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(C) The electric field is given by $E = 5x \hat{i}$.The potential difference between two points is given by $V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d

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$-10$

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(C) The electric field is given by $E = 5x \hat{i}$.The potential difference between two points is given by $V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r}$.Since the electric field is only in the $X$-direction, the $Y$-axis (where $x=0$) is an equipotential line. Thus, the potential at point $A(0, 5)$ is the same as the potential at the origin $C(0, 0)$.$V_A = V_C$.Now, we calculate the potential difference between $C(0, 0)$ and $B(2, 0)$:$V_B - V_C = -\int_{0}^{2} E_x dx = -\int_{0}^{2} 5x dx$.$V_B - V_C = -5 \left[ \frac{x^2}{2} ight]_{0}^{2} = -5 \left( \frac{4}{2} - 0 ight) = -5(2) = -10 	ext{ V}$.Since $V_A = V_C$, we have $V_B - V_A = -10 	ext{ V}$.

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