If I have a relation between the electric field and the radius, can I calculate the relation between the scalar potential and the radius using: $\phi = \int \nabla \phi = - \int \vec{E}$?

$$\nabla \phi = - \vec{E}$$ $$\phi = \int \nabla \phi = - \int \vec{E}$$?

If so, how do I fix the vector to magnitude conversion for the equation below?

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

$\phi = - \int \vec{E} = -\int \frac{\lambda}{2\pi \epsilon_0 r} dr = -\frac{\lambda}{2\pi \epsilon_0} ln(r)$?

When calculation $- \int \vec{E}$, but $E = \frac{\lambda}{2\pi \epsilon_0 r}$ is not a vector.


To be precise, we have for an electrostatic field

$$\phi(\vec r_1)-\phi(\vec r_2)=\int_{C_{12}}\vec E(\vec r')\cdot d\vec r'$$

where $C_{12}$ is any contour that begins at $\vec r_1$ and ends at $\vec r_2$.

For $\vec E(\vec r) =\hat r E_r(r)=\hat r\frac{\lambda}{2\pi \epsilon_0\,r}$, we have

$$\phi(\vec r)-\phi(\vec r_0)=\int_{\vec r}^{\vec r_0} \hat r\frac{\lambda}{2\pi \epsilon_0 r}\cdot \hat r dr=-\frac{\lambda}{2\pi \epsilon_0}\log(r/r_0)$$

where we took the integration path along the straight line radial path from $r_0$ to $r$. There is no contribution to the integral from any path of constant $r$ since the electric field has only a radial component (i.e., $\vec E \cdot \hat \theta r d\theta =0$).

Inasmuch as the scalar potential is relative to within an additive constant, we may write

$$\phi(\vec r)=-\frac{\lambda}{2\pi \epsilon_0}\log r \tag1$$


We can easily verify that

$$\vec E=-\nabla \phi \tag 2$$

Note that in $(1)$, we tacitly set $\phi(\vec r_0)=\frac{\lambda}{2\pi \epsilon_0} \log r_0$. Regardless of this choice, the relationship $(2)$ holds.

  • $\begingroup$ log in equation 1 is the natural logarithm? Why 2 times $\hat r$ in the integral? So, what was your conclusion? I can say $\phi = -\int \frac{\lambda}{2\pi \epsilon_0 r} dr = -\frac{\lambda}{2\pi \epsilon_0} ln(r)$?? $\endgroup$ – PatrickWin Jun 21 '15 at 20:31
  • $\begingroup$ Yes, the natural logarithm is abbreviated "$\log$" throughout. I don't see a factor of $2$ in the integral. And yes, we conclude that $\phi=-\frac{\lambda}{2\pi \epsilon_0}\log r$. Please let me know how I can improve my answer. I just want to give you the best answer I can. $\endgroup$ – Mark Viola Jun 21 '15 at 20:39
  • $\begingroup$ I meant the $\hat r$ in the integral $\int_{\vec r}^{\vec r_0} \hat r\frac{\lambda}{2\pi \epsilon_0 r}\cdot \hat r dr$. The first $\hat r$ is a vector in the direction of the radius 'r' with unit length to make the a vector out of the magnitude of the electric field. Is the $\hat r dr$ to takes only the part of the vector that is in the same direction as r, but the entire vector is in the direction of r, so $cos(\theta)=cos(0)=1$. So I can also calculate it the way is done in my question, when taking this step into account? $\endgroup$ – PatrickWin Jun 22 '15 at 9:51

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