Recent content by ZEROBRAINCAPACITANCE

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...

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...

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!

It's not of much help, but here it is.

Makes sense, yeah, thanks!

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?

Do look at the way they've derived this here: https://courses.lumenlearning.com/physics/chapter/19-2-electric-potential-in-a-uniform-electric-field/ They seem to equate -V to Va-Vb stating that they're both equal in magnitude. Well.. why bother? Why not just straight up work with -V?

But that would mean there's a possibility of a negative field strength as well for when we move a charge against the field?

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.

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?

Alright, I know I'm being a bit of a bother here, but this is what's written in my book. 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...

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...

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...

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...

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...