Voltage vs Potential Energy Formulas

[[V]]

I am having a hard time figuring out the signs for some of these formulas:

First of all, U=Potential Energy
$$V=\frac{U}{q}$$
$$\int F dr = \int Eq dr = - Work$$
Since,
$$Work=K=-U$$
Therefore
$$\int Eq dr=+U$$
$$U=\frac{QqK}{r}$$

Therefore,
$$V=\frac{KQ}{r}$$

BOTH U & V have a positive slope of 1/r.

So far, everything checks out. But when I want to find the ΔV across a distance of a parallel plate capacitor, something seems to break down...

$$\int E dr = V$$
Since E is constant here:
$$E\int dr = V$$
I take the derivative WRT to 'r'
$$E=\frac{dv}{dr}$$
$$dV=E * dr$$

What is the potential difference between 20cm and 40cm in the uniform 3000 (V/m) electric field?

The answer is -600V. I don't fully understand why it is negative!

The equation I just drived, ΔV=EΔr gives me a positive value!
How do you justify this negative number in the end? Please explain with calculus terminology if possible.

Are all my assumptions up until this point correct?

 

[tiny-tim]

Hi [V]!

You've lost a minus …

U = -Qqk/r (= ∫_-∞^r Qqk/x² dx) ​

 

[[V]]

Ahh! Cant be!
$$\int_i^f F dr = (-\frac{KQq}{r_{f}})-(-\frac{KQq}{r_{fi}})$$

$$W=-(\frac{KQq}{r_{f}}-\frac{KQq}{r_{fi}})$$

$$\Delta U=-W$$

$$\Delta U=\frac{KQq}{r_{f}}-\frac{KQq}{r_{fi}}$$

$$U=\frac{KQq}{r}$$

Correct?

 

[tiny-tim]

Hi [V]!

Sorry, you're right, I was getting confused with gravitational potential.

The error was in your equation …

$$\int E dr = V$$

… potential energy = minus work done by a conservative force

electric potential = minus work done per charge

so V = -∫ E dr ​

 

[[V]]

Thank you! :)

Soo
$$V = -\int E dr$$

If I derive both sides, I get

$$\frac{dV}{dr}=-E$$

$$E=\frac{-\Delta V}{\Delta r}$$

With this equation, my other problem seems to work.
However, now this presents another problem!

Using this relation, I want to find the Capacitance of a parallel plate capacitor.

Given that

$$E=\frac{Q}{\epsilon_0A}$$

And

$$C=\frac{Q}{\Delta V}$$

Therefore, using the relation I just proved in the previous stepped:

$$\frac{-\Delta V}{\Delta r} = \frac{Q}{\epsilon_0A}$$

Solve for Q

$$Q=-\frac{\epsilon_0A\Delta V}{\Delta r}$$

$$C=\frac{Q}{\Delta V}=-\frac{\epsilon_0A}{\Delta r}$$

then just replace variables to match my literature...

$$C=-\frac{\epsilon_0A}{d}$$

But Waiit!

My textbook says it is positive!
They seem to be using this relationship to derive this equation:

$$E=\frac{\Delta V}{d}$$

They use different variables, but why is it positive? I thought I just proved earlier that it should be :

$$E=-\frac{\Delta V}{d}$$

 

[tiny-tim]

hi [V]!

Using this relation, I want to find the Capacitance of a parallel plate capacitor …

let's measure all displacements from the -ve to the +ve plate …

then D and E are negative, D = -Q/A, E = -Q/ε₀A …

the potential difference from the -ve to the +ve plate is V = -∫ E.dx = ∫ Q/ε₀A dx = xQ/ε₀A

(to put it in more general terms, E goes from +ve to -ve, so the work done from the -ve to the +ve plate must be negative, and the potential difference must be positive)