# Problem with a sphere and a charge outside

I have something puzzling me...

Consider the standard problem of a grounded sphere and a charge $q$ outside: Do I need to describe it? Alright! Consider a grounded sphere, $V=0$, with radius $a$ centered at the origin of the coordinate system. Then bring to it a charge $q$ from infinity and place it at position $\mathbf x_1 = x_1\mathbf n$, where $\mathbf n$ is some unit vector. The problem is solved by considering an image charge $q'$ placed at position $\mathbf x_2 = x_2\mathbf n$, where the image values are: $$x_2 = \frac{a^2}{x_1},\quad\quad q' = -\frac{a}{x_2}q$$

The sphere was initially uncharged, but now, because $q$ is there, it got a charge of $q'$. This can be seen by drawing a gaussian outside the sphere, and one shall get $q'$ given the field is as if there exists two charges $q$ and $q'$.

For the record, the potential is: $$\Phi(\mathbf x) = \frac{q}{4\pi\epsilon_0}\left[ \frac{1}{|\mathbf x - x_1 \mathbf n|} - \frac{a}{x_1|\mathbf x - \frac{a^2}{x_1} \mathbf n|} \right]$$

By the way, why is there a charge $q'$ on the sphere? We started with a sphere with zero charge on it. From where did $q'$ came from?

Now, consider an insulated, charged (with $Q$) conducting sphere. We can build the result by superposition, like, if the charge $q$ didn't exist, the potential would simply be: $$\Phi(\mathbf x) = \frac{Q}{4\pi\epsilon_0}\frac{1}{|\mathbf x|}$$

But now, once we bring the charge $q$ from infinity to its place, by superposition, it the new potential will be the sum of them: $$\Phi(\mathbf x) = \frac{q}{4\pi\epsilon_0}\left[ \frac{1}{|\mathbf x - x_1 \mathbf n|} - \frac{a}{x_1|\mathbf x - \frac{a^2}{x_1} \mathbf n|} \right] \quad+\quad \frac{Q}{4\pi\epsilon_0}\frac{1}{|\mathbf x|}$$

However, that is not the correct answer. What is wrong with this reasoning?

In the middle of writing this question, I figured out the answer. But for future reference for myself, and in case other people has the same question, I decided to answer my own question.

There was a deceptively simple problem with that reasoning: the sphere was supposed to be insulated. If the potential were as suggested, a gaussian integral outside sphere, but without $q$ inside, would yield a charge of $Q+q'$. That is impossible by conservation of charge, given no charge could have been created/destroyed, given the sphere is insulated. To keep the charge constant at $Q$, we need to offset the $q'$ by actually adding $Q-q'$ of charge. Now, applying gauss law, we shall have the charge on the sphere always constant: $(Q-q') + q' = Q$.

Insulated, charged, conducting sphere: $$\Phi(\mathbf x) = \frac{q}{4\pi\epsilon_0}\left[ \frac{1}{|\mathbf x - x_1 \mathbf n|} - \frac{a}{x_1|\mathbf x - \frac{a^2}{x_1} \mathbf n|} \right] \quad+\quad \frac{Q - q'}{4\pi\epsilon_0}\frac{1}{|\mathbf x|}$$

Insulated (uncharged), conducting sphere: $$\Phi(\mathbf x) = \frac{q}{4\pi\epsilon_0}\left[ \frac{1}{|\mathbf x - x_1 \mathbf n|} - \frac{a}{x_1|\mathbf x - \frac{a^2}{x_1} \mathbf n|} \right] \quad+\quad \frac{-q'}{4\pi\epsilon_0}\frac{1}{|\mathbf x|}$$

• The charge is constant and equal to $Q$.
• If $q'$ is not there, the charge is still $Q$.
• The potential changes: as $q$ gets closer, $V$ from the sphere gets farther from zero.

Grounded, conducting sphere: $$\Phi(\mathbf x) = \frac{q}{4\pi\epsilon_0}\left[ \frac{1}{|\mathbf x - x_1 \mathbf n|} - \frac{a}{x_1|\mathbf x - \frac{a^2}{x_1} \mathbf n|} \right]$$

• The charge not constant and equal to $q'$.
• Where it got its charge? Simple. The sphere is grounded. It got it from ground.
• Charge was needed to be taken from ground to keep $V=0$ on the sphere.