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I was reading an article about resistivity when I faced this equation:

$$V_P-V_O=\int_P^O E.dr$$

The equation is in one dimension and $P$ and $O$ are points on the $x$ axis.

The Electric field is oriented in the $+x$ direction ans has the formula $c\frac1r$ where $c$ is a coefficient and $r$ is the distance form the origin of the coordinate system.

The vector $\vec r_P - \vec r_O$ has the same direction as that of the electric field.


Now, Why do we put $P$ below the integral and $O$ above it? Can someone explain the equation for me?

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  • $\begingroup$ Because by definition $V_P - V_O$ is the line integral of $(-E)$ from $O$ to $P$. If you add a minus sign to $E$, then $O$ is below and $P$ is above. $\endgroup$ – velut luna Apr 21 '16 at 5:17
  • $\begingroup$ @Mathaholic so to obtain the potential difference from point $A$ to point $B$ all I need is to take the line integral of $-E$ from $B$ to $A$. Why $-E$ ? $\endgroup$ – AHB Apr 21 '16 at 5:30
  • $\begingroup$ It is just a convention. But if you define it as the line integral of $E$ instead of $-E$, then positive charges flow from lower potential to higher potential, negative charges flow from higher potential to lower potential. It's perfectly ok if everyone can accept this. $\endgroup$ – velut luna Apr 21 '16 at 5:34
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There are two equivalent definitions for the potential difference between two points.

  • The potential difference between two points is the work done by an external force in moving unit positive charge from one point to the other.
  • The potential difference between two points is the minus work done by the electric field in moving unit positive charge from one point to the other.

Inverting the limits in your equation $V_P-V_O=\int_P^O \vec E \cdot d\vec r$ gives $V_P-V_O=-\int_O^P \vec E \cdot d\vec r$.

So the right hand side of the equation is minus the work done by the the electric field in taking unit positive charge from position $O$ to position $P$.

Writing your equation as $V_P-V_O=\int_O^P \left( -\vec E \right)\cdot d\vec r$ gives you the work done by the external force $-\vec E$ in taking unit positive charge from position $O$ to position $P$.

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