What is Voltage? Elaborated Question Explained

[rcgldr]

voltage is what you call conserved energy.

Voltage is potential energy per unit charge, it's not potential energy. It's not useful for predicting the amount of work done unless the current (charge flow / unit time) or associated capacitance is included.

Back to my previous analogy, a 1 gram weight at 30 meters above sea level has the same gravitational potential as a 1 kilo-gram weight, but the potential energies are not the same.

Similarly, a 1 micro farad capacitor and a 1 farad capacitor might have the same potential, 12 volts, but the 1 farad capacitor has a lot more potential energy stored in it.

 

[rcgldr]

$$\Delta V = \int_a^b \vec{E} \cdot d\vec{s}$$

I don't know calculus that well (I'm a freshman in high school), but isn't that just "Es" or Ed? (Electric field times distance)

It's a dot product between the vector E and the small distance moved by the vector$\Delta s$. Only a change in the direction of the field matters, movement perpendicualr to the field doesn't change the voltage.

$$\Delta V = \int_a^b \vec{E} \cdot d\vec{s} = \int_a^b E cos(\theta) ds = \int_{r_a}^{r_b} E dr$$

where r is the distance from the field source to a point in the field.

For a point source
$$E = k q / r^2$$
$$\Delta V = k q \int_{r_a}^{r_b} dr / r^2 = - k q (\frac{1}{r_b} - \frac{1}{r_a})$$

For an infinite line source
$$E = k q / r$$
$$\Delta V = k q \int_{r_a}^{r_b} dr / r = k q (ln(r_b) - ln(r_a))$$

For an infinite plate source
$$E = k q$$
$$\Delta V = k q \int_{r_a}^{r_b} dr = k q ((r_b) - (r_a))$$

Since change in potential energy from point a to b = - work done moving from point a to point b

$$\Delta E_p = E_{pa} - E_{pb}$$

[tex]\Delta V = V_a - V_b[/itex][/tex]

 

[AUMathTutor]

Jeff Reid:

A clear and elegant exposition of the relevant facts.

(if you already know what you're talking about, and have studied the things before, and already know the answer to the question being asked, ...)