Is the electric field a Force? Coloumb force that act between two charges is equal to :
\begin{gather} F=KQ_{}Q_{0}/r^2 \end{gather}
And this Coloumb force has units of Newton .
But the electric field is defined as
\begin{gather} E=F/Q_{0}=KQ/r^2 \end{gather}  with units of $(N/C)$
The interpretation of electric field is that the force that act on $1$ Coloumb of charge
But is this interpretation correct?
Like how it is the force and has units of $(N/C)$ ? It should have the units of Newton
and The electric field will be no more than like a coloumb force that has the test charge $Q_{0} = 1$, it's like saying that $E=F=KQQ_{0}/r^2$ but with condition that $Q_{0}=1$,,this will be the result if the interpretation is correct
I'am so confused about the interpretation of it and how it's defined like how it will be the force and has units of $(N/C)$ ? And also the field doesn't depend on charge, if so, how we state in our definition the charge must be 1 coloumb?
 A: The electric field generated by a charge  is not a force. A force is only experienced when another electric charge is located in the field. Then both charges experience an equal and opposite force per Newton’s third law.
Similarly the gravitational field (force per unit mass} is not a force. The gravitational field $E_g$ produced by mass $M$
$$E_{g}=\frac{GM}{r^2}$$
only produces a force $F_g$ when another mass $m$ is placed in the field where $F_g$ is
$$F_{g}=\frac{GMm}{r^2}$$
Think of the field (electric or gravitational) as the potential effect (potential for a force) on a charge or mass being placed in the field, whereas a force is the actual effect on a charge or mass  being placed in the field.
Hope this helps.
A: The electric field is a vector magnitude that at each point in space has a magnitude and a direction. The direction is given by the direction in which a positive charge would move if we place it at that point. The magnitude is given by the force that the field generates on a charge placed at that point, that is, the force per unit charge. The field exists even if we do not place any charge in it, and when we place it, it generates a force that is equal to the value of the field at that point multiplied by the placed charge.
This is the same as saying that a speed is the same as a distance, because 25 meters is the same as a speed of 25m/s if I set the time to 1 second. This does not make sense. The velocity tells me how much I move forward for each second I add to my trajectory. The field tells me how much force it makes for each charge I add.
A: If the vector field property is not clear, consider a completely different subject.  Consider pressure and force within an inflated balloon.  The compressed air has pressure greater than standard air pressure throughout the balloon, but the pressure only creates force at the inner surface of the balloon.  The pressure does not have a specific direction within the balloon.  The force created on the surface is outward, perpendicular to the balloon's inner surface.  The pressure only creates force where it has a surface to press against.  Likewise, the electric field only creates electric force only where it has an electric charge to push on or pull on.
A: The force, $F$, (unit $\rm N$), on a charge $Q$, (unit $\rm C$), placed a a point where the electric field strength is $E$, (unit $\rm N/C)$, is given by the equation
$F$ (unit $\rm N$) = $E$ (unit $\rm N/C) \times Q$ (unit $\rm C$)
Note that if the charge happens to be one coulomb then the numerical value of the electric field strength is the force, in newtons, on the charge.
A: The "electric field" is a mathematical object that is a function that takes as its input a point in space and gives as output a vector. Stated another way, $E: \mathbb R^3 \rightarrow \mathbb R ^3$ (although if we're being precise, the domain is an affine space, while the codomain is a vector space). It is the gradient of the electric potential $V: \mathbb R^3 \rightarrow \mathbb R$.
The important point here is that the electric field is treated as a property of space. You put in a point, and you get out a vector. The position is the only input to the function. Given an electric field, you know what the value at particular location is without having to refer to anything else. In particular, it is defined in such a way that we don't need to know how much charge is in a particular location to know what the electric field acting on charges in that location is. To do this, we divide the force experienced by a charge by the quantity of charge, giving the electric field in units of N/C. So, no, it's not a force.
Similarly, the electric potential field (which, as state above, the electric field is the gradient of), does not have units of J, but rather units of J/C, or volts. When we talk about the electric potential at a point, we are talking about a property of that location in space, and so it is the same for all charges, and so we divide out by charge.
You can compare this to gravity. It would make no sense to ask "What is the gravitational force at this point?" because the strength of the gravitational force is different for different masses. Instead, we have to ask what the force per mass is.
A: An electric field is not itself a force, but it exerts a force on a test charge. This force is proportional to the field strength $E$ at the location of the test charge and also proportional to the size of the test charge $Q$. So
$F = kEQ$
where $k$ is a constant of proportionality. If we measure $E$ in units such that a field strength of $1$ unit exerts a force of $1$ newton on a charge of $1$ coulomb (or a force of $2$ newtons on a charge of $2$ coulombs etc.)  then the constant of proportionality $k$ becomes $1$, which is convenient, and we have
$F = EQ$
Notice that this allows us to calculate the force on any test charge as long as we know the field strength and the size of the test charge - there is nothing special about a test charge of $1$ coulomb.
So, if we are working in SI units, $E$ is measured in newtons per coulomb, which we can show is equivalent to volts per metre.
A: Most of the other answers are missing the point that this is simply about an imprecise use of language.
As you acknowledge in the question, the electric field by definition is the force per charge, i.e. $E=F/q$. Which means by definition it has units of $N/C$. You could also say this as "the force per unit charge", but saying the electric field is "the force on a unit charge" is not correct. The force on a unit charge is the product of the force per unit charge and the unit charge itself, i.e. you'll put the same number on it, but with units of $N$ instead of $N/C$.
Example: Where we measure an electric field of $10\,N/C$ the force on a unit charge would be $F=E*q=F/q*q=10\,N/C*1\,C=10\,N$ whereas the force per unit charge will just be $10\,N/C$, the electric field.
A: The force per unit charge is the "density" of the electric force. Therefore it is not a force exerted on a particular body and it cannot be used directly in Newton's 2nd law. Likewise the per capita income is not the income of a particular individual so that the individual can be taxed!
A: The other answers are excellent technical answers. If by chance you are looking for a visual, I think of fields as terrain (rocky, muddy, a stream, etc.). The force associated with the field is what you stumble over, what slows you down, or what diverts your path as you cross the field. It is an imperfect analogy, but it does the job.
I think of Potential Energy as how challenging the terrain is. Electrical Potential is odd. V = energy/charge, so a giant can more easily ignore the effects of terrain.
Feel free to tear my answer to shreds if you have a better, practical visual - a way you could explain it to my mom.
