If we put a charge inside the cavity of a hollow conducting sphere, the charge will create an electric field and so the negative charges will move close to the cavity. This will create a shortage of negative charges at the surface of the sphere, creating an overall positive charge at the surface. My question is: now the surface has positive charge distribution, and around the cavity there is a negative charge distribution. How can the electric field still be zero in this case? I draw a sketch to illustrate what I mean:

red arrows indicate the electric field, taken from: https://youtu.be/240W4EIWxiA


2 Answers 2


The electric field in the cavity is not zero. Inside the cavity it is as you have drawn it.

I recommend that you use a textbook to check the argument for the absence of an electric field in a cavity in a conductor. You will see that it fails if there are charges suspended in the cavity.

Your diagram also seems to show a field inside the conductor itself. That would mean that free charges would be moving. So we wouldn't have a steady state.

Your radially inward arrows inside the conductor show the field contribution due to the induced charges on the inner and outer surfaces of the conductor. But there will also be an outward-pointing field contribution inside the conductor due to the charge suspended in the cavity. In the steady state these two field contributions inside the conductor cancel to zero.

  • $\begingroup$ What "conspiracy" (as Walter Lewin would say) is this that the field's contributions cancel out? If we approximate the spheres to a capacitor, the electric field inside is always the same and equal to $\frac{\rho}{\epsilon_0}$, but the electric field caused by the charge is $k\frac{q}{r^2}$, it depends on the distance ($r^2$). How can a constant field be equal to a distance-dependent field? $\endgroup$
    – arpg
    Feb 15, 2022 at 12:19
  • $\begingroup$ But with a spherical capacitor the field due to the charges on the surfaces isn't uniform but radial, as you'd see by imagining more red arrows on Farcher's lovely diagram! [If the conducting shell is very thin, you might not notice the non-uniformity – but you wouldn't notice the non-uniformity of the cavity charge's field either!] But my main point about the field inside the conductor is that the induced charges arrange themselves so that the resultant field inside the conductor is zero. It must be zero in the steady (electrostatic) state or the free charges would (still) be moving! $\endgroup$ Feb 15, 2022 at 12:43
  • $\begingroup$ Indeed, I get the argument. If there's a field inside the conductor, the charges are not still, and therefore they're not static. In order to achieve an electrostatic state, the field inside the conductor must be zero. But the electric field inside the conductor caused by the spheres is radial, yes, but its magnitude is also constant, let's say it is $\left\|\vec{E} \right\| = E$. $\endgroup$
    – arpg
    Feb 15, 2022 at 14:58
  • $\begingroup$ If there's a point (let's call it "1") in the conductor such that $E=k\frac{q}{r_1^2}\Rightarrow \vec{E}_{net}=\vec{0} (1)$, if we choose a different point (let's call it "2") at a different distance from the charge ($r_2\neq r_1$), (1) wouldn't be satisfied since the equality only works for $r_1$. I can't see what possible rearranging the charges could do to surpass this. $\endgroup$
    – arpg
    Feb 15, 2022 at 15:00
  • 1
    $\begingroup$ "I thought the electrical field was constant inside every capacitor, because we can say it's pretty much an infinite plane." This is for plane parallel plate capacitors. Do you know Gauss's law? You can apply it to a charged spherical surface and it tells you that, outside the surface, the field is the same as if the charged spherical surface were replaced by an equal charge at the centre of the sphere. That's assuming that nothing spoils the symmetry outside the surface. An equally and oppositely charged concentric spherical shell preserves the symmetry. $\endgroup$ Feb 15, 2022 at 18:04

Your diagram is not complete.

The $+q$ charge induces negative charges on the inside of the conducting shell and an equal number of positive charges on the outside of the spherical shell as shown in the digram below.

enter image description here

The $+q$ charge produces a radial electric field (blue) and the induced charges produce a field (red) inside the conductor exactly in opposition to the radial field produced by the charge $+q$.

The net result (green) is a radial field inside the shell, no electric field within the conducting shell, and a radial field outside the conducting shell.

  • $\begingroup$ I ask you the same question I asked to @philipwood. How is the green field exactly the same as the red one in that area? $\endgroup$
    – arpg
    Feb 15, 2022 at 12:21
  • $\begingroup$ @arp As like with a parallel plate capacitor there is no field produced outside the conductor by the induced charges. The blue and the red fields cancel each other out inside the conductor. $\endgroup$
    – Farcher
    Feb 15, 2022 at 17:18

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