0
$\begingroup$

In the textbook ''introduction to Electrodynamics'' by Griffiths and also in textbook '' Matthew-N.-O.-Sadiku-Elements-of-Electromagnetics''

both books defined the electric field strength that it's: force that a unit positive charge experiences when placed in an electric field which is the force per unit charge''

written mathematically $E=F/q$

but in Serway textbook he defines the electric field as

''the electric field vector at a point in space is defined as the electric force acting on a positive test charge placed at that point divided by the test charge.''

so, what is the precise definition? is the electric field the force that a 1C of charge experiences or is it the ratio between the electric force and the charge?

and if the electric field strength doesn't depend on the test charge how it doesn't depend, and we merely stated in our definition that the electric field strength is the force that act on 1C of charge? according to Griffiths definition

$\endgroup$
5
  • 10
    $\begingroup$ Can you imagine a situation in which these definitions are not equivalent? $\endgroup$ Commented Oct 28, 2022 at 15:30
  • 5
    $\begingroup$ I don't see any material difference between Serway's definition and those of the other books. Seeking a "precise" definition of anything is a futile exercise. $\endgroup$
    – Bob D
    Commented Oct 28, 2022 at 15:32
  • 2
    $\begingroup$ Note: a "test charge" is supposed to be an intuitive term, not a rigorous one; the field E is defined by what force a charge would observe at a particular location, so our "test charge" is the fictional point charge that doesn't impact anything else but can sense the electric field. It's a mental model, not something to take to heart or use in calculations. It's another way of saying "force per charge" at a particular location. $\endgroup$ Commented Oct 28, 2022 at 15:45
  • 2
    $\begingroup$ Physicist often say "per unit something something" when they mean these kinds of ratios. It's a figure of speech that is not necessary. We know what we mean, which in this case is the same as F/q. $\endgroup$ Commented Oct 28, 2022 at 15:45
  • $\begingroup$ @DuncanHarris what then differ the electric field from a usual Coulomb force if we define the electric field strength as the Force that a $1$ unit of charge experience we should not bother ourselves by writing that $E=F/Q$ we could state that it's the force $F=E$ by condition that the charge that experience this force due to a source charge $q$ is $1$ unit value of charge and we should not say that the electric field depend on the source charge only we have to take in account the $1$ unit value of charge that is experiencing the force $\endgroup$
    – Mans
    Commented Oct 28, 2022 at 16:26

5 Answers 5

1
$\begingroup$

It's rather unclear what you are asking, but perhaps it is worthwhile to point out a basic physical assumption. In the presence of an electric field, the force experienced by a test charge on an object at a point in space is a vector. That vector always points in the same direction for positive charges and the opposite direction for negative charges. The magnitude of the force is proportional to the amount of charge.

As a consequence of these physical assumptions, combined with mathematical properties of vectors, it follows that the force $F$ that is experienced by a charge $q$ is equal to $q$ times the force $E$ that is experience by a charge of unit value $1$: that is, $F=qE$, and therefore $E=F/q$.

$\endgroup$
2
  • $\begingroup$ well, if we define the electric field strength as $E$ the force that is experienced by a charge of unit value 1 why we don't say that the Electric field strength has the units of Newton and it doesn't differ from any electric force it's just states that the charge that experincing the electric force is unit value 1 $\endgroup$
    – Mans
    Commented Oct 28, 2022 at 16:06
  • $\begingroup$ So your question is really about the units of the electric field versus the units of the force field? $\endgroup$
    – Lee Mosher
    Commented Oct 28, 2022 at 17:03
1
$\begingroup$

You need to understand the idea of a limit: $$ {\bf E} = \lim_{q \rightarrow 0} \frac{\bf f}{q} $$ where $\bf f$ is the force experienced by a charge $q$. Thus we have that the field is equal to the force per unit charge, but we don't actually put one unit of charge (e.g. one Coulomb) in our experimental apparatus, because that would be a huge charge and very likely it would disturb whatever other charges caused the field we are trying to investigate (whose strength we are aiming to specify). Instead, therefore, we imagine an amount of charge tending to zero, so it does not disturb anything, and we will find the force on it also tends to zero, with their ratio tending to a constant. That constant (vector) tells us the size and direction of the electric field at the location where this test charge is placed.

$\endgroup$
1
  • 1
    $\begingroup$ I'm glad someone finally brought up the definition I have always known i.e. a limit as the test change -> 0. I thought this was fairly standard - both my E&M books use it. $\endgroup$ Commented Nov 2, 2022 at 20:03
1
$\begingroup$

"is the electric field the force that a 1C of charge experiences or is it the ratio between the electric force and the charge?"

The former gives the basic idea, but it's technically defective, for at least two reasons...

(a) 1 coulomb is far too large a charge to be the net charge on a very small 'test' body suitable for determining the electric field at a 'point'. Not only would a real test body disintegrate under mutual repulsion of its parts, but a 1 coulomb test body, if it could be made, would displace the very charges responsible for setting up the electric field that it is supposed to be investigating!

(b) The units of electric field strength are not N but NC$^{-1}$. We can then write $\vec F=q\vec E$ and obtain a force in N when we substitute a value for $q$ given in correct S.I. form as the product of a number and its unit, e.g. $3.0\times10^{-9}\ $C.

The ratio definition (on which the unit of N C$^{-1}$ is based) is preferable. It relies on the experimental finding that for a small testing charge, $q$, at a given location, the quotient $\vec F/q$ gives a vector of constant value independent of $q$. But it should be stated that the (test) charge, indeed any charge $q$ used in the equation $\vec E=\vec F/q$ or $\vec F=q\vec E$, must not be too large.

$\endgroup$
0
$\begingroup$

The phrase "per unit $x$" is used in physical definitions because it's convenient and familiar (once you've done some physics stuff for a while). Density is the mass per unit volume, the electric field is the electric force per unit charge, the gravitational field is the gravitational force per unit mass, and so on. This doesn't mean you actually have a unit of $x$, it's just because in defining a ratio, having a $1$ (unity) in the denominator makes it easier.

Example: a solid block has a mass of 13kg, and a volume of 2.5L. The density is therefore 13kg/2.5L, but you would more likely say 5.2 kg/L (i.e. 5.2 kg/1 L), or 5.2 kg per unit volume.

For the electric field, we could put everything in terms of force, but it's immediately clear that because this force depends on charge, it's easier to work with force per charge $E=F/q$ at each point in space. Then when we're interested in the electric force on a charge $q_0$, we don't calculate $F$ directly, we calculate $E$, then use $F=qE$. Makes life easier. The fundamental reason is electric force depends linearly on charge, so it's better to divide the dependence on charge out when doing calculations.


Addendum: Here's how it makes calculations easier. The Coulomb force between two charges, $q_1,q_2$, separated by a distance $R$, is defined without reference to an electric field as:

$$ F = k\frac{q_1 q_2}{R} $$

Therefore if $q_1$ is at position $(0,0,0)$, and $q_2$ is at $(x_0,y_0,z_0)$, then $R = \sqrt{x_0^2+y_0^2+z_0^2}$ and $$ F = k \frac{q_1 q_2}{\sqrt{x_0^2+y_0^2+z_0^2}} $$

Whatever $q_2$ has for position and charge, we can always say $$ E(x,y,z) = k\frac{q_1}{\sqrt{x^2+y^2+z^2}} $$ Which is called "the electric field of charge $q_1$", or maybe more exactly, the electric field contribution of $q_1$. Now we know that if $q_2$ is located at $(x_0,y_0,z_0)$, we first calculate $E$ at this location, then find the force by multiplying the field by the charge of $q_2$.

$$ \begin{align} E(x_0,y_0,z_0) = E_0\\ F = q_2 E_0 \end{align} $$ This is the same equation as above. What effect would $q_1$ have on a different charge in a different location? Simply calculate $E$ and multiply by the charge. If you have a system of 20 charges, sum up the electric field contributions of each, then you get the total electric field; calculate $E$ at some location, and you know what the force would be for a charge $q$.

$\endgroup$
3
  • $\begingroup$ so, the Serway definition is more precise because if we define the electric field strength as the force that a unit charge experince so then the electric field doesn't differ from any coloumb force and should have the units of newton and instead of writing that $E=F/q$ we could easily write that ''$E=F$ by condtion that the charge that experince the force is a unit charge of 1 Coulomb am I thinking right ? defining it as the force per unit charge we merely mean that $E=F/Q$ and I think that what Griffiths meant to say $\endgroup$
    – Mans
    Commented Oct 28, 2022 at 16:15
  • $\begingroup$ because if we say that the electric field doesn't depend on the affected charge ''the charge that experince the force'' how could we say that and we merely stated that the filed is the force that a $1$ unit of charge would experince $\endgroup$
    – Mans
    Commented Oct 28, 2022 at 16:19
  • $\begingroup$ Why do you think you need 1C of charge, or 1 unit of charge, for the definition? Instead define it as the force on any charge, divided by that charge. The "per unit charge" is not something to latch onto. $\endgroup$ Commented Oct 28, 2022 at 16:42
0
$\begingroup$

Serway's definition is more precise because he mentions "positive test charge" which then defines the direction of the field vector. But other than that both definitions are basically identical.

Your two interpretations are equivalent when you say: "is the electric field the force that a 1C of charge experiences or is it the ratio between the electric force and the charge?" What you achieve by deviding a force with a charge is exactly the force that 1 unit of charge would experience.

To your final question, "and if the electric field strength doesn't depend on the test charge how it doesn't depend", the field strength does indeed not depend on the test charge. Because when you have defined it as a force-per-charge, then from now on when a test charge is place nearby it will feel an electric force equal to the charge times the field strength - the charge may be different and thus the force will vary, but the electric field strength is constant.

$\endgroup$
4
  • $\begingroup$ well numerically it well equals to the force that acts on a unit charge but what I'm asking about is different. defining it as the force that a unit charge experiences (physically not numerically) then it doesn't differ from a usual Coulomb force it well equal as saying $F=KQ.1/r^2$ and should have the units of Newton the precise one to define it as Electric force per charge and then it will have units of $N/C$ $\endgroup$
    – Mans
    Commented Oct 28, 2022 at 17:20
  • $\begingroup$ @Mans I do understand your point now. The correct definition of electric field strength is electric force per unit charge at a point, or $$E=F/q,$$ thus with units of $\mathrm{N/C}$. This is equivalent to the force that a 1C-charge would experience, but you are correct that although equivalent these two definitions are not identical. Only the first one is correct, because otherwise we are defining field strength as being equal to a force and with force units, which is not what we want to achieve. It is often stated in that manner, because that is an intuitive approach to the definition. $\endgroup$
    – Steeven
    Commented Oct 28, 2022 at 17:46
  • $\begingroup$ so, then the ratio definition is the precise one that was written in Serway textbook which is: ''the electric field vector at a point in space is defined as the electric force acting on a positive test charge placed at that point divided by the test charge''. is that right ? $\endgroup$
    – Mans
    Commented Oct 28, 2022 at 18:08
  • $\begingroup$ @Mans Yes, I agree. But why would you say that the definition from the other two books is less precise? What is their exact wording? $\endgroup$
    – Steeven
    Commented Oct 28, 2022 at 18:14

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.