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Jun 11, 2020 at 9:33 history edited CommunityBot
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S Dec 21, 2017 at 11:20 history suggested andselisk CC BY-SA 3.0
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Oct 25, 2013 at 13:23 vote accept Juan
Oct 25, 2013 at 12:26 history edited Qmechanic
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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
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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