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Question

If we place a charge, say Q, ON the surface of a sphere, will there be a flux through the sphere?

My Understanding of the Concept

I'm studying high school physics, so I have learnt Gauss's Law and according to it if a charge is placed inside the sphere, a flux is produced. And if the charge is placed outside the sphere, the net flux through the sphere is 0 as entering flux equals the leaving flux.

So while practicing a few workout exercises, I encountered this particular problem:

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80 μC/m^2. So, what is the total electric flux leaving the surface of the sphere?

So over here I assumed the answer will be zero as the charged enclosed within the surface of the sphere is 0 so flux becomes 0 from the Gauss's formula that says flux=q(enclosed)/epsilon The reason due to which I assumed charge inside is 0 is because the question says surface charge density, which basically is (Net Charge)/Area, rather than volume charge density.

Please kindly help me with this

Hrishi
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2 Answers2

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The problem statement asks for the net flux leaving the sphere, not through the sphere as you initially stated.

Therefore you would imagine a closed surface outside the sphere, e.g., another imaginary sphere. Then, per Gauss' law the net flux produced by the charged sphere across the surface of the imaginary sphere is

$$\phi=\frac{Q}{e}$$

where $Q$ is the net charge on the sphere and $\epsilon$ is the electrical permittivity of the space.

In this case

$$Q=\pi d^{2}\sigma$$

Where $d$ is the diameter of the charged sphere and $\sigma$ is the charge density on the sphere.

Hope this helps

Bob D
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  • The surface area of a sphere is $\frac{1}{4} \pi d^2$? – Michael Seifert Dec 08 '21 at 14:57
  • @MichaelSeifert Mike thanks. It is $4\pi r^2=\pi d^2$ – Bob D Dec 08 '21 at 14:59
  • Bob, in my opinion the net flux leaving the sphere is half of yours \begin{equation} \phi= \dfrac{1}{2}\dfrac{Q}{\epsilon} \tag{01}\label{01} \end{equation} – Frobenius Dec 10 '21 at 22:17
  • @Frobenius The charge enclosed by a surface that surrounds (is outside of) the charged sphere is the surface area of the charged sphere time the surface charge density. Divide that by the permittivity. Where do you get the half from.? – Bob D Dec 10 '21 at 22:28
  • Think of a point charge $:q:$ on the $:xy-$plane of a system $:xyz$. The flux upwards ($z+$) is in a solid angle $:2\pi:$ so $:1/2(q/\epsilon)$. Similarly the flux downwards ($z-$) is in a solid angle $:2\pi:$ so $:1/2(q/\epsilon)$. So, any infinitesimal or point charge $:q:$ on the spherical surface produces flux $:1/2(q/\epsilon):$ outwards (leaving) and flux $:1/2(q/\epsilon):$ inwards the sphere. – Frobenius Dec 10 '21 at 22:29
  • I don’t know what you mean. They’re asking for the total flux leaving the surface of the sphere. To me that means the flux across a closed surface (not a plane) enclosing the charged sphere. Then the flux is total charge on the sphere divided by the permitivity per Gauss’s law – Bob D Dec 11 '21 at 08:42
  • Ok, Bob, thanks for your attention. Let leave it as it is. I hope either me or you will realize in the future where is the right and where is the wrong. – Frobenius Dec 11 '21 at 09:42
  • No problem. But I don’t think it’s a matter of who’s right or wrong. It just seems we are answering different questions. The total flux would be Q/2e, for example, across any infinite plane outside the sphere because it’s the projection of the surface of the sphere on the plane. But if it’s the total flux leaving the sphere we want to know we apply Gauss’s law which is Q/e. Anyway, take care – Bob D Dec 11 '21 at 13:08
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If the charge is uniformly distributed on the surface of the sphere, place a spherical Gaussian anywhere outside the surface to get the outward flux (which will appear to be from a point charge).

R.W. Bird
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