer

(B) The electric potential decreases in the direction of the electric field. Since the electric field $\vec{E}$ is in the positive $x$-direction,the potential decreases as $x$ increases. Point $A$ is at $x = 0$ and point $B$ is at $x = +1 \, cm$. Since $A$ is at a smaller $x$-coordinate than $B$,$V_A > V_B$. Points $A$ and $C$ lie on the same equipotential line (a line perpendicular to the electric field lines),therefore $V_A = V_C$. Thus,the correct relationship is $V_A > V_B$ and $V_A = V_C$.

$A$ uniform electric field pointing in the positive $x$-direction exists in a region. Let $A$ be the origin,$B$ be the point on the $x$-axis at $x = +1 \, cm$,and $C$ be the point on the $y$-axis at $y = +1 \, cm$. Then the potentials at the points $A$,$B$,and $C$ satisfy

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