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In our physics textbook, we learned in applications of Gauss Law that the intensity of a field inside a hollow charged conducting sphere, is 0.

Given as flux inside such a sphere, $\phi$, is 0, so

$$ \phi = \textbf{E.A} $$

$$ \phi \neq 0 $$

$$ \textbf{A} \neq 0 $$

$$ \therefore \textbf{E} = 0 $$

All the explanations I read online, (like this) only imply the case of the conducting sphere.

But there is an exercise question in the textbook, which i will quote verbatim so as to remove any ambiguity from my point of view.

Is $\textbf{E}$ necessarily zero inside a charged rubber balloon if the balloon is spherical? Assume that charge is distributed uniformly over the surface.

Now, I wasn't able to come up with an answer myself so I checked with a solution manual (Assume , that it is not credible), and I found it solved for the rubber sphere (balloon) the exact same way.

Was that right, is the field inside a rubber sphere , 0?

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  • $\begingroup$ If you place rubber balloon in an electric field the field is not zero inside but the question make a very specific condition that the charge is distributed uniformly over the surface of the balloon and that will make field inside the balloon identically zero. $\endgroup$ – hsinghal Aug 31 '16 at 17:46
  • $\begingroup$ Answer the question, what is the difference between rubber balloon and conducting sphere, first. Essentially, what is the difference between rubber and metal (ex.). Then you will see why the field inside the conducting sphere is always zero, while for dielectric or semiconductor - no. $\endgroup$ – MsTais Aug 31 '16 at 18:06
  • $\begingroup$ @hsinghal Why is this condition necessary? Why is it not necessary for metals? Is it not necessary for metals? $\endgroup$ – Azhar Mehmood Sep 1 '16 at 16:36
  • $\begingroup$ In the case of metals it is automatically satisfied $\endgroup$ – hsinghal Sep 1 '16 at 18:06
  • $\begingroup$ @hsinghal Like this? $\endgroup$ – Azhar Mehmood Sep 2 '16 at 16:12
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Question specifically mentioned that the charge is distributed on the surface. So applying the Gauss law:

ϕ=EA=Q/ϵ

Select the Gaussian surface inside the sphere and apply the Gauss law. Then Q becomes zero and hence E is zero, when the charge distribution is symmetric over the surface.

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  • $\begingroup$ Though this says the same thing as the answer of @CountTo10, it doesn't address my confusion of why conducting matters $\endgroup$ – Azhar Mehmood Sep 1 '16 at 13:38
  • $\begingroup$ OK , I get that, but what will happen of the charge distribution is not symmetric? And why are metals not affected? ( Are they not affected by non symmetry of surface charges?) Or Am I missing something fundamental? $\endgroup$ – Azhar Mehmood Sep 1 '16 at 16:35
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    $\begingroup$ Un-even charge distribution also affects conducting spheres. There is a question asked before regarding this. Search for "Electric field inside a conductor when charges are asymetrically distributed ". $\endgroup$ – Kosala Sep 1 '16 at 17:34
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For generality, assume the sphere is an insulator, has a radius R, and has a charge Q distributed uniformly throughout the sphere, what then is the electric field as a function of r, the radial distance?

As the comment by hsinghal says, the question stipulates that the charge is on the outside, so then in the actual question, the field inside, as with a conductor, would be zero.

Forget the stipulation for the moment that the charge is spread over the outside of the sphere, just to find out how a charge inside an insulating body would behave.

If we compare the field emitting from a point charge Q to the electric field existing outside the insulating sphere. Draw a Gaussian surface around the rubber/insulating shell.

Applying Gauss's Law will produce, in both situations,

$E = kQ/r^2$

So, outside a sphere of charge Q the field is identical to that from a point charge.

Now we have to find the electric field at the centre of the object.

What is the electric flux through a interior Gaussian sphere r < R, and how much charge is enclosed by the sphere?

The volume of the Gaussian sphere is:

$V_g = (4/3)\pi r^3$

The charge enclosed by the Gaussian surface is the charge per unit volume multiplied by the volume of the sphere.

The uniform charge per unit volume $\rho $ in the insulating sphere is its total charge (Q) divided by its total volume.

$\rho =\frac {Q}{(4/3)\pi R^3 }$

The charge enclosed by the gaussian surface is then:

$q_{enc} = Qr^3/R^3$

The flux through the gaussian surface is $EA = 4\pi r^2E$

Applying Gauss' Law:

Net flux = $\frac {Q }{\epsilon_0}$

enter image description here

For the electric field inside the sphere we get:

$E =\frac {Qr}{(4\pi \epsilon_0R^3)}$

So, internally the field is proportional to $r$, and externally it's proportional to $1/r^2$. At the boundary, (the insulating shell), the two equations give the same value.

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  • $\begingroup$ Very detailed explanation , thanks. But I don't quite understand the part where you apply Gauss' Law to find the Net Flux. Isn't it $\frac{Q}{\epsilon}$ $\endgroup$ – Azhar Mehmood Sep 1 '16 at 8:40
  • $\begingroup$ Hi Azhar, Yes it is, my apologies, I was working through this at 2.30 am, so I was half asleep. I want to go through the whole thing again, it's been a while since I covered this and I was trying to do it from memory, as a refresher for myself. I want to check my battered copy of Halliday & Resnick. $\endgroup$ – user108787 Sep 1 '16 at 10:12
  • $\begingroup$ So, is it 0, or is it not? $\endgroup$ – Azhar Mehmood Sep 1 '16 at 15:54
  • $\begingroup$ This is all good , but isn't this only for when the charge is inside? $\endgroup$ – Azhar Mehmood Sep 1 '16 at 16:19
  • $\begingroup$ Yes , I see my dumb mistake, I asked the question wrong. $\endgroup$ – Azhar Mehmood Sep 1 '16 at 16:22

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