2
$\begingroup$

enter image description here

We were doing some problems on electric fields and my teacher discussed this one :

Prerequisite: The electric field at a radial distance r inside a uniformly charged sphere of charge density ρ is given by $\frac{ρ\vec r}{3ε_o}$

Q) There are two oppositely charged spheres of uniform charge density ρ(pink) and -ρ(green). We fuse them together so that the vector joining O to O' is given by $\vec a$. Find the electric field at any point P inside common region .

Solution) Due to being oppositely charged, the common fused portion becomes overall uncharged. We calculate the electric field at P(in figure) due to left and right spheres individually. $$\vec E_O=\frac{ρ\overrightarrow r}{3ε_o}$$ $$\overrightarrow E_{O'}=\frac{-ρ\overrightarrow {r'}}{3ε_o}$$ Since : $$\overrightarrow a+\overrightarrow {r'}=\overrightarrow r$$ We can conclude: $$\overrightarrow E_O+\overrightarrow E_{O'}=\frac{ρ\overrightarrow {a}}{3ε_o}$$

I had a doubt that why are we using a relation derived for uniformly charged sphere for the uncharged portion in which P is located. The point P is not inside charged portion of the sphere, so the formula should not be applicable at that point. I hope you guys understand that.

I asked this to my teacher and his reply was "Result itself is the explanation" . Am I missing something here ? And how can we explain a result whose explanation is the result itself ?

I have come here as a last resort hoping that I could get a satisfactory answer.

PS: This is not a homework question or off-topic. I don't understand a conceptual thing, so moderators please have mercy on me.

$\endgroup$
1
  • 1
    $\begingroup$ As a note: I agree this isn't an off topic question, but typically question closure comes from regular users, not moderators. $\endgroup$ – BioPhysicist Apr 29 '20 at 17:22
2
$\begingroup$

This is just application of the superposition principle. The field due to both spheres is just the sum of the field produced by each sphere individually. The fact that the overlap region is uncharged is irrelevant for this to be true.

Where the uncharged portion would matter is if you were wanting to calculate the field directly from the total charge distribution of the system. Then you would need to take into account the contribution of the field from the non-overlapping regions, and the overlapped region would not contribute to the field. But this is a less-efficient way to solve the problem. Since electric fields superimpose, you can exploit that here.

$\endgroup$
4
  • $\begingroup$ Suppose I removed one of the spheres now, since the material isn't conducting, the uncharged portion would remain uncharged. Then, would the electric field at point P still be the same ? P is not on the inner surface of that uncharged arc , if it was, then i understand that the field would still be given by $\frac{{\rho}\overrightarrow r }{3e}$.In this scenario, I think it would change. Now bring back the other sphere. I hope you get the point I'm trying to make. $\endgroup$ – Physicsa Apr 29 '20 at 17:57
  • 1
    $\begingroup$ @Physicsa I don't understand. If you take one of the spheres away then $P$ is inside of the other sphere, so $P$ is now in a location with charge. With only one sphere the field is described by the equation you have given for the field due to one sphere. $\endgroup$ – BioPhysicist Apr 29 '20 at 18:05
  • $\begingroup$ Ok, I think i understand my mistake now. I was assuming as if all the charged has vanished from that area. The net charge is 0 but charge due to individual spheres is still present. Thank You! $\endgroup$ – Physicsa Apr 29 '20 at 18:12
  • 1
    $\begingroup$ @Physicsa Technically either view is correct if you are just looking at that system. But yes, if you take a sphere away, the uncharged region would not remain uncharged. $\endgroup$ – BioPhysicist Apr 29 '20 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.