4
$\begingroup$

I have a questions about the electric field lines. enter image description here

Well in the basic learning, we know that:

  1. The electric field lines extend away from a positive charge
  2. They move forward a negative charge

Let's take parallel plates, which make a uniform electric field.If we take the basic learning, which I mentioned, in accounts, it's very easy to understand that this is the direction of a positive charge object if we put the object between the plate (move away the positive charged plate and toward negatively charged plate).

The problem is if we put a negative charged object between. Isn't the electric field reversed ?

Improve this question
$\endgroup$

3 Answers 3

Reset to default
6
$\begingroup$

The direction of the field is defined to be the direction of the force on a positively charged test particle. Positive charges always move away from other +ve charges and towards -ve charges.

As @Charlie says, it is a convention, like driving on the right (or left), or which pin on a plug is "live". So that everyone can agree on the result of a calculation, we all have to define it the same way. It could be defined the other way round, but it isn't. And we can't have both - that would be confusing.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ Oh I get it. So (I just want to be sure that I misunderstood). The conclusion is the electric field lines are describe to go away from positive charges and stop at negative charges. This statement is the standard and complete one despite being reversed. You say we can't have both because of the fact that things would go confusing. Am I correct !? $\endgroup$
    – PandoraU.U.D
    Commented Oct 24, 2016 at 3:36
  • $\begingroup$ @PandoraU.U.D that is correct. It is convention. $\endgroup$
    – dimes
    Commented Feb 7, 2021 at 22:21
2
$\begingroup$

The electrostatic field is a smart way to work with the Coulomb's law. We know that a charge $Q$ located in $P$ will produce a force $\textbf{F}_e = \frac{Qq}{4\pi\epsilon_0\|\textbf{PM}\|^3}\textbf{PM}$ on a charge $q$ located in $M$. However, this expression of $\textbf{F}_e$ is not very convenient when you work with a continuous distribution of charges, that's why we prefer to work with what we call the electrostatic field, no more than a mathematical object which facilitate the maths behind physics. We define the electrostatic field created by the charge $Q(P)$ as the field $\textbf{E}(M)$ that, when multiplied by $q$, gives the force of the charge $Q$ on a charge $q$ located in $M$, ie. $$\textbf{E}(M) = \frac{Q}{4\pi\epsilon_0\|\textbf{PM}\|^3}\textbf{PM}$$

Using this definition of the electrostatic field, we can notice that it always goes away from positive charges, and toward negative charges.

A field line is defined as a line that is always tangent to the field, and is oriented by the field. Since the electrostatic field is always directed away from positive charges and toward negative charges, field lines must go away from positive charges and toward negative ones.

However, we can't say that field lines are oriented by the motion of any charged object: indeed, as you pointed out, the electromagnetic force acting on a charge $q$ depends on the signe of $q$ since this force is $\textbf{F} = q \textbf{E}$. The electrostatic field doesn't change, it is only the electrostatic force that is reversed by replacing a positive charge by a negative one.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Oh I understand your explanation. Thank you :). But I mean how do we know the electric field always directs always from a positive charge and toward a negative charge. I understand how the theory work. But still don't know how this matter is figured out. $\endgroup$
    – PandoraU.U.D
    Commented Oct 22, 2016 at 16:02
  • 1
    $\begingroup$ I guess that the answer is "convention": you get correct results as long as you chose a convention (electric field from positive to negative charge or viceversa) and you stick to the one you chose. One could have chosen the opposite one (redefining what needs to be redefined) and still gets right results. $\endgroup$
    – Charlie
    Commented Oct 22, 2016 at 16:21
  • $\begingroup$ @Charlie Even if the orientation is caused by conventions (why does a proton have a positive charge ?), it is not directly a convention. The convention was choosen when Coulomb stated his law, and the orientation of the electrostatic field can be understand as a consequence of Coulomb's law. $\endgroup$
    – Spirine
    Commented Oct 22, 2016 at 16:28
  • 2
    $\begingroup$ @Spirine The Coulomb's law, vectorially speaking, just tells you that two charges of the same sign repulse each other and two charges with opposite sign attract each other. When you define the electric field from the electrostatic force you have to choose a convention. $\endgroup$
    – Charlie
    Commented Oct 22, 2016 at 16:37
  • $\begingroup$ Oh thank you everyone for helping me a lot. So in conclusion (just that I understand well). The sum is that: $\endgroup$
    – PandoraU.U.D
    Commented Oct 24, 2016 at 3:33
0
$\begingroup$

It's arbitrary. The physical properties of the "positive" and "negative" charge types play no role. We have chosen to use positive test charges by convention. Why? Because, since the electric field at a point is described by a vector, we can describe things in terms of negative and positive. It turns out that when using the equation $\mathbf{F} = q\mathbf{E}$, where $\mathbf{F}$ and $\mathbf{E}$ are vectors, if $q$ is allowed to be positive or negative, and we think of the test charges in the electric field as being positive, it is easier to understand and talk about the electric field model. Remember that fields are just used as a model, albeit a very helpful and relevant model.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.