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(A) The electric field is given as $\vec{E} = E_0 \hat{i}$,where $E_0 > 0$.$1$. For points $A(0,0)$ and $B(1,0)$: The potential difference is $V_B - V

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$V_A > V_B$ and $V_A = V_C$

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(A) The electric field is given as $\vec{E} = E_0 \hat{i}$,where $E_0 > 0$.$1$. For points $A(0,0)$ and $B(1,0)$: The potential difference is $V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r} = -\int_0^1 E_0 dx = -E_0$.Since $E_0 > 0$,$V_B - V_A  V_B$.$2$. For points $A(0,0)$ and $C(0,1)$: The potential difference is $V_C - V_A = -\int_A^C \vec{E} \cdot d\vec{r}$. Since the electric field is in the $x$-direction and the displacement $d\vec{r}$ is in the $y$-direction,$\vec{E} \cdot d\vec{r} = 0$.Therefore,$V_C - V_A = 0$,which implies $V_A = V_C$.Combining these,we get $V_A > V_B$ and $V_A = V_C$.

$A$ uniform electric field exists in the positive $x$-direction. Let $A$ be the origin. Point $B$ is at $x = +1 \ cm$ on the $x$-axis and point $C$ is at $y = +1 \ cm$ on the $y$-axis. Which of the following relations is correct regarding the potentials at $A, B,$ and $C$?

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