Using method of images on ungrounded spheres

Question

I have recently learnt how to calculate the induced charges on a grounded sphere using the method of images. However, in all of the examples I did, the conductor itself is grounded, so I am wondering what would happen if the conductor is ungrounded. Since the electric potential on the surface is still constant, would the final result be different? Is it only a matter of "reference levels" ($V=0$) for the ungrounded case?

Answer 1

The method of images for a grounded sphere of radius $R$ centred at $z=0$ and a charge $q$ at $z=a$ is given by the image charge $$q'=-\left(\frac{R}{a}\right)q$$ at $z=R^2/a$. This setup ensures that the surface of the sphere remains equipotential.

What we can do here for an ungrounded sphere is exploit the principle of superposition: consider a superposition of (1) a charge $q$ and a grounded sphere and (2) a sphere at potential $V_0$.

Setup (1) is solved by $q'$, while setup (2) is achieved by a charge $q''$ placed at the centre of the sphere $z=0$, such that $$ V_0 = \frac{1}{4\pi\epsilon_0}\frac{q''}{R},\quad q'+q''=0. $$ The second equality is true because the sphere is neutral, so $$ q'' = -q' = \left(\frac{R}{a}\right)q . $$

You could refer to Griffiths' Introduction to Eletrodynamics chapter 3 for similar introductory problems.

Answer 2

In 3D, the electric potential converges to a finite value at infinity (provided the charge distribution is sufficiently localised). Grounding a conductor means that its potential is set to be equal to the potential at infinity. It is therefore not a matter of convention as it is the difference of two potentials (the one of the conductor and the one at infinity) that is at stake here. For convenience, you can take that these common value is zero.

In the case of the conducting sphere, the grounding of the sphere is still not a mere convention and has physical effects. For the ungrounded sphere, any outside charge induces a virtual charge within the sphere. When you charge the sphere, you'll need to add an extra virtual charge at its center equal to this extra charge. You can derive this by the principle of superposition and checking that this constructed potential satisfies the correct boundary conditions.

Hope this helps.