Animated Solution for Physics - Electrostatics: Consider the situation shown in the figure. The capacitor $A$ has a charge $q$ on it whereas $B$ is uncharged. The charge appearing on the capacitor $B$ a long time after the switch is closed is

The Mystery of the Unmoving Charge: A Tale of Bound Electrons

Have you ever looked at a circuit diagram and felt your intuition screaming one thing, only to find out the physics dictates something entirely different? This problem is a classic example of an intuition trap.

At first glance, closing a switch between two capacitors seems like an open invitation for charge to flow and equalize. But as we will see, electrons don't just move because a path exists; they move because they are pushed or pulled.

Let's carefully break down the initial state of our system. We have two capacitors, $A$ and $B$. Capacitor $A$ is fully charged with a charge $q$. This means its left plate holds a charge of $+q$, and its right plate holds a charge of $-q$.

Capacitor $B$, on the other hand, is completely uncharged.

Now, look closely at the outer wires. The wire connected to the left plate of $A$ just ends. The wire connected to the right plate of $B$ also just ends. These plates are electrically isolated from the rest of the universe.

When we close the switch $S$, we connect the right plate of $A$ to the left plate of $B$. It feels like we've opened a floodgate.

These two connected plates now form a single, isolated central conductor. Before the switch was closed, the total charge on these two plates was $-q$ (from $A$) plus $0$ (from $B$). By the law of conservation of charge, the total charge on this new central conductor must remain $-q$.

But does this $-q$ charge spread out?

Here is where the magic of electrostatics comes into play. Remember that isolated left plate of $A$? It holds a charge of $+q$, and because it's isolated, that charge is trapped. It cannot flow away.

This trapped positive charge acts like a powerful anchor. It exerts a strong electrostatic attraction on the negative electrons located on the right plate of $A$.

For the electrons to flow through the switch and populate capacitor $B$, they would have to fight against this intense attractive force. They would have to leave the comforting proximity of the positive charges.

In physics, systems always seek the lowest possible energy state. The lowest energy configuration here is for the negative charges to stay exactly where they are—as close to the positive charges as possible.

Because the negative charge is bound by the isolated positive charge, it will not flow through the switch.

Therefore, the charge on capacitor $B$ remains exactly zero. The switch was closed, the path was there, but the electrostatic forces said, "Stay put."