The task is to calculate the voltage between points $M$ and $N$ if the electric field vector is known to be $\vec{E}=\frac{V_0\cdot x^2}{a^3} \cdot \vec{i} + \frac{V_0 \cdot y}{a^2} \cdot \vec{j}$, where $V_0$ and $a$ are constants. I've provided a simple sketch below.

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Formula for calculating voltage is: $$U_{MN}=\int_M ^N {\vec{E}\cdot d\vec{l}}$$

We are going to integrate from $M$ to $P$ and then from $P$ to $N$, because this way we integrate along one axis and consider only $x$ or $y$ component of $\vec{E}$ because one of them will be perpendicular to the path of integration and thus make no influence.

$U_{MN}=\int_M ^N {\vec{E}\cdot d\vec{l}} = \int_M ^P {{\vec{E}\cdot d\vec{l}}} + \int_P ^N {\vec{E}\cdot d\vec{l}}$

The first integral I understand.

$\int_M ^P {\vec{E} \cdot d\vec{l}}=\int_M ^P {\vec{E_x} \cdot d\vec{l}} =\int_M ^P {\frac{V_0\cdot x^2}{a^3} \cdot \vec{i}}\cdot d\vec{l} =\int_M ^P {\frac{V_0\cdot x^2}{a^3} \cdot \vec{i}}\cdot dx \cdot$ $\vec{i} = \int_a ^{4a} {\frac{V_0 \cdot {x^2}}{a^3}} \cdot {dx} = \dots $

What I do not understand is the second integral.

$$\int_P ^N \vec{E}\cdot d\vec{l} = $$ $$= \int_P ^N \vec{E_y} \cdot d\vec{l}$$

We are integrating from $P$ to $N$, so it should be that $d\vec{l} = dy \cdot -{\vec{j}}$, because the path $\vec{PN}$ is in the opposite direction from $\vec{j}$. Minus sign goes in front of the integral, and then we can swap integral limits(which destroys the minus sign):

$$= \int_P ^N {\frac{V_0 \cdot y}{a^2} \cdot \vec{j} \cdot dy \cdot -\vec{j}}$$ $$= \int_{\sqrt{7}a} ^{7a} {\frac{V_0 \cdot y}{a^2} \cdot dy}$$

and now we are integrating from $N$ to $M$ in the direction of $d\vec{y}$. When we get the final result, all we need to do is to perform a simple addition.

In the solution of this problem it says to change the sign of the integral AND swap the limits of integration. Then, $d\vec{l} = d\vec{y}$. Their integral turns out to be $$- \int_{\sqrt{7}a} ^{7a} {\frac{V_0 \cdot y}{a^2} \cdot dy}$$ and I don't understand why. I think they ignored that $d\vec{l}$ goes in the opposite direction of $\vec{j}$, but this is just a wild guess.

So, what interests me is how did the final integral become negative?

  • $\begingroup$ How do we know that choosing an other path we will find the same value for the voltage ??? $\endgroup$ – Frobenius Apr 14 at 23:17

There's a problem in this equation: $$ d\vec{l}=dy \cdot -\vec{j} $$ here $dy$ needs a minus sign. It is easier to see this writing the path.

The path of integration is parameterized by: $$ \vec{l}=\vec{P}+(\vec{N}-\vec{P})\frac{y-y_P}{y_N-y_P} = \vec{P}+\hat{j}(y-7a) \\ y_p \le y \le y_N $$

Therefore $$ d\vec{l}=\hat{j}dy $$ and $$ \int_P^N E_y \cdot d\vec{l} = \int_P^N E_y \cdot \frac{d\vec{l}}{dy}dy = \int_{7a}^{\sqrt{7}a} \frac{V_0y}{a^2}\hat{j} \cdot \hat{j}dy = -\int_{\sqrt{7}a}^{7a} \frac{V_0y}{a^2} dy $$

Sometimes I find useful to change the variable because using the coordinates as parameter of integration can be confusing and lead to this problems.


If this is an exercise for path integration then it's Mathematics. But as Physics problem it's important to note first that the given electric field is irrotational \begin{equation} \boldsymbol{\nabla\times}\mathbf{E}\boldsymbol{=}\left(\dfrac{\partial E_y}{\partial x}\boldsymbol{-}\dfrac{\partial E_x}{\partial y}\right)\mathbf{k} \boldsymbol{=0} \tag{1}\label{1} \end{equation} so it comes from a potential $\phi\left(x,y\right)$ \begin{equation} \mathbf{E}\left(x,y\right)\boldsymbol{=-}\boldsymbol{\nabla}\phi\left(x,y\right)\boldsymbol{=-}\left(\dfrac{\partial \phi}{\partial x}\mathbf{i}\boldsymbol{+}\dfrac{\partial \phi}{\partial y}\mathbf{j}\right) \tag{2}\label{2} \end{equation} hence the voltage between two points is the potential difference which is independent of the path of integration of $\mathbf{E}\left(x,y\right)$.

In our case the expression of $\mathbf{E}\left(x,y\right)$ \begin{equation} \mathbf{E}\left(x,y\right)\boldsymbol{=}E_x\,\mathbf{i}\boldsymbol{+}E_y\,\mathbf{j}\boldsymbol{=}\left(\dfrac{V_0 x^2}{a^2}\right)\mathbf{i}\boldsymbol{+}\left(\dfrac{V_0 y}{a^2}\right)\mathbf{j} \tag{3}\label{3} \end{equation} is very simple, we could guess $\phi\left(x,y\right)$ and so to find the voltage between any two points without integration.


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