# field lines in a grounded conductor next to a charge

let us imagine we have a charge and a conducting grounded sphere at some distance. The sphere will be charged but at zero potential. However, if I try to compute the potential using the integral of the electric field from infinity, following a line that inlcudes the centers of both the sphere and the charge, I would get a result different than zero, unless: 1) the field line changes direction at some point (I cannot imagine how), or 2) the electric field along that line is zero

Is 1 or 2 correct, or is there a different explanation? Thanks

The usual approximation here is that the two spheres are so far apart that we can consider the conducting sphere to have a uniformly charged surface. If this is the case, then its potential is given by two contributions and, as you say, must sum up to zero: $$\frac{q}{4\pi\epsilon_0} \frac{1}{d} + \frac{q_s}{4\epsilon_0} \frac{1}{R} = 0$$

where $$q$$ is the charge, $$q_s$$ is the charge on the conductor, $$R$$ its radius and $$d$$ the distance between the two. From here you can find the charge that is transferred from the ground to the sphere:

$$q_s = -\frac{qR}{d}$$

If you want to compute the electric field, you can use use Coulomb law and superpose the two contributions. We can fix the reference frame so that the first charge is in the origin and the centre of the conducting sphere is in $$\vec{s} = (d, 0, 0)$$. Then, the field at any point (but the interior of the conducting sphere, where it is $$0$$), is given by:

$$E = \frac{q}{4\pi\epsilon_0} \frac{\hat{r}}{r^2} + \frac{q_s}{4\pi\epsilon_0} \frac{(\vec{r} - \vec{s})}{|\vec{r} - \vec{s}|^3} = \frac{q}{4\pi\epsilon_0} \frac{\hat{r}}{r^2} - \frac{qR}{4\pi\epsilon_0 d} \frac{(\vec{r} - \vec{s})}{|\vec{r} - \vec{s}|^3}$$

From this expression is easy to see that the field does change direction somewhere. In the limit of $$d \ll R$$, if we consider the line that connects the two charges, we can imagine this crossover point as being quite close to the position of the first charge.

Now, remember that, if you want to integrate it over a line that goes from infinity to $$\vec{s}$$ (while passing through the origin), you'll have at least one issue: the potential would be infinite at the position where the first charge is. You can easily get rid of this issue by also considering the first sphere as a conductor (or as an uniformly charged insulator).