Hello everyone! I have a question here and it has to do with a problem regarding Gauss's Law, which I encountered today and I felt that the problem was somewhat confusing, in the sense that it had an answer which didn't really seem right. I'll try to explain it as best as I can and then, give my argument. I will, most likely, give arguments which are qualitative and quantitative, so that you guys could understand my line of reasoning and, if necessary, correct me. So, here's the question;

Suppose that there is a charge which is positive and denoted by Q. It is placed at the origin. What I will do now is that I will cover it with a sphere, which is non-conducting and it is centered on the origin as well. It has a radius given by R. Find the electric flux through an area of 4(pi), which is on the sphere itself. That is, imagine the surface of the sphere and consider a circular hole which has an area of 4(pi).

The answer to this is Q/e, where e is the permittivity of free space. I don't understand this, because my answer was Q/eR^2.

Now, the way I actually found this answer is through simple reasoning about surfaces but I'm going to bring an analogy into this first and then extend the case to the sphere. Suppose that I have a cube instead, which has a total surface area given by A. At the center of the cube, there is a charge Q. The total flux is given by Q/e, as is the case for any closed surface. The flux through each side is Q/6e, which makes sense, because when the area decreases, we would expect the flux to decrease.

Now, for the question, the area of the small element is given by 4(pi) and the area of the entire sphere is given by 4(pi)R^2. Therefore, the conclusion is that there are a total of R^2 of these area elements and this is a pure number, as it is a ratio of the areas. So, it is legitimate to say that the flux is Q/eR^2.

My second argument is this; suppose that I say that the small area element is artistically wrapped in such a way that it forms a smaller sphere of surface area 4(pi) and it encloses the charge Q within it. Simply by directly applying Gauss's law, we find that the the total electric flux is still given by Q/eR^2, if you do the calculations right.

So, I've given both a quantitative and a qualitative argument about why my answer should make sense but I'm still not too sure about the validity of the arguments. I've asked a couple of seniors and they're just as confused as I am about the question. Please do provide insight into this and state why my arguments are wrong, if they are. Thank you, in advance! :)

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TopNewestPerhaps you have forgot to mention the area of the smaller piece and the logic is right given the symmetry. That is to say that flux through a surface area \(A\) of the sphere would be \( \dfrac{A}{4 \pi R^2} \). But the flux through the entire sphere would be \(\dfrac{Q}{\epsilon_0 } \).

Is this what you wanted to know or something else? – Sudeep Salgia · 2 years, 1 month ago

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The surface area of that little portion of the sphere of radius R is 4(pi). So, the thing is, I'm saying that there are R^2 number of those pieces and the flux through each one of them is Q/eR^2. If we add all that up, it equates to Q/e, as it should.

What i actually want to know is if I am correct, in terms of my answer for the question and I also want to know if I've reasoned it out well enough. The post here is based on the question at the start of it, which has a wrong answer, according to the conditions of the question.

Oh, I've also made the relevant amendments. I was pretty tired when I typed this out so I missed that. Sorry! – Abhijeet Vats · 2 years, 1 month ago

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The reason I articulated this way was because "having \(R^2\) parts " is intuitive only if \(R^2\) is an integer or at the most a rational. This argument makes things clear and more sound. Also it makes it clear that the symmetry is till the smallest surface area on a sphere unlike that on a cube where it cannot be smaller than the size of the face.

Hope this helps. – Sudeep Salgia · 2 years, 1 month ago

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– Abhijeet Vats · 2 years, 1 month ago

Oh, okay, I get way in which I should have articulated it. Thank you so much for the help!Log in to reply