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From Griffiths' Introduction to electrodynamics:

enter image description here

In this problem, Griffiths says that we will set the potential inside and on the surface of the conductor to be zero since its an equipotential surface anyway, but then he claims that, due to 'symmetry' the entire xy plane is at potential zero aswell.

I tried to make sense out of this and noticed that the potential due to the field $ \vec E = E_0 \hat z$ in the xy plane does not change, but there's still the electric field of the induced charges on the sphere, why did he not account for it? Thanks

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1 Answer 1

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By symmetry the electric field lines are perpendicular to the $xy$ plane
That means that no work will be done moving a "test" charge along the plane with in turn means that the plane is an equipotential.
As the $xy$ plane hits the surface of the conductor which is at zero potential so must be the $xy$ plane.

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  • $\begingroup$ What about the E field of the induced charge on the conductor, it should be radial, or does it also cancel out in the x and y directions? $\endgroup$
    – khaled014z
    Commented Jul 2, 2019 at 9:24
  • $\begingroup$ @khaled014z But it will also be symmetrical about the $xy$ plane. $\endgroup$
    – Farcher
    Commented Jul 2, 2019 at 11:55
  • $\begingroup$ @Farcher Can you explain why by symmetry the electric field lines are perpendicular to the $xy$ plane, please? I couldn't understand it. $\endgroup$
    – S.H.W
    Commented Dec 31, 2019 at 20:00
  • $\begingroup$ The region above the xy-plane is no different to the region below the xy-plane other than the sign of the induced charges and the direction of the electric field. At the xy-plane the electric field lines cannot have a “kink” in them, ie they need to be smooth. $\endgroup$
    – Farcher
    Commented Dec 31, 2019 at 20:43
  • $\begingroup$ @Farcher Is it possible to prove that by other methods? I mean the method of images or superposition principle for example. I really couldn't understand many arguments that involve symmetry in physics. $\endgroup$
    – S.H.W
    Commented Jan 1, 2020 at 7:42

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