# How to find the lines of force from electric field?

I know that the line of force at any point in electric field is tangent to the field there however how can I quantitatively find it? As an example if the electric field in a region is given by $\vec E=x\hat i+y\hat j$ how can I find the equation of line of force?

• If I understand you correctly, you're asking how to draw field lines given an expression for the vector field. This is really more of a mathematics question than a physics question, so you might try rephrasing it and asking it over there if you don't get a good response here. Commented May 24, 2017 at 15:13
• @MichaelSeifert I was asking for the equation that represents field lines given an expression for the vector field and not how to draw field lines. Commented May 25, 2017 at 11:46

Farcher's answer talks about the direct method to find the electric field. I am going to talk about a more elegant method which uses the concept of flux. But it can be used only when the problem has symmetry in it.

The main idea is that the flux through all surfaces is a constant, as long as the field lines that enclose the edge of the surfaces are the same.

Let us take a case of a point charge $$q$$ situated at the origin. We shall attempt to calculate the equation of a field line that comes out at an angle $$\theta$$ with the $$x$$-axis. But due to the symmetry in the probe, all such field lines enclose a conical surface ABC as shown in the figure:

Now the flux through the solid angle BC is given by $$\phi = \frac{q}{2\epsilon_{0}}(1-\cos(\theta))$$. But this is independent of the distance of the field line from the charge AB, which implies that it always makes an angle $$\theta$$ with the origin. Hence,the equation of the field line is given by $$y=\tan(\theta)x$$
Notice that your problem can be reduced to this problem since the field due to a point charge is given by $$E=Q\frac{x\hat{i} + y\hat{j}}{4\pi\epsilon_0 r^3}$$

Let me show you another problem (Taken from the book by SS Krotov).

An electric field line emerges from a positive point charge $$q_1$$ at an angle $$\alpha$$ to the straight line connecting it to a negative point charge $$—q_2$$. At what angle will the field line enter the charge $$-q_2$$?

When we look at the field lines close to $$q_1$$ (as shown in figure), the effect of $$—q_2$$ is negligible. Hence, the flux due to the field lines that make an angle $$\alpha$$ is
$$\frac{q_1}{2\epsilon_{0}}(1-\cos(\alpha))$$.
Suppost the field lines enter $$—q_2$$ at an angle $$\beta$$, the flux enclosed will be
$$\frac{q_2}{2\epsilon_{0}}(1-\cos(\beta))$$
Since the 2 flux are equal, we get $$\sin(\frac{\beta}{2}) = \sin(\frac{\alpha}{2})\frac{q_1}{q_2}$$

Here is another problem which uses the same concept.

2 point charges with equal magnitude and same sign are kept a distance 6 cm apart. Find the minimum distance between 2 field lines which leave both the charges making an angle 45 degrees with the line joining the 2 charges.

Try to solve this on your own.

Please comment if you did not understand.

• The method Farcher used is a smart way to get the job done but it is impressive that you are able to find the answer with a completely different take on it though I may not be able to appreciate it entirely as I am not sure if I understood it right. Pardon me if it is silly but I did not get how you came to the conclusion that $y=\tan\theta$ and how do you solve the problem I stated from there? Commented May 25, 2017 at 11:33
• Also I lost it here "When we look at the field lines close to q1, the effect of —q2 is negligible. Hence, the flux due to the field lines that make an angle α is" what did you mean by effect of -q2? the field lines enter -q2 right? so why would there be any effect from -q2? Commented May 25, 2017 at 11:40
• @rahulrj ,Don't worry if your question is silly or not, no one's going to loose anything from you asking it :-). The field line when it starts at an angle $\theta$ with the x axis, goes as a straight line, and since the slope is given by $\tan(\theta))$, I used the equation of line passing through origin $y=mx$. If you did not understand any specific point, please mention it and I'll try to edit my answer to make it more understandable Commented May 25, 2017 at 16:42
• @rahulrj If you try to draw the field lines, then they emerge from $q_1$ and then go on a curved path after which it enters $-q_2$. So in the large picture, the electric field around each charge is not uniformly distributed due to the effect of each other. But if you keep on magnifying onto a charge, then what will happen is that the the field around that charge will become more and more uniformly distributed around the charge. It would look as if there were no other charge in the vicinity. Commented May 25, 2017 at 16:46
• @rahulrj I have added an image and also explained why your problem can be reduced to the problem with a point charge Commented May 25, 2017 at 17:02

Suppose that at position $A$ $(x,y)$ the electric field is $\vec E = E_{\rm x} \hat i + E_{\rm y} \hat j$.

You want to move to a new position $B$ $(x+dx,y+dy)$ such that line $AB$ is along the same direction as the electric field vector at position $A$ which means that $\dfrac{dy}{dx} = \dfrac {E_{\rm y}}{E_{\rm x}}$.

In your example you do not have to do an integration.

• Is it not $E_x$ in the denominator? Commented May 25, 2017 at 11:53
• @rahulrj It is indeed and shows one of the problems which can arise when cutting and pasting. Thank you for spotting my error which I have now corrected. Commented May 25, 2017 at 11:55

Each field line ${\bf x}(\alpha)$ is a solution to the coupled ordinary differential equations $\frac{d{\bf x}}{d\alpha} = {\bf E}({\bf x})$, so you need to solve that ODE in order to find the field lines, which are also called "integral curves."

First of all, electric lines of force are not quantitative. they are just imaginary lines and have nothing to do with quantity (except density).

for finding electric field lines, find unit vectors of electric field at every point.