Potential Gradient for individual charges and parallel plates?

In summary: No, I mean, when you're considering the potential gradient of a point charge, you assume that we place a test charge anywhere in the vicinity of a point charge and observe its behavior. What we observe is that once the test charge is allowed to freely move, it tends to repel away from the point charge, thereby moving towards an area of lower potential and thus having a lower electric potential energy. This loss in electric potential energy is then represented by the negative sign which appears in the formula of the potential gradient, i.e E = (-)V/r. Now, we know what that negative sign represents when we're just considering point charges.But what about the potential gradients of parallel
  • #1
ZEROBRAINCAPACITANCE
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In my book, the potential gradient for a charge placed anywhere in space is defined as: E = -V/r

HOWEVER, for parallel plate (capacitors) the potential gradient is defined as E = V/d (V being the potential difference). How come there's no negative sign for the potential gradient of the parallel plates?

My book derives it as follows: E = -V/r, but V=V2-V1 (V2 being the lower potential), so E=-(V2-V1)/r and -(V2-V1) = V1-V2, which is the potential difference and r = d, so E = V/d.

HOW? How did they even equate V = V2-V1, when V is just electric potential, fixed for a specific point in space.
 
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  • #2
Which book is that? If it is true what you wrote, I'd look for another better book as quickly as possible ;-)).

The Coulomb potential of a point charge ##q## reads (in SI units)
$$\phi(\vec{x})=\frac{q}{4 \pi \epsilon |\vec{x}|} \; \Rightarrow \; \vec{E}=-\vec{\nabla} \phi(\vec{x}) = \frac{q}{4 \pi \epsilon_0 |\vec{x}|^3} \vec{x}.$$

For a parallel-plate capacitor you don't have point charges but (neglecting edge effects) homogeneous surface charge densities of opposite sign on the two plates. The solution of the Maxwell equations (not too close to the edges of the plates) is a constant electric field ##\vec{E}=\text{const}## (pointing from the positively charged plate to the negatively charged one and is perpendicular to these plates). The potential is ##\phi(\vec{x})=-\frac{q}{A \epsilon_0} \vec{n} \cdot \vec{x}##, where ##\vec{n}## is the surface-normal vector pointing from the positive to the negative plate, i.e., ##\vec{E}=-\vec{\nabla} \phi=\frac{q}{A \epsilon_0} \vec{n}##. Here ##q## is the charge on the positive plate (##-q## the one on the negative plate) and ##A## is the area of the plates.
 
  • #3
vanhees71 said:
Which book is that? If it is true what you wrote, I'd look for another better book as quickly as possible ;-)).

The Coulomb potential of a point charge ##q## reads (in SI units)
$$\phi(\vec{x})=\frac{q}{4 \pi \epsilon |\vec{x}|} \; \Rightarrow \; \vec{E}=-\vec{\nabla} \phi(\vec{x}) = \frac{q}{4 \pi \epsilon_0 |\vec{x}|^3} \vec{x}.$$

For a parallel-plate capacitor you don't have point charges but (neglecting edge effects) homogeneous surface charge densities of opposite sign on the two plates. The solution of the Maxwell equations (not too close to the edges of the plates) is a constant electric field ##\vec{E}=\text{const}## (pointing from the positively charged plate to the negatively charged one and is perpendicular to these plates). The potential is ##\phi(\vec{x})=-\frac{q}{A \epsilon_0} \vec{n} \cdot \vec{x}##, where ##\vec{n}## is the surface-normal vector pointing from the positive to the negative plate, i.e., ##\vec{E}=-\vec{\nabla} \phi=\frac{q}{A \epsilon_0} \vec{n}##. Here ##q## is the charge on the positive plate (##-q## the one on the negative plate) and ##A## is the area of the plates.
But, we know that in case of a point charge, the negative sign indicates a loss in potential energy (assuming you traverse in the direction of the field) or that the direction in which the potential increases is opposite to the direction of the electric field. So, what would the negative sign in the parallel plate potential gradient represent?
 
  • #4
I don't understand the question. An electric field is always directed away from positive and pointing towards negative charges and the sign of the potential has to be correspondingly. See the equations for the Coulomb field of a point charge given in my previous posting.

The potential is the potential of the electric field not potential energy. The potential energy of a test charge ##q'## is given by ##V(\vec{x})=q' \phi(\vec{x})## and it's sign is such that like-sign charges are repelled from and unlike-sign charges attracted to each other, as it should be.
 
  • #5
vanhees71 said:
I don't understand the question. An electric field is always directed away from positive and pointing towards negative charges and the sign of the potential has to be correspondingly. See the equations for the Coulomb field of a point charge given in my previous posting.

The potential is the potential of the electric field not potential energy. The potential energy of a test charge ##q'## is given by ##V(\vec{x})=q' \phi(\vec{x})## and it's sign is such that like-sign charges are repelled from and unlike-sign charges attracted to each other, as it should be.
No, I mean, when you're considering the potential gradient of a point charge, you assume that we place a test charge anywhere in the vicinity of a point charge and observe its behavior. What we observe is that once the test charge is allowed to freely move, it tends to repel away from the point charge, thereby moving towards an area of lower potential and thus having a lower electric potential energy. This loss in electric potential energy is then represented by the negative sign which appears in the formula of the potential gradient, i.e E = (-)V/r. Now, we know what that negative sign represents when we're just considering point charges.

But what about the potential gradients of parallel plates? What does the negative sign, in the formula E = (-)V/d represent?
 
  • #6
I think I'm really confusing stuff up.

What I want to ask is, why is there no - sign in the formula, E = V/d when you're considering parallel plates, but there is a - sign when you're considering E = -V/r when you consider a point charge.

For reference, you can visit this site to get a clue of what I'm talking about: https://courses.lumenlearning.com/p...ectric-potential-in-a-uniform-electric-field/

They equate -V to Va-Vb and then derive the formula E = V/d with no - sign. WHY do they do that? Why not just consider -V and write E = -V/d? IF THIS IS POSSIBLE, then what would this - sign represent? The same thing it represents in case of point charges?
 
  • #7
It is very important to understand the concept of potentials. A potential of a vector field is a scalar function whose gradient gives the vector field,
$$\vec{E}=-\vec{\nabla} \phi.$$
Here ##\phi## is the potential of the electrostatic field (non-static fields are more complicated and cannot be represented by a scalar potential).

If you have a constant electrostatic field ##\vec{E}=\text{const}##, of course its potential is ##\phi=-\vec{x} \cdot \vec{E}##, because then ##-\vec{\nabla} \phi=\vec{E}##.

If you have a test charge ##q## in an electrostatic field, the potential of the force is
$$V(\vec{x})=q \phi(\vec{x}).$$
Indeed you rightly get
$$-\vec{\nabla} V=-q \vec{\nabla} \phi=q \vec{E},$$
i.e., the correct force on the point charge in the electrostatic field.

Further if you have given an electrostatic field, it must have a potential (at least in simply-connected regions of space), because one of the static Maxwell equations reads ##\vec{\nabla} \times \vec{E}=0##. Then you can calculate the potential within a simply connected region of space time by calculating the line integral connecting a fixed point ##\vec{x}_0## and the variable point ##\vec{x}##. Since ##\vec{\nabla} \times \vec{E}=0## in this region the value of the potential is independent of the path, i.e., it only depends on the boundary points of the integration path:
$$\phi(\vec{x})=-\int_{C(\vec{x}_0,\vec{x}} \mathrm{d} \vec{x}' \cdot \vec{E}(\vec{x}').$$
Of course you get the potential energy of a test charge in this field by
$$V(\vec{x})=-\int_{C(\vec{x}_0,\vec{x}} \mathrm{d} \vec{x}' \cdot \vec{F}(\vec{x}')=q \phi(\vec{x}).$$
 
  • #8
Alright, I know I'm being a bit of a bother here, but this is what's written in my book.
WhatsApp Image 2021-02-20 at 4.50.26 PM.jpeg


Does this make any sense, what-so-ever, at all, to you?

ALSO, I'm not really comprehending what this relation even means. If you consider a potential gradient of a point charge, you'd say it is the rate of change of electric potential w.r.t the distance a test charge covers. Perfectly understandable as the distance covered would either be from infinity to a point within an electric field or vice versa. It represents the electric field between the point in infinity and the point in the electric field.

BUT, when it comes to the potential gradient due to two parallel plates, what does E = -V/d represent? What does it EXPLAIN? I'm not really at that stage where I can understand calculus in physics, so pls do pretend that I'm an 8 year old xD
 
  • #9
ZEROBRAINCAPACITANCE said:
BUT, when it comes to the potential gradient due to two parallel plates, what does E = -V/d represent? What does it EXPLAIN?
If it is easier to intuitively understand, you might compare it to gravitational potential on a hill.

You have a man who masses 100 kilograms on the top of a 20 meter hill. His potential energy is 100 kg x 20 meters x 9.8 m/s2 = 19,600 joules.

The "E field" here would be 9.8 joules per kilogram per vertical meter.
 
  • #10
jbriggs444 said:
If it is easier to intuitively understand, you might compare it to gravitational potential on a hill.

You have a man who masses 100 kilograms on the top of a 20 meter hill. His potential energy is 100 kg x 20 meters x 9.8 m/s2 = 19,600 Joules.

The "E field" here would be 9.8 Joules per kilogram per vertical meter.
So, basically, it's the electric field between the two plates or for any other potential difference, the electric fields between those two points having said potential difference. Right?
 
  • #11
I think that you have it, yes.

The way I like to think of it as a "force field" and a "potential field". The potential field is the [path] integral of the force field. The force field is the gradient of the potential field.

Edit: We stick in a minus sign so that the "potential" gets more positive to reflect increasing potential energy stored against the force rather than increased energy released by the force. So the potential is actually the additive inverse of the integral and the force is the additive inverse of the gradient. Thanks to @kuruman for the gentle heads up on that point.

If you have not encountered vector calculus yet, a "gradient" is like a derivative. But vector-valued.
 
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  • #12
Alright, now since that's out of the way, you wouldn't seem to know if or if not we're supposed to add the - sign, would you? This has been bothering me for hours.
 
  • #13
I see no hint in the problem statement about a direction for the coordinate axis. A positive field strength would point in the direction of the force on a positive test charge. Which would be the direction of increasing kinetic energy and decreasing electrical potential energy.

Line your coordinate axis up so that the force on a positive test charge points in the positive direction and you have a positive-pointing field strength.
 
  • #14
But that would mean there's a possibility of a negative field strength as well for when we move a charge against the field?
 
  • #16
ZEROBRAINCAPACITANCE said:
But that would mean there's a possibility of a negative field strength as well for when we move a charge against the field?
Sure. But it is a vector field. It would be more proper to say that it has a direction rather than a sign.

The potential field is a scalar field. Its value at any point is a scalar. It can have a sign.

The gradient of a scalar field is a vector field. Its value at any point is a vector. It has a direction.
 
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  • #17
jbriggs444 said:
I see no hint in the problem statement about a direction for the coordinate axis. A positive field strength would point in the direction of the force on a positive test charge. Which would be the direction of increasing kinetic energy and decreasing electrical potential energy.

Line your coordinate axis up so that the force on a positive test charge points in the positive direction and you have a positive-pointing field strength.
The main problem with such apparently simplified treatments (it's in fact oversimplified) is that you get easily confused. It's an example, where misunderstood didactics is in the way of understanding. One should start with clear concepts and treat scalars as scalars and vectors as vectors. Then together with a bit intuition in which direction the vectors (here the electric field and forces on a point charge) should point all the trouble with signs is gone.

Take the example with the constant electric field. You have
$$\vec{E}=E_1 \vec{e}_1+ E_2 \vec{e}_2 + E_3 \vec{e}_3,$$
where ##\vec{e}_1##, ##\vec{e}_2##, and ##\vec{e}_3## are three vectors of length 1 which are perpendicular to each other and form in this order a right-handed Cartesian coordinate system (if you point with the thumb of your right hand in direction of ##\vec{e}_1## with the index finger in direction of ##\vec{e}_2##, then your middle finger points in direction of ##\vec{e}_3##).

With the given basis you can now also just use the components of the electric field to uniquely describe it. Usually it's written as a column vector:
$$\vec{E}=\begin{pmatrix} E_1 \\ E_2 \\ E_3 \end{pmatrix}.$$
Now, if you have a scalar field ##\phi(\vec{x})## by definition the components of its gradient wrt. to this basis is
$$\vec{\nabla} \phi = \begin{pmatrix} \partial_1 \phi \\ \partial_2 \phi \\ \partial_3 \phi \end{pmatrix},$$
where
$$\partial_1 \phi=\frac{\partial \phi}{\partial x_1},\; \ldots.$$
Now for our constant field we can easily check that it has a potential, i.e., we look for a scalar field ##\phi## such that ##\vec{E}=-\vec{\nabla} \phi##. The minus sign here is a convention, but it's a useful one such that for forces in the corresponding energy-conservation law, the potential energy is added to the kinetic energy. That's why usually one writes this minus sign in the relation between a vector field and its potential (if it exists).

To find ##\phi## we start with the first component. From ##\vec{E}=-\vec{\nabla} \phi## we get
$$\partial_1 \phi=-E_1.$$
Since ##E_1=\text{const}## (by assumption, this integrates to
$$\phi(x_1,x_2,x_3)=-x_1 E_1 + \phi_2(x_2,x_3),$$
because in integrating with respect to ##x_1## you can always add some function which is constant wrt. ##x_1##. For the 2nd component we find
$$\partial_2 \phi=\partial_2 \phi_2=-E_2 \; \Rightarrow \; \phi_2 = -x_2 E_2 + \phi_3(x_3).$$
And once more for the 3rd component you finally get
$$\phi_3=-x_3 E_3+\phi_0,$$
where ##\phi_0## is an arbitrary constant. Plugging everything together we get
$$\phi=-x_1 E_1 - x_2 E_2 -x_3 E_3+\phi_0=-\vec{x} \cdot \vec{E} + \phi_0.$$
The constant ##\phi_0## is arbitrary. Potentials are usually only defined up to such an arbitrary additive constant. What's really physical in the description here is ##\vec{E}## and because ##\vec{E}=-\vec{\nabla} \phi## the constant drops out again.
 
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  • #18
Here are my own two bits of wisdom. The textbook description is poorly written and it is small wonder why @ZEROBRAINCAPACITANCE is confused. One reads
##"##Let the charges on the plates are (sic) ##+Q## and ##-Q## when the potential difference is ##V##. If the positive plate is at potential ##V_1## and the negative plate is at potential ##V_2##, then the electric field strength between plates is $$E=\frac{-\Delta V}{\Delta r}=\frac{-(V_2-V_1)}{d}=\frac{(V_1-V_2)}{d} = \frac{V}{d}"$$What I consider confusing:
1. The opening sentence calls the potential difference ##V## when the symbol is also used to denote electrostatic potential in the equation below (##V_1,~V_2##). Furthermore, the starting potential difference ##\Delta V## is transformed to what is apparently electric potential ##V## or is that the difference as originally stated?

2. A difference, denoted by the symbol ##\Delta##, is defined as "What comes later minus what comes earlier". When time is the independent variable, that is easy to figure out. With distance ##r## one has to define first which way ##r## is increasing and then take the difference accordingly. Here, nothing is said about the direction of increasing ##r##. ##\Delta r## is replaced with a positive quantity ##d##, but for taking the difference in the numerator how does that explain the choice ##(V_1-V_2)## in the numerator? Why not the other way around?

How to fix it:
Far be it for me to rewrite this textbook but here it is
"In this particular case, we are looking for the strength (magnitude) of the electric field, a positive quantity. Thus, we write $$E=\left |\frac{\Delta V}{\Delta r}\right|=\frac{V}{d}.$$Note: Symbol ##V## has two uses: (a) to denote electrostatic potential in which case it is a function of position, e.g. ##V(r)=\frac{kq}{r}##; (b) to denote the positive difference of electrostatic potential between two fixed points in space, in which case it is also called voltage", e.g. ##V=IR##. In the equation above, ##V## is the voltage.
 
  • #19
kuruman said:
Here are my own two bits of wisdom. The textbook description is poorly written and it is small wonder why @ZEROBRAINCAPACITANCE is confused. One reads
##"##Let the charges on the plates are (sic) ##+Q## and ##-Q## when the potential difference is ##V##. If the positive plate is at potential ##V_1## and the negative plate is at potential ##V_2##, then the electric field strength between plates is $$E=\frac{-\Delta V}{\Delta r}=\frac{-(V_2-V_1)}{d}=\frac{(V_1-V_2)}{d} = \frac{V}{d}"$$What I consider confusing:
1. The opening sentence calls the potential difference ##V## when the symbol is also used to denote electrostatic potential in the equation below (##V_1,~V_2##). Furthermore, the starting potential difference ##\Delta V## is transformed to what is apparently electric potential ##V## or is that the difference as originally stated?

2. A difference, denoted by the symbol ##\Delta##, is defined as "What comes later minus what comes earlier". When time is the independent variable, that is easy to figure out. With distance ##r## one has to define first which way ##r## is increasing and then take the difference accordingly. Here, nothing is said about the direction of increasing ##r##. ##\Delta r## is replaced with a positive quantity ##d##, but for taking the difference in the numerator how does that explain the choice ##(V_1-V_2)## in the numerator? Why not the other way around?

How to fix it:
Far be it for me to rewrite this textbook but here it is
"In this particular case, we are looking for the strength (magnitude) of the electric field, a positive quantity. Thus, we write $$E=\left |\frac{\Delta V}{\Delta r}\right|=\frac{V}{d}.$$Note: Symbol ##V## has two uses: (a) to denote electrostatic potential in which case it is a function of position, e.g. ##V(r)=\frac{kq}{r}##; (b) to denote the positive difference of electrostatic potential between two fixed points in space, in which case it is also called voltage", e.g. ##V=IR##. In the equation above, ##V## is the voltage.
Can one say that the potential gradient can also be negative, provided that we try to move the charge from a position of lower potential to a position of higher potential using an external force? Also, what's the reason behind the negative sign vanishing?
 
  • #20
Yes, and you don't need an external force. The gradient of the potential at a particular point in space is a vector that points in the opposite direction as the local electric field. See e.g. post #16 by @jbriggs444.
 
  • #21
kuruman said:
Yes, and you don't need an external force. The gradient of the potential at a particular point in space is a vector that points in the opposite direction as the local electric field. See e.g. post #16 by @jbriggs444.
Makes sense, yeah, thanks!
 
  • #22
The problem seems to be that this book is of this kind of "calculus free physics" which is a contradictio in adjecto. There is no physics without calculus. It's as if you want to write plays like Shakespeare and forbidding to use English as a language. Through away this book and take a good introductory physics book using calculus. If you want to understand physics you have to learn quite some deal of mathematics to be able to communicate about it adequately (multi-variable calculus and linear algebra is sufficient for quite a lot of physics).
 
  • #23
ZEROBRAINCAPACITANCE said:
In my book, the potential gradient for a charge placed anywhere in space is defined as: E = -V/r

ZEROBRAINCAPACITANCE said:
BUT, when it comes to the potential gradient due to two parallel plates, what does E = -V/d represent?
Untitled.jpg

Kind of confused me, according to the picture you posted, it says E=V/d, I can’t see the minus sign.

Anyway, back to the topic, I think this equation is just trying to show that if you want to calculate the electric field in the direction pointing from V1 to V2, then you should use V2-V1 as the ΔV, and this makes it look consistent with the famous equation ## \vec E = - \vec \nabla \phi ## .
 
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  • #24
Is there a figure with the plates? We cannot answer that question without knowing the geometry the book assumes. Again, look for a better book. Electric fields are vectors!
 
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  • #25
vanhees71 said:
Is there a figure with the plates? We cannot answer that question without knowing the geometry the book assumes. Again, look for a better book. Electric fields are vectors!
WhatsApp Image 2021-02-21 at 7.25.23 PM.jpeg

It's not of much help, but here it is.
 
  • #26
alan123hk said:
View attachment 278383
Kind of confused me, according to the picture you posted, it says E=V/d, I can’t see the minus sign.

Anyway, back to the topic, I think this equation is just trying to show that if you want to calculate the electric field in the direction pointing from V1 to V2, then you should use V2-V1 as the ΔV, and this makes it look consistent with the famous equation ## \vec E = - \vec \nabla \phi ## .
That's the exact thing I was confused about! WHY isn't there a negative sign? Why derive E = V/d when you could've just stuck with -V/d? ALSO, how did ΔV change into V2-V1 when it's actually the ELECTRIC POTENTIAL, AND NOT the electric potential difference!
 
  • #27
ZEROBRAINCAPACITANCE said:
That's the exact thing I was confused about! WHY isn't there a negative sign? Why derive E = V/d when you could've just stuck with -V/d? ALSO, how did ΔV change into V2-V1 when it's actually the ELECTRIC POTENTIAL, AND NOT the electric potential difference!
I thought I explained that to you in post #18. There is no negative sign because you are looking for the strength or magnitude of the electric field which is always positive. Please reread the post about V and ΔV and let me know if there is something you don't understand.
 
  • #28
kuruman said:
I thought I explained that to you in post #18. There is no negative sign because you are looking for the strength or magnitude of the electric field which is always positive. Please reread the post about V and ΔV and let me know if there is something you don't understand.
Well, when you consider the equation, E = -V/r, that too represents the strength of the electric field, but the - sign just indicates that the potential gradient increases in a direction opposite to the electric field. I guess since the electric field remains constant between two parallel plates, your reason for it remaining positive seems sensible.

Also, the thing is, when you're using the formula E = -V/r to derive E = V/d, the important thing to realize is that the V in -V/r, is actually just the electric potential, a number assigned to any point in space relative to infinity. The V in V/d is the potential difference, between two points within the electric field. How does one automatically equate V in -V/r to V2-V1 is something I don't understand. It's the quantity each symbol represents in this specific case, which confuses me.
 
  • #29
Take the figure you posted. What's lacking is a clear notation. So the first thing is to draw a Cartesian coordinate system. Let's put the origin on the plate connected to the +pole and make the ##z##-direction perpendicular to this plate pointing to the other plate. Now it's clearly set up.

To calculate the field we neglect fringe effects, i.e., we consider the plates as very large compared to the distance, i.e., we look for the field not too close to the edges. Then by symmetry the potential must be a function of ##z## only: ##\phi(\vec{x})=\phi(z)##. Now within the plates you have ##\vec{\nabla} \cdot \vec{E}=-\Delta \phi=0##. This implies that ##\phi(z)=A z + B## with integration constants ##A## and ##B##. We can arbitrarily choose ##B=0## since the potential is anyway only determined up to a constant. Further you know that there's a potential difference ##\phi(0)-\phi(d)=V##, where ##d## is the distance between the plates, because ##V## is the voltage of the battery and the plus pole is at the higher voltage (more precisely at the higher electromotive force, but that's a detail only relevant if you want to analyze how a battery works electrochemically). This determines ##A##: ##\phi(0)-\phi(d)=-A d=V## So you have ##A=-V/d## and thus ##\phi(z)=-V z/d##. The electric field is ##\vec{E}=-\vec{\nabla} \phi=\vec{e}_z V/d##. So specifying clearly a coordinate system and just doing the calculations makes the determination of the correct sign very easy, and it should also be intuitively clear that the sign is correct, because the electric field points from the positively charged plate to the negatively charged plate, and that's what we indeed got from our analysis of the potential.
 
  • #30
I personally use the simplest method below to understand the equation of the capacitor electric field vector.

Potential Difference 1.jpg
 
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  • #31
ZEROBRAINCAPACITANCE said:
Well, when you consider the equation, E = -V/r, that too represents the strength of the electric field, but the - sign just indicates that the potential gradient increases in a direction opposite to the electric field.
The equation that you want me to consider is the source of your confusion. The definition of the electric field, as other have already said, is (in one dimension ##r##) ##E=-\dfrac{dV}{dr}## not what you have. When you say E=-V/r, you equate the ratio of small differences to the ratio of the function ##V## to independent variable ##r## and then you wonder where the difference ##\Delta V## came from. When the field is uniform as between parallel capacitor plates, the ratio is constant whichmeans that the ratio of small differences is the same as the ratio of large differences in which case you can write ##E=-\dfrac{dV}{dr}=-\dfrac{\Delta V}{\Delta r}.##

Now here comes the tricky part. The denominator is a displacement and has magnitude and direction. The magnitude is the plate separation ##d##. Let's pick a direction from positive plate 1 to negative plate 2. Then the potential difference in the numerator is ##\Delta V=V_2-V_1## and this is a negative number because the negative plate is at a lower potential. Then ##E=-\dfrac{V_2-V_1}{d}## is a positive number which means that the E field points in the direction of the displacement that we chose. At this point we choose a symbol for the positive difference in the numerator whichever way the difference is taken, ##|V_2-V_1.## We call this positive difference "voltage" and we give it the same symbol ##V## as the potential function. It's not a very good choice but it is what it is. To avoid confusion we have to understand the context under which symbol ##V## is used because it can stand for two different things, the function ##V(r)## and the difference between two points in space at which this function is evaluated.

Your confusion stems from being unable to separate the two contexts when you see ##V## used in the same equation with different contexts. I don't blame you for this and I hope that you will get used to it after you have digested the duality of ##V##.

As I was typing this I noticed that @alan123hk posted and says essentially the same things.
 
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  • #32
kuruman said:
As I was typing this I noticed that @alan123hk says essentially the same things and includes a diagram

Yes, we must first assume that the electric field is uniformly distributed, because we want to find the electric field in the direction pointing from V1 to V2 or x=0 to x=d, so the amount of change should be V2-V1 and d-0.
 
  • #33
kuruman said:
The equation that you want me to consider is the source of your confusion. The definition of the electric field, as other have already said, is (in one dimension ##r##) ##E=-\dfrac{dV}{dr}## not what you have. When you say E=-V/r, you equate the ratio of small differences to the ratio of the function ##V## to independent variable ##r## and then you wonder where the difference ##\Delta V## came from. When the field is uniform as between parallel capacitor plates, the ratio is constant whichmeans that the ratio of small differences is the same as the ratio of large differences in which case you can write ##E=-\dfrac{dV}{dr}=-\dfrac{\Delta V}{\Delta r}.##
This is the bit I don't understand.

I'm beginning to understand everything. HOWEVER, I still am confused about how V turned into a difference. I'll try to keep this simple, so I understand it better. When we derived the formula, E = -V/r (ignoring whatever symbol we use, it maybe V or anything else), we said that V represented the electric potential at a point in space due to a point charge. So, in the formula, V was the electric potential, meaning the difference between the potentials of a point in the field to a point in infinity. Now how did you change the difference of potentials at infinity and at a point within an electric field, to the difference in potentials WITHIN an electric field.

I'm sorry if I seem a bit annoying, it's just that the book I'm getting my information from is a course book, which is like a 100 years old, has NO CALCULUS, no mention of any in-depth information about anything, at all. The only thing I learned was that once they finally derived E = -V/r, they mentioned V to be the electric potential at a point in space w.r.t a point at infinity. So, what I understand is that, sure, V can be a difference, but only a difference where one of the points must be at infinity. So, how come this difference now includes BOTH points to be within an electric field?
This is how I interpreted the formula. So, I guess now you can better understand where I'm messing up...?
 
  • #34
Are you familiar with the idea of the slope at a certain point? If you are not sure, read this Wikipedia article. Bear in mind that the definition of the electric field is the negative of the slope of the curve depicting the electric potential as a function of position. The negative sign is consistent with the idea that the electric field points from a region of high potential to a region of low potential. Also bear in mind that if the slope is constant (like a flat inclined plane) then the electric field is constant as is the case for a parallel plate capacitor.

Maybe these ideas will be clearer to you after you read the article. If not, ask again.
 
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1. What is potential gradient?

Potential gradient is a measure of the rate of change of electric potential with distance. It is a vector quantity that indicates the direction and magnitude of the change in electric potential.

2. How is potential gradient related to individual charges?

Potential gradient is directly proportional to the magnitude of the charge. This means that as the charge increases, the potential gradient also increases. However, the direction of the potential gradient is dependent on the sign of the charge.

3. What is the potential gradient between two parallel plates?

The potential gradient between two parallel plates is constant and equal to the electric field strength between the plates. This means that the potential gradient is directly proportional to the voltage difference between the plates and inversely proportional to the distance between them.

4. How does the potential gradient change with distance from a charged object?

The potential gradient decreases as the distance from a charged object increases. This is because the electric field strength decreases with distance, leading to a decrease in the potential gradient.

5. What is the unit of potential gradient?

The unit of potential gradient is volts per meter (V/m). This unit indicates the change in electric potential (volts) per unit distance (meter).

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