An electric field $\vec{E} = 10x\,\hat{i}$ (SI units) exists in a region. The potential difference $V_O - V_A$ between the origin ($O$) and the point $x = 2\ m$ ($A$) is

1. $V_O - V_A = -\displaystyle\int_{A}^{O} \vec{E}\cdot d\vec{r} = \int_{O}^{A} E\,dx$ with $A$ at $x=2$. 2. $V_O - V_A = \displaystyle\int_{0}^{2} 10x\,dx = [5x^2]_0^2 = 20$... but $V_O-V_A = -\int_0^2 E\,dx = -20$ V. 3. Since the field points along $+x$, the origin is at **lower** potential than $A$, giving $V_O - V_A = -20$ V. _Source: Samacheer Kalvi Class 12 Physics Vol.1, Unit 1 'Electrostatics', Evaluation Q.9_