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A straight chain consisting of n identical metal balls is at rest in a region of free space as shown. In the chain, each ball is connected with adjacent balls by identical conducting wires. Length l of a connecting wire is much larger than the radius r of a ball. A uniform electric field E pointing along the chain is switched on in the region. Find magnitude of induced charges on one of the end bal

I got that the end balls have equal and opposite charges and in between all are neutral. But then i dont get how to calculate the charges in the balls.

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  • $\begingroup$ Think of them as capacitors. $\endgroup$ – harshit54 Oct 19 '18 at 17:51
  • $\begingroup$ CApacitors in series. $\endgroup$ – harshit54 Oct 19 '18 at 17:56
  • $\begingroup$ The answer given is b. If i proceed in this way i dont get the correct answer. Why are they capacitors in first place? $\endgroup$ – Rituraj Tripathy Oct 19 '18 at 18:02
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    $\begingroup$ You said that the middle spheres will be uncharged, why is that? $\endgroup$ – harshit54 Oct 19 '18 at 18:04
  • $\begingroup$ Because i think the chain behaves as a rod and all charges accumulate at the ends. Could you refer me to a method how i can find charge distribution in a conductor under given electric fields ? $\endgroup$ – Rituraj Tripathy Oct 19 '18 at 18:06
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I don't think b (or any other answer) is correct.

This solution appears to be based on the assumption that the charges are induced in the end balls only and those charges alone cancel the field in the whole wire. Based on this assumption, the charges on the end balls have to increase with the length of the wire, so that their fields could reach further and further.

In reality, the charges would be induced all along the length of the wire, contributing mostly to the local field cancellation. According to this approach, the charges on the end balls will be mostly contributing to the field cancellation near to the ends of the wire and those charges won't have to increase with the length of the wire.

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  • $\begingroup$ For reducing the potential of the system, wouldn't the charges be induced in the end balls only? $\endgroup$ – harshit54 Oct 21 '18 at 19:15
  • $\begingroup$ @harshit54 The potential of the system would not be just reduced - it would be have to be equalized over the whole length of the wire and the charges in the wire would be distributed to achieve that equalization. The charges on the two end balls alone would not be able to create such equalization, because the field between them is not uniform, while the original field E, which is being cancelled, is uniform. The induced charges always end up on the surface of a conductor, but they don't have to be concentrated exclusively at its tips. $\endgroup$ – V.F. Oct 21 '18 at 19:25
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So we know that an isolated sphere has a capacitance of $C = 4\pi\epsilon_0 r$ and two of them in series will have a total capacitance of $2\pi\epsilon_0r$.

The potential difference is $E\cdot distance$ = $E(n-1)l$.

Also $Q=CV$.

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    $\begingroup$ I did the same way too but i think the method here is wrong. If one refers to the way the formula for capacitance of spheres is derived, you can notice why this is wrong. Or may be i am wrong here so please help me out of this mind lag! $\endgroup$ – Rituraj Tripathy Oct 19 '18 at 18:25
  • $\begingroup$ If the charge on one sphere is q, we know that the potential of the one sphere is kq/r and the other is - kq/r. The potential difference of the two is 2kq/r. Now find the net capacitance of the system. $\endgroup$ – harshit54 Oct 19 '18 at 18:33
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    $\begingroup$ That's not how potential differences work $\endgroup$ – Aaron Stevens Oct 20 '18 at 11:08
  • $\begingroup$ Isn't that how you calculate capacitance of any capacitor by calculating the potential difference between the two objects holding the opposite charges. $\endgroup$ – harshit54 Oct 20 '18 at 11:29
  • $\begingroup$ Right... But your previous comment isn't the right way to find the potential difference between the spheres. That would be the potential at a point that is a distance $r$ from both spheres. $\endgroup$ – Aaron Stevens Oct 20 '18 at 21:53

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