# How to divide areas in electric field lines based on field strength?

A negative charge is surrounded by four positive charges. They are all of the same strength. The electric field lines are plotted below. I am looking for the property of the drawn 'red lines' that originate from the negative charge.

Is there anyway that I can explicitly find the lines that originate from the center negative charge, and divide each pair of neighboring positive charges? (e.g. for the top-left and bottom-left chargers, they each has an electric field line bending to the left, and a neighboring electric field line that ends at the negative charge. I need to find the 'red line' such that divide the space between the two electric lines pointing left equally, as well as the space between the two electric lines pointing to the negative charge)

Also, when the charges are of different strength, the 'red lines' will change to curves, and what will the properties of these curves be?

The red lines represent equipotential lines, which have the additional property of having an electric field of zero.

The electric field at a point on the red line can be defined as

$$\mathbf E_{red}=\sum_{i=1}^{4}\mathbf E_i$$

$$\mathbf E_i=\frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i^2}\cdot{\mathbf{\hat{r}_{0i}}}$$

Let's take the case where all charges are the same and spread out in a way as shown in your figure. If we look at the pairs of charges, the values will be the same $E_1=E_2$, only the vector $\mathbf{\hat{r}_{01}}=-\mathbf{\hat{r}_{02}}$ will be opposite. If we do a vector addition we get:

$$\mathbf E_{red}=0$$

This has another side effect. As we want to find the potential field (e.g. voltage) we can do another step.

$$\mathbf E=-\nabla V$$

Without calculating this, we see that the first derivate of the potential is zero on the red line. If the first derivate is zero, we have a point that is either a local minimum or maximum of a function. A maximum (either positive or negative) wouldn't make much sense, it is a local equilibrium point. Therefore the potential of this line is zero too. Mathematically this can be done if you calculate the integral(for example for a point $k$):

$$V=\int_0^\infty E_{red}(k) \mathrm{d}r=0-0=0$$

As the electric field in both the point $k$ and in infinity($\infty$) are zero, so the potential of the point $k$ has to be zero.

Look at your figure again, aren't the red lines mirrors? There is a method that uses this principle to calculate (for example) the capacitance of conductors on power lines towards ground. It is the method of image charges

• Thank you so much for the explanation. However, I am a little confused about the 'equipotential line'. It is clear that electric fields caused by charge 1 and 2 cancel out at the midpoint k, but there are still two positive charges and a negative charge to the right side of point k. How do they cancel out? If I decompose the electric field into x-direction and y-direction, certainly the y-direction strength is always zero on that line. But how about x-direction? Thank you so much for your time. – mike Aug 27 '15 at 7:05
• Actually, I did this in MATLAB. for i = 1:5 { r = (X-x(i)).^2 + (Y-y(i)).^2 + 0.01; U(:,:,i) = q(i)./r; UU = UU + U(:,:,i);} end Where, q(i) is the strength of charge i, x(i) and y(i) are simply coordinated. X and Y are the 'meshgrid' generated in MATLAB. – mike Aug 27 '15 at 7:05
• Actually, I did this in MATLAB. for i = 1:5 { r = (X-x(i)).^2 + (Y-y(i)).^2 + 0.01; U(:,:,i) = q(i)./r; UU = UU + U(:,:,i);} end Where, q(i) is the strength of charge i, x(i) and y(i) are simply coordinated. X and Y are the meshgrid generated in MATLAB. Then I looked the values of UU at y=4, they are not zeros. I do, however, pick the smallest UU value, and plotted two stremlines starting from a little bit to its left and to its right. This is how I get the left horizontal red line. – mike Aug 27 '15 at 7:12