Consider for example a positive charge placed between a positive charge and a negative charge. Initially the test charge is placed closer to positive charge.

Why is path followed by the test charge not same as the electric field lines?

  • $\begingroup$ Hello and welcome to PSE. I have edited your question to pinpoint (what I consider to be) the main issue, and have also answered the same. In case you think this drastic edit deviates from your original question, please feel free to use the edit link below your post and roll back my edit. Cheers :) $\endgroup$
    – 299792458
    Commented May 28, 2017 at 2:41

1 Answer 1


Why is path followed by the test charge not same as the electric field lines?

If the test charge corresponds to a massive particle, the path actually traced out differs from the actual direction of the electric field line, due to inertia.

Assume for illustration that the electric field ${\vec E}$ curves in space, i.e. deviates from a straight line. As this test charge particle moves an infinitesimal distance along the direction of ${\vec E}$, inertia of motion dictates that it would tend to move along that direction itself, even as the ${\vec E}$ for the next infinitesimal element changes direction. Thus, for the second infinitesimal path length, the problem becomes a case of initial velocity along one direction, and the instantaneous acceleration ${\vec a} = q{\vec E}/m$ pointing in a slightly different direction. It is easily imaginable that the trajectory for this infinitesimal path length is not exactly along the direction of ${\vec E}$ itself. The same argument can be readily extended to subsequent infinitesimal length elements.

Thus, any massive particle, left all by itself in an electric field ${\vec E}$ that curves in space, will never trace out the direction of electric field ${\vec E}$. (Please note that there is no such fallacy if electric field points in a straight line, and we launch the test particle along this very straight line!)

Thus, while we may physically like to perceive electric field line as the path taken${}^{1}$ by a test charge particle when launched in the field, in general, there are only two cases in which this procedure is valid (in general, even for curved electric fields):

  • The mass of the test charge particle tends to $0$. : This way, there would be no inertia, and hence, we avoid the logical fallacy.

  • Instead of leaving the test charge particle free to move all by itself, there is external involvement. : For example, an "experimenter" would leave the test charge free to move ONLY for an infinitesimal length element, and trace out the path that it takes. Thereafter, the experimenter "holds" the test particle in place, bringing it to rest, and hence, getting rid of the initial velocity for its motion through the next infinitesimal length element. Then again, leaving it free, the path gets traced out. The process can be indefinitely extended to trace out the full path, and hence, the complete direction of electric field (line).

Either way, the bottom-line is, the determination of the electric field line is not that straightforward. Even as a thought experiment, it is smeared with logical pitfalls, and mathematical abstractions!

$^1$ Of course, as the answer establishes, this perception is not strictly true unless one reads the fine print and incorporates any of the additional conditions discussed in the answer. So we are better off visualizing an electric field line as "a line whose tangent at every point points along the direction of the force on a (test) charge", as this would be an accurate description − via this statement, we don't claim that the test charge actually moved.

Edit courtesy : Comment by Farcher, "Footnote" courtesy : John Rennie, in chat.

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    $\begingroup$ "while we define electric field line as the path taken by a test charge particle when launched in the field" Having shown this is not so unless a constraint is added might it be better to use "a line whose tangent is along the direction of the force on a charge" After all this is how one would trace out a line of force. $\endgroup$
    – Farcher
    Commented May 28, 2017 at 6:38
  • $\begingroup$ @Farcher - Yes, you are right. I had used the "definition" which was more suited for my purpose from the point of view of the answer :P But yes, your suggestion is very valid, so I'll find a way to logically incorporate that into the answer too (soon). Thanks :) $\endgroup$
    – 299792458
    Commented May 29, 2017 at 4:25

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