Timeline for Potential energy in a field and potential energy between two objects? (Electrostatics and Gravitation)
Current License: CC BY-SA 4.0
13 events
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| Jan 12, 2020 at 22:47 | comment | added | Shootforthemoon | Ahh, I see. Thank you very much for all the help! | |
| Jan 12, 2020 at 22:43 | comment | added | Philip Wood | Well done! My $\Delta \text{PE}$ was the amount by which the PE at $r_a$ is greater than that at $r_b$, so and (for charges of the same sign) is positive if $r_a<r_b$, that is $r_b>r_a$. And indeed a positive amount of work is done by the electrostatic force as the separation increases from $r_a$ to $r_b$. | |
| Jan 12, 2020 at 22:15 | comment | added | Shootforthemoon | Eventually I got $$ \Delta (\text {PE})=\frac{q q_0}{4 \pi \epsilon_0} \left(\frac{1}{r_a} - \frac{1}{r_b} \right)=\frac{q q_0}{4 \pi \epsilon_0} (\frac{r_b-r_a}{r_a r_b})\approx q_0E(r_b-r_a).$$ But maybe a silly question: shouldn't it be $(r_a-r_b)$ since we are evaluating the work done by the electrostatic force from $a$ to $b$? | |
| Jan 12, 2020 at 21:24 | comment | added | Shootforthemoon | Yes, of course I wrote it wrong XD. Thanks again! | |
| Jan 12, 2020 at 21:17 | comment | added | Philip Wood | The difference in potential energy between the charges at separations $r_a$ and $r_b$ isn't what you've written. It is $$\Delta (\text {PE})=\frac{q q_0}{4 \pi \epsilon_0} \left(\frac{1}{r_a} - \frac{1}{r_b} \right).$$ You can indeed arrange this to get $(r_b - r_a)$ on the top. Give it a go! | |
| Jan 12, 2020 at 19:30 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Jan 12, 2020 at 19:23 | comment | added | Philip Wood | Yes, by integration as 𝑚 goes from infinity away from 𝑀 to 𝑟 from the centre of 𝑀(assumed spherically symmetric) we get $$\text{GPE}=-\frac{GMm}{r}=-\frac{GMm}{r^2}\times r = -mgr$$ This is the grav PE taking zero PE at infinite separation. Note that $-mgr$ is just a neat way of writing $-\frac{GMm}{r}$; it ISN'T where $-\frac{GMm}{r}$ comes from. This comment replaces my previous comment, which was poorly argued. I've added the correct treatment of "passing from $r$ to $h$" to my answer (above). | |
| Jan 12, 2020 at 19:22 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Jan 12, 2020 at 12:03 | comment | added | Shootforthemoon | Thanks very much! | |
| Jan 12, 2020 at 11:45 | vote | accept | Shootforthemoon | ||
| Jan 12, 2020 at 11:26 | comment | added | Shootforthemoon | Thanks! now I see it better. But if we take the gravitational potential energy for an object and the Earth: we know that locally $mgr=-\frac{GMm}{r}$. But how do we pass from $r$ to $h$ (or $y$)? | |
| Jan 12, 2020 at 11:24 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Jan 12, 2020 at 11:19 | history | answered | Philip Wood | CC BY-SA 4.0 |