Timeline for Homework question about electric field between two spheres
Current License: CC BY-SA 3.0
15 events
when toggle format | what | by | license | comment | |
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Jun 11, 2020 at 9:33 | history | edited | CommunityBot |
Commonmark migration
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S Dec 21, 2017 at 11:20 | history | suggested | andselisk | CC BY-SA 3.0 |
OCRed image, post-processed figure
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Dec 21, 2017 at 10:37 | review | Suggested edits | |||
S Dec 21, 2017 at 11:20 | |||||
Oct 25, 2013 at 13:23 | vote | accept | Juan | ||
Oct 25, 2013 at 12:26 | history | edited | Qmechanic♦ |
edited tags
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Oct 25, 2013 at 12:24 | answer | added | Alfred Centauri | timeline score: 2 | |
Oct 25, 2013 at 12:24 | comment | added | John Rennie | @Juan: I've added an answer with the details | |
Oct 25, 2013 at 12:23 | answer | added | John Rennie | timeline score: 3 | |
Oct 25, 2013 at 11:06 | history | edited | Juan | CC BY-SA 3.0 |
replaced Q with V
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Oct 25, 2013 at 10:44 | comment | added | Juan | @JohnRennie: sorry to say but I don't think that I understand your comment. If I do this, I believe that I find that the E-fields are equal. That is, $F_1∝q_1E_1$ then $F_2/F_1=(q_2^2/r_2^2)/(q_1^2/r_1^2)=1$. This doesn't make any sense. | |
Oct 25, 2013 at 10:37 | answer | added | Rijul Gupta | timeline score: 0 | |
Oct 25, 2013 at 10:30 | comment | added | Juan | @rijulgupta: Why is the electric field $E_2 > E_1$? What's the physics that explains why $E_2 > E_1$? I want to say that the surface charge density is greater for sphere 2 than sphere 1 since sphere 1 is larger than sphere 2. But I am not sure. | |
Oct 25, 2013 at 9:55 | comment | added | John Rennie | The field strength is the force on a charge of 1 Coulomb, and this is given by the usual law for the force between two charges. If you write down the force on your test charge for a charge $Q_1$ at a distance $r_1$, do the same for sphere 2, then use the expression $Q_1/r_1 = Q_2/r_2$ you can calculate the ratio of $F_1$ to $F_2$. | |
Oct 25, 2013 at 9:21 | comment | added | Rijul Gupta | What do you mean by not seeing this physcally ? | |
Oct 25, 2013 at 3:27 | history | asked | Juan | CC BY-SA 3.0 |