How to find the lines of force from electric field?

Question

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?

Answer 1

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.

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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 $E=Q\frac{x\hat{i} + y\hat{j}}{4\pi\epsilon_0 r^3}$

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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 $\sin(\frac{\beta}{2}) = \sin(\frac{\alpha}{2})\frac{q_1}{q_2}$

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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.

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Please comment if you did not understand.

Answer 2

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.

Answer 3

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."

Answer 4

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.