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How can we write the equations of a line of force between two charges, say $q$ and $q'$?

As an example, you may consider the simpler case of two opposite charges $+q$ and $-q$, and focus on the field line emerging out of $+q$ by making an angle of $\theta$ and reaching $-q$ with an angle $\phi$ with respect to the $x$-axis (see picture below).

enter image description here

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  • $\begingroup$ @sammygerbil I hope I can get a satisfactory answer to it! $\endgroup$ – Aneek Sep 2 '16 at 19:47
  • $\begingroup$ What is $+ve$ $x-axis$? And what are $\theta$ and $\phi$ measured with respect to? $\endgroup$ – Bill N Sep 2 '16 at 19:49
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    $\begingroup$ It was rather a random question sounds like you have not tried to solve it. If you have tried, please show your effort, however poor you think it is. You've said you hope to get a satisfactory answer, but how much do you really want one? What have you tried already? We respect effort here, and despise laziness. $\endgroup$ – sammy gerbil Sep 2 '16 at 19:54
  • $\begingroup$ @BillN $\theta$ and $\phi$ are measured with respect to positive direction of the x-axis. Just like you measure the slope of a line. $\endgroup$ – Aneek Sep 2 '16 at 20:03
  • $\begingroup$ @sammygerbil By random question, I mean, that it randomly popped into my head, rather than seeing it in any kind of exercise book. And yes, I've given a try, but I just can't figure out what the first step should be. There isn't any application of Gauss' Law, nor any coulombic force. $\endgroup$ – Aneek Sep 2 '16 at 20:04
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The electric field lines are defined as being tangent in every point to the electric field in that point.

Therefore, calling $\boldsymbol r(s)$ the "trajectory" of a field line, with $s$ a parameter telling us at which point of the line we are, $\boldsymbol r(s)$ simply follows the equation

$$ \frac{d \boldsymbol r(s)}{d s} = \boldsymbol E(\boldsymbol r(s)). \tag 1$$

In your example case the electric field is given by

$$ \boldsymbol E(\boldsymbol r) = \frac{q}{4\pi \epsilon_0} \left[ \left( \frac{x}{\left[ x^2 + y^2 \right]^{3/2}} + \frac{R-x}{\left[ (x-R)^2 + y^2 \right]^{3/2}} \right) \hat{\boldsymbol x} \\ + \left( \frac{y}{\left[ x^2 + y^2 \right]^{3/2}} - \frac{y}{\left[ (x-R)^2 + y^2 \right]^{3/2}} \right) \hat{\boldsymbol y} \right], $$ making the solution of the system of differential equations (1) quite non trivial even in this simple case. I for one am not sure if this can be solved analytically (I had Mathematica have a try but to no avail).

If are interested in numerically verifying that this equation is true and see how the actual curve looks like, and know how to use Wolfram Mathematica, you can try the following code:

Manipulate[
 With[{
   sol = NDSolve[
     {
      x'[s] ==
       A (x[s]/(x[s]^2 + y[s]^2)^(3/2) + (
            R - x[s])/((x[s] - R)^2 + y[s]^2)^(3/2)),
      y'[s] ==
       A (y[s]/(x[s]^2 + y[s]^2)^(3/2) -
          y[s]/((x[s] - R)^2 + y[s]^2)^(3/2)),
      x[0] == 0.01,
      y[0] == 0.01 Tan[\[Theta]],
      WhenEvent[
       Abs[x'[s]] > 10^6, "StopIntegration"
       ]
      },
     {x, y}, {s, 0, 20}
     ]
   },
  ParametricPlot[
   {x[s], y[s]} /. sol,
   {s, 0, sol[[1, 1, 2, 1, 1, 2]]},
   PlotRange -> {{0, 2}, {-1, 1}}
   ]
  ],
 {{A, 0.1}, 0.001, 1, 0.01, Appearance -> "Labeled"},
 {{R, 2}, 0.001, 4, 0.001, Appearance -> "Labeled"},
 {{\[Theta], Pi/4}, -Pi/2, Pi/2, 0.001, Appearance -> "Labeled"}
 ]

enter image description here

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  • $\begingroup$ Thanks for your kind effort glS (I guess that's not your name though!). So it seems that there is no fixed formula or equation of them. I mean, it will take days to solve that equation. And I am not quite familiar with Wolfram Mathematica. Anyways, thanks for your contribution! $\endgroup$ – Aneek Sep 5 '16 at 13:53
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We can find the equation for the line which makes theta angle with positive charge through these steps. I have tried to make each step clear.enter image description here

enter image description here

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I'll give you the start. You need to finish the work.

Start with an understanding of how the lines are constructed. The lines are paths through a vector field that are resultant electric force from the particles (number does not matter) in the system. You then connect between the charges by starting a small distance from one of the particles and move a small amount in the direction of the electric field and then take successive steps. Eventually you will arrive at either your calculation boundary or another particle. The equations can be derived in this mathematical method but there are going to be some errors that will vary with the size of the step you are taking.

But if you have some more math skills you can calculate the vector field and calculate the differential geometric space associated with it. Then the equations can be directly calculated using the geodesic equation through that curved space. That method would yield a more accurate result but it would be fun to calculate.

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  • $\begingroup$ Thanks for your kind effort, but I'm not yet familiar with geodesic equations. Seems I have to learn that first! :( $\endgroup$ – Aneek Sep 2 '16 at 20:23
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Let a charge $+q$ be at the point $(a, 0)$ and a charge $-q$ be at the point $(-a, 0)$. Then the electric field at a point $(x, y)$ is \begin{equation}\tag{e1}\label{e1} \vec{E} = q\vec{r}\left(\frac{1}{r_1^3} - \frac{1}{r_2^3}\right) - qa\hat{e}_x\left(\frac{1}{r_1^3} + \frac{1}{r_2^3}\right), \end{equation} where $\vec{r} = x\hat{e}_x + y\hat{e}_y$, $\vec{r}_1 = \vec{r} + a\hat{e}_x$ and $\vec{r}_2 = \vec{r} - a\hat{e}_x$. The equation of lines of force is \begin{equation}\tag{e2}\label{e2} \frac{\partial y}{\partial x} = \frac{y}{x + a\frac{r_2^3 + r_1^3}{r_2^3 - r_1^3}}. \end{equation} We will now solve equation \eqref{e2}. We first rearrange it as \begin{equation} \left((r_2^3 - r_1^3)x + a(r_2^3 + r_1^3)\right)\frac{\partial y}{\partial x} = (r_2^3 - r_1^3)y. \end{equation} Multiplying both sides by $y/(r_1^3 r_2^3)$, \begin{equation} \left(\frac{x + a}{r_1^3} - \frac{x - a}{r_2^3}\right)y\frac{\partial y}{\partial x} = \frac{y^2}{r_1^3} - \frac{y^2}{r_2^3}. \end{equation} We substitute $y^2 = r_1^2 - (x + a)^2$ in the first factor on the right hand side and $y^2 = r_2^2 - (x - a)^2$ in the second factor to get \begin{equation} \left(\frac{x + a}{r_1^3} - \frac{x - a}{r_2^3}\right)y\frac{\partial y}{\partial x} = \frac{1}{r_1} - \frac{(x+a)^2}{r_1^3} - \frac{1}{r_2} + \frac{(x-a)^2}{r_2^3} \end{equation} or, \begin{equation}\tag{e3}\label{e3} \frac{1}{r_1} - \frac{1}{r_2} - \frac{(x+a)}{r_1^3}\left((x + a) + y\frac{\partial y}{\partial x}\right) + \frac{(x-a)}{r_2^3}\left((x - a) + y\frac{\partial y}{\partial x}\right) = 0. \end{equation} We use the derivatives of $r_1$ and $r_2$ with respect to $x$ \begin{eqnarray*} \frac{\partial r_1}{\partial x} &=& \frac{x + a}{r_1} + \frac{y}{r_1}\frac{\partial y}{\partial x} \\ \frac{\partial r_2}{\partial x} &=& \frac{x - a}{r_2} + \frac{y}{r_2}\frac{\partial y}{\partial x} \end{eqnarray*} in equation \eqref{e3} to get \begin{equation} \frac{1}{r_1} - \frac{x+a}{r_1^2}\frac{\partial r_1}{\partial x} - \frac{1}{r_2} + \frac{x-a}{r_2}\frac{\partial r_2}{\partial x} = 0, \end{equation} or \begin{equation} \frac{d}{dx}\left(\frac{x + a}{r_1}\right) - \frac{d}{dx}\left(\frac{x + a}{r_1}\right) = 0, \end{equation} from which we readily get \begin{equation}\tag{e4}\label{e4} \frac{x + a}{r_1} - \frac{x - a}{r_2} = C, \end{equation} where $C$ is a constant, as the solution of the differential equation (e2). This is also the solution given in article 63 of 'The Mathematical Theory of Electricity and Magnetism' by Sir James Jeans (5th edition).

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