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What is the use of this equation and Gauss's Law enter image description here

As an example, for this problem enter image description here

I was able to to use the equation to have the exact same result without the use of Gauss's law as in the answer. However, for the same problem but for a charge inside the sphere like thisenter image description here I can not use the equation to obtain the same result.enter image description here So when to choose one over another?

Trần Hoàng Quân
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It is used when there are symmetries, but let me explain. Gauss' law is applicable everywhere, but sometimes it isn't helpful. To apply it, you need to invent any Gauss' surface around the charges, and applying the law you will get: $$\oint_S E·dS=\frac{q}{\epsilon_0}$$ However what you get is the result of the integral, not the electric field, as it is probably different at every point of the surface. So, if we have a symmetric surface like a sphere you know that the electric field will be constant along the surface. Then, the integral become: $$E·4\pi R^2=\frac{q}{\epsilon_0}$$ and you will get the electric field. However, if the surface is irregular, like a geoid, you will see that Gauss' law won't help you work out the electric field. To sum up, as in your problem you have a sphere, Gauss' law can definitely be used, so maybe you made a calculation mistake.

Marc Barceló
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  • I have added how I used the equation for problem B. My idea is that I treat the part inside the sphere as a smaller sphere with a radius that equals r, then I applied the equation. Clearly, I didn't get the desired result. Did I make a calculation mistake or is that impossible to use the equation for this kind of problem. – Trần Hoàng Quân Apr 05 '22 at 07:07
  • The calculations from the web are correct. Do you understand them? They have used Gauss' Law – Marc Barceló Apr 05 '22 at 07:20
  • Yes, I understand them, I just don't get why I can not use the equation for the electric field mentioned above for problem B while it is fine for problem A. In my mind, the only difference between the two problems is that in problem B, the radius of the sphere is a variable equal and in problem A, it is a constant. – Trần Hoàng Quân Apr 05 '22 at 07:31
  • Okey, I understand the question. The problem changes as what produces an electric field is only the charge inside the surface. So, you can use the first equation as long as you change the quantity Q as the charge inside the surface is not the entire charge. – Marc Barceló Apr 05 '22 at 07:45