After having just learned about electric fields I am trying to write down an equation for the net electric field at a point. Though aware of Gauss's law, I want to try to solve this problem with what I already know.

Say I have a function $Q : \mathbb{R}^3 \to \mathbb{R}$ giving me the charge at any given point in $2D$ space. I want to write down an equation $\vec{E} : \mathbb{R}^3 \to \mathbb{R}^3$ describing the net electric field induced by a point charge on any given point in $\mathbb{R}^3$.

But I'll start with a simpler idea. Let $q_1$ and $q_2$ be two point charges at $\vec{r_1}$ and $\vec{r_2}$ respectively. I want to find an equation for the components of the force of $q_1$ on $q_2$. Here's my problem. Coloumb's law $$\vec{F} = k\frac{q_1q_2}{r^2}\hat{r}$$ only gives the radial force. I need general equation to describe the components of $\vec{F}$ along $\hat{i}$ and $\hat{j}$, and $\hat{k}$. So I must somehow get the radial force components along the regular $\hat{i}$ and $\hat{j}$ axes.

My first intuition was to use the equation of projection (which, incidentally, I just learned in Calc III yesterday). Then, I would write down the new force vector $\vec{F_t}$ as projected onto the unit vector for the x axis.

$$\vec{F_t} = \frac{\vec{F} \cdot \hat{i} }{ \vert \hat{i}\vert^2}\hat{i} = \vec{F} \cdot \hat{i} $$

Substituting in $\vec{F}$ from Coloumb's law:

$$\vec{F}_t = \left(k\frac{q_1q_2}{r^2} \hat{r}\right) \cdot \hat{i}$$

With the field coming from $q_1$, the radial vector $\hat{r}$ from $\vec{r}_2$ to $\vec{r}_2$ can be expressed as

$$\hat{r} = \vec{r}_2 - \vec{r}_1$$

And then finally:

$$\vec{F}_t = k\frac{q_1q_2}{r^2} (\vec{r}_2 - \vec{r}_1) \cdot \hat{i}$$

Then, dividing by $q_2$, we can obtain the general electric field equation between two points $\vec{r}_1$ and $\vec{r}_2$:

$$\vec{E}(r) = k\frac{q_1}{r^2} (\vec{r} - \vec{r}_\text{ref}) \cdot \hat{i}$$

This equation for field relative to a point charge is really powerful. Choosing a point $r_2$, I can integrate $\vec{E}$ over a bounded region of $Q$ to find the net field at $r_2$.

What I am looking for

Very simple. I'm hoping someone can verify my derivation for the field equation ($\vec{E}$) is correct. We didn't discuss this is in class. Purely for my own enjoyment.


1 Answer 1


After thinking about it a bit I realized what I had been doing was wrong. The solution is a great triviality.

First, the electric field equation is along $\hat{r}$ which is a radial vector to any other point $\vec{r}^\prime$

$$ \vec{E} = k\frac{q}{r^2}\hat{r}$$

If the point particle is located at $\vec{r}_p$, the radial vector from $\vec{r}_p$ to some $\vec{r}$ is given by:

$$\hat{r} = \vec{r} - \vec{r}_p$$

Substituting this into the equation for $\vec{E}$:

$$ \vec{E} = k\frac{q}{r^2}\hat{r} = k\frac{q}{\left|\vec{r} - \vec{r}_p\right|^2}(\vec{r} - \vec{r}_p)$$

Now, given an arbitrary function for charge density $\lambda(\vec{x})$ we can define the differential field at a point as $d\vec{E} = k\frac{\lambda (x)}{r^2}\hat{r} d\vec{x}$.

Then, integrating over the surface on which $\lambda (\vec{x})$ is defined:

$$\vec{E}(r_p) = \oint d \vec{E} = \oint k\frac{\lambda (\vec{x})}{r^2}\hat{r}(\vec{r_p}) d\vec{x}$$

where $\hat{r}(\vec{r_p})$ is a radial vector to an arbitrary point $r_p$

For example, this image shows the electric field induced on a point $\vec{r_p} = 4\hat{i} + 2\hat{j}$. Summing these vectors for some infinite or finite region yields the net field at $\vec{r_p}$. If the integral doesn't converge...then I don't know what that would mean physically.

enter image description here


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