Can I calculate the scalar potential from the electric field using: $\phi = \int \nabla \phi = - \int \vec{E}$

Question

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.

Answer 1

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

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

NOTE:

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.