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According to the equation , $$E = kQ/r^2$$

If the source charge is negative electric field produced by the charge must also be negative. My teacher said electric field can never be negative, it'll either be positive or zero. Online sources pointed out that since electric field is a vector when doing calculation we only report the magnitude.

Another doubt was with electrostatic force is,why is it always positive? According to columb's law force is directly proportional to modulus of product of charges. Can't it be negative like attractive and repulsive forces.

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Try to ask yourself the question: what does it mean that anything is "negative"? The term "negative" has no physical meaning in itself before we define it to mean something.

  • How does a negative number (scalar) make physical sense? What does $-2\;\mathrm{kg}$ or $-10\;\mathrm{apples}$ mean? We can choose to understand it as the loss of an amount when it fits the context.

  • How does a negative arrow (vector) make physical sense? What does $-\vec F$ or $-\vec v$ or $-\vec E$ mean? We can choose to define it as the opposite of the vector, meaning the same vector in the opposite direction.

And so, a negative vector - or more precisely: the negative of a vector - has been defined to mean: The same vector in the opposite direction.

Now that we have a chosen definition, we can use any vector quantity with signs. Forces, velocities and also fields, including electric fields, are represented by vectors. A negative electric field just means: a field pointing/pushing opposite to what a positive field would do.

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  • $\begingroup$ About the last sentence: then what is a positive field? $\endgroup$
    – lalala
    Jun 8, 2019 at 3:42
  • $\begingroup$ @lalala That depends on the context, on the type of field. In the case of electric fields, the positive field direction is always along with the direction it pushes a positive electric charge. By definition. $\endgroup$
    – Steeven
    Jun 8, 2019 at 6:05
  • $\begingroup$ I know what a positivd direction is. But this doesnt define positive field. Also how would you distinguish this from the 'negative" field. To be explicit: the fiel Ez=1 all othe componentd 0. Is this a positive or negative field? $\endgroup$
    – lalala
    Jun 8, 2019 at 7:03
  • $\begingroup$ @lalala The point of my answer is that the notion of "negative" just means "opposite", so to say, in this case. There is no such thing as a negative field - it doesn't make physical sense. We still say it now and then, and when we say it, what we imply is that a negative field is a field where the direction at every point is opposite to that of a positive field, whatever that positive field is. Defining a positive field is a bigger task. But having that, defining a negative field is easy: it is just the same field but pointing opposite at every point. $\endgroup$
    – Steeven
    Jun 10, 2019 at 10:16
  • $\begingroup$ Good answer esp that second bullet. But I think a negative number can have real meaning. If y is how far up you are, then y<0 is below ground. If people are bowling and the machine reports ball speed across the centerline (halfway to the pins), we could bowl for decades without a negative number. Then some weirdo sneaks in the back and rolls the ball at us. Anyway just opinions $\endgroup$
    – Al Brown
    Aug 2, 2021 at 21:23
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An electric field is a vector field, which assigns a vector to each point in space. A vector itself cannot be negative or positive (unless we consider the one-dimensional case where a sign is meant to designate the direction). Arguing about the sign of the electric field vector generally makes no sense.

Unfortunately, your question is ambiguous, but I can consider three different ways of interpreting the post.

  1. In the three-dimensional case, what your equation should be written as is $$ \vec{E}(\vec{x}) = \frac{kQ}{r^{2}} \hat{r} $$ where $\hat{r}$ is the unit vector pointing from the charge $Q$ to the point in space $\vec{x}$. Clearly, this does not have a sign. Instead, it has a direction along with a magnitude (with one exception which is that if its magnitude is zero, then the direction is not well-defined).

  2. The magnitude of a vector $\vec{v} = v_{1}\hat{x} + v_{2}\hat{y} + v_{3}\hat{z}$ is $|\vec{v}| = \sqrt{v_{1}^{2} + v_{2}^{2} + v_{3}^{2}}$ in the 3D case. So in the case of the electric field, we find $$ |\vec{E}(\vec{x})| = \frac{k|Q|}{r^{2}}, $$ which is indeed always nonnegative. Therefore, if you are talking about the magnitude of the electric field vector, then it must be nonnegative.

  3. The only case where you could possibly say that the electric field is negative is if you're considering a one-dimensional scenario. In 1D, a vector only has one component and we write $\vec{v} = v_{1} \hat{x}$. Unfortunately, people tend to use a bit of abuse of notation by identifying $\vec{v}$ with the value $v_{1}$ (which is justified because this is the only value that determines $\vec{v}$ to begin with in 1D once we fix the unit basis vector $\hat{x}$). In that case, $v_{1}$ can be positive, negative, or zero. However, the interpretation of the sign is not meant to correspond to any "amount" of anything. Instead, the sign is meant to signify the direction of the vector (left or right or zero). Lastly, note that the magnitude of the vector ends up being the absolute value $|v_{1}|$ (this is becaue $|\vec{v}| = \sqrt{v_{1}^{2}} = |v_{1}|$).

Of the three cases, I doubt #3 is the one that is relevant, and equating $\vec{v}$ with $v_{1}$ is bad practice, since it could lead to confusion. There's much more to say about this, but I won't go much further on this idea in this post.

It's likely that your teacher had #2 in mind, in which case he would have been correct. However, you did not specify that you were talking about the magnitude of the electric field vector, which means the ambiguity of your question still stands.

If you are considering the electric field vector $\vec{E}(\vec{x})$, then the question of its sign makes no sense, because vector quantities in three-dimensions don't have any sign. They only have either a non-zero magnitude and a direction OR a zero magnitude and an undefined direction.

You can analogize this by visualizing an arrow that represents the vector. Clearly an arrow has a length and a direction, but the question of whether an arrow is positive or negative makes no sense. It's length, however, is a nonnegative real number.

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I think positive and negative is irrelevant once you introduce the vector formalism. Instead, contemplate the direction in which the electric field vector point at a point in space? That's a more relevant and important question.

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An electric field is a vector field, it has both magnitude and direction. In vector theory, the magnitude is the "size" of the vector and, like spatial sizes, is always positive. For example you can measure 100 mm or 100 V/m backwards but a size of -100 mm or -100 V/m has no meaning. To fully describe such a vector, we say that it has a positive magnitude but in the backwards direction (rotated by pi radians). So, when talking of the magnitude as that of a vector field and ignoring the angle, it is always positive.

But of course in reality, the sign of the electric field is arbitrary. We might equally well treat electrons as positive and protons as negative. All the vectors in our sums would then point the other way, because they are all still positive. The maths would still work fine. So thinking of the field created by an electron as negative just reflects our arbitrary choice of polarity for the units of charge.

Thinking of the electron's field as negative is thus is not wrong as such, but when we are writing down the maths we always have in mind the vector formalism - where treating it as negative is wrong.

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The electric field is a vector quantity. For the vector $ {\bf v}$ you can write this as $|{\bf v}|$ multiplied by a unit vector, say ${\bf n}$, that is $${\bf v} = |\bf{ v}| {\bf n}.$$

It is conventional to take the magnitude as a positive number, but if you take it as a negative number you can write $${\bf v} = -|\bf{ v}| (-{\bf n})$$ and the field points in the opposite direction.

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  • $\begingroup$ Is electrostatic force always positive, well it is according to columb's law. Or as it is written in the other answer it can be effectively represented by a unit vector? $\endgroup$
    – susan J
    Mar 6, 2018 at 12:55
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Electric Field can never be negative . As electric field is Force experienced by charge divided by magnitude of charge . So in magnitude , we take mod of charge. So even in case if charge is negative , then due to mod it becomes positive . Hence Electric Field also becomes positive . So , electric field is always positive .

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No, it can't. Electric field is a physical entity, it's not a number, it's not a vector or a vector space. We can choose to quantify electric field using vectors/vector fields in which case the sign change changes the direction (sense) but in general $Electric\ field \ne Vector\ field$ and the sign is not intrinsic.

Electric fields can alternatively be quantified using a scalar field like potential or even a charge distribution which generates the field.

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  • $\begingroup$ The electric field is a vector field, and it is a vector. Now one can debate the definition of a “negative vector” and thus of a negative electric field, but I fail to see how one can debate the vectorial nature of $\vec E$, which has a direction and a magnitude at every point in space. Additionally electric fields cannot be uniquely quantified using a potential since the potential is not unique. $\endgroup$ Feb 12 at 15:30
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The electric field can never be negative even if the charge is negative. The direction of the electric field is changed (reversed) when the sign of the (source) charge is made opposite. This indicates that the electric field is a vector.

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