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Could someone please explain the difference between potential, potential energy, potential difference (in all types, like electrical/gravitational etc.)?

Take potential for example: "The value of the potential at a point in space gives the amount of work that needs to be done to bring that object from infinity to that point" What does that even mean?

If we have an object in the Earth's gravitational field, isn't the potential within the object increasing as you take it away from the centre of mass, not if you bring that object from a place at infinity, with zero potential?

Also, without the dependance on formulas, how do I understand this intuitively:

At infinity, the potential energy of something is zero.

But when we think about it, as you raise an object higher, further from the surface of the Earth, it is said that the potential energy increases, and at the surface, you don't have any potential energy, as when it falls, it's converted into kinetic energy.

I'm not sure where my understanding is failing, please help.

Thank you.

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Potential is the potential energy per unit mass (gravitational) or charge (electrostatic). Potential energy at a point $\mathbf{r}$ is the negative of the work done in moving the particle from a reference point $\mathbf{r}_{ref}$ to $\mathbf{r}$ $$U = - \int_{\mathbf{r}_{ref}}^{\mathbf{r}} \mathbf{F} \cdot \mathbf{dr}$$ If you choose $\mathbf{r}_{ref} = \infty$, then $$U (r) = - \int_{\infty}^{\mathbf{r}} \left( - \frac{GMm}{r^2} \hat{r} \right) \cdot \hat{r} dr $$ $$ = - \frac{GMm}{R}$$ If you don't choose infinity as your reference point, then you get an additional constant to the right hand side of $U(r)$. That constant is conveniently zero most of the time when we pick our reference point to be infinity, which is why we usually do so. The equation $U = mgh$, is an approximation at low $h$, when you pick your reference point to be the ground. It is not accurate for high h.

However, you were partly correct. The potential energy does increase as you increase height in both formulas. In the formula derived from integration, as $R$ increases, $U(r)$ becomes less and less negative, and hence it increases ($-5$ is bigger than $-10$).

Potential difference is the difference in potential between two arbitrary points $\mathbf{r}_1$ and $\mathbf{r}_2$, $U(r_1) - U(r_2)$. You could hence say that $mgh = U(h) - U(ground)$ for low $h$.

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