Why E = σ/2ε₀ for an infinite sheet but σ/ε₀ just outside a conductor? Same Gauss law, different answer

7 answers & replies

• [+][–]Arjun Nair12thSample· 29d ago

  Different pillbox. For the sheet, field exists on both sides, so flux passes out through two end caps: $\Phi = 2EA$. For a conductor in electrostatics the field inside the metal is zero, so the pillbox sticking into the metal has flux only through the outer cap: $\Phi = EA = \sigma A/\varepsilon_0$.

  And the $\sigma$ means different things. In the sheet case it's the density of that one thin sheet. In the conductor case it's the density of the surface right there, and the rest of the charge in the whole system also contributes to that field. Total field is the vector sum of everything, which is why the number looks different.

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• [+][–]Siddharth RaoDropperSample· 29d ago

  Your coaching test answer. Thin conducting plate with total charge $Q$: the charge splits equally onto the two faces. Each face has density $\sigma_f = \dfrac{Q}{2A}$. Just outside each face, using the conductor result: $E = \frac{\sigma_f}{\varepsilon_0} = \frac{Q}{2A\varepsilon_0}$ So the first option. If you define $\sigma = Q/A$ (total charge per unit area of the plate, as if it were a non-conducting sheet), this is $\sigma/2\varepsilon_0$, same as the non-conducting sheet. So it's the same physics in the end, just hiding behind different definitions of σ. I lost marks on exactly this last year.

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  • [+][–]Nisha Verma12thSample· 29d agoAuthor

    The "same physics, different definitions of σ" part finally made it click. Writing a note: always check what σ means in the question.

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• [+][–]Harsh Gupta12thSample· 29d ago

  Inside a uniformly charged thin shell, the field is zero only at the centre because of cancellation, and it becomes non-zero as you move off-centre. Shell has symmetry about the centre only.

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  • [+][–]Meera Patel11thSample· 29d ago

    I would disagree, Harsh. By Gauss law, take a spherical Gaussian surface of any radius $r < R$ inside the shell. It encloses no charge, and by symmetry $E$ is the same magnitude everywhere on that surface and radial, so $E \cdot 4\pi r^2 = 0$, giving $E = 0$. This is true at every point inside, not just the centre. The reason is the cancellation of contributions from the nearer and farther parts of the shell, which exactly balance for an inverse-square law.

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    • [+][–]Harsh Gupta12thSample· 29d ago

      Fair, I was only thinking about the centre. Meera wins, as usual 😔 Field is zero everywhere inside.

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• [+][–]Kavya NaikDropperSample· 29d ago

  For the line charge, same style: cylinder of radius $r$ and length $l$, flux only through curved surface: $E\cdot 2\pi r l = \lambda l/\varepsilon_0$, so $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$.

  My quick list to stop mixing up:

  • Line: $\lambda / (2\pi\varepsilon_0 r)$
  • Non-conducting sheet: $\sigma / 2\varepsilon_0$
  • Conductor surface: $\sigma / \varepsilon_0$
  • Shell outside: $kQ/r^2$, inside: 0 and then I just ask "which Gaussian surface and where does the flux go?"

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