Potential Gradient for individual charges and parallel plates?

[kuruman]

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.

 

[alan123hk]

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.

 

[ZEROBRAINCAPACITANCE]

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...?

 

[kuruman]

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.