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I am reading Griffiths' Introduction to Electrodynamics.

On page 364, example 8.2 (4th edition), he calculates the force on the northern hemisphere of a ball with total charge $Q$ spread uniformly.
In this question this is done using Maxwell's stress tensor.
(For those who don't have the book, the same calculation can be found on these lecture notes, on page 9)

He calculates the force on the upper part, the bowl, and the force on the lower part, the disk.

What I don't understand is why won't the stress tensor vanish on the disk.
My intuition is that the whole system has spherical symmetry, therefore there should be no force on the disk (since we can reflect across the xy-plane).
Note that on this formulation of the question he doesn't ask for the force due to the southern hemisphere, as he does in problem 2.47 (clearly in this case there should be force on the disk).

Where am I wrong?
Thanks, Shay

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  • $\begingroup$ Consider a sphere full of a pressurised gas. This is a spherically symmetric system, but if you cut the sphere along the equator the pressure of the gas would make the top half fly up and the bottom half fly down. So there is a net upwards force on the top half, just like in the charged sphere. The spherical symmetry just means the net force on the top half is equal and opposite to the net force on the bottom half. $\endgroup$ – John Rennie Apr 9 '15 at 15:43
  • $\begingroup$ Great analogy, thanks. That does it. Why didn't post it as an answer? $\endgroup$ – Shay Ben Moshe Apr 9 '15 at 16:13
  • $\begingroup$ Hi Shay, I thought it was a bit too brief for a good answer, but if it works for you I'll post it. $\endgroup$ – John Rennie Apr 9 '15 at 16:20
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Consider a sphere full of a pressurised gas. This is a spherically symmetric system, but if you cut the sphere along the equator the pressure of the gas would make the top half fly up and the bottom half fly down. So there is a net upwards force on the top half, just like in the charged sphere. The spherical symmetry just means the net force on the top half is equal and opposite to the net force on the bottom half.

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