Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is unchar

Q: Why doesn't the charge just distribute equally between the two capacitors like it usually does?

Charge only distributes equally if the circuit allows for a complete path where both plates of the capacitors can balance their charges. Here, the outer plates are isolated, preventing the necessary charge redistribution.

Q: Doesn't closing the switch create a potential difference that forces charge to move?

Closing the switch connects the two inner plates, making them an equipotential surface. However, because the outer plates are isolated, moving any charge would create a massive opposing electric field, so the system is already at its lowest energy state with no charge transferred.

Q: What if the right plate of B was grounded?

If the right plate of B was grounded, it could pull positive charge from the earth. In that case, charge would indeed flow from A to B until their potentials balanced out!

Q: Is the electric field inside capacitor B zero?

Yes. Since no charge flows to its left plate, and its right plate is isolated and uncharged, both plates have zero charge. Thus, the electric field inside capacitor B remains zero.

Q: Why is the self-capacitance of a single plate important here?

Because the right plate of B is isolated, the left plate of B acts almost like a single isolated conductor. Its self-capacitance is very small, meaning even a tiny charge would cause a huge spike in potential, which acts as a barrier to charge flow.

Question

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

Options:

(A) zero

(B) $q/2$

(C) $q$

(D) $2q$

Sigma Percentile · Accepted Answer

### The Illusion of the Closed Switch

Imagine you are an electron sitting comfortably on the right plate of capacitor $A$. You are part of a negative charge $-q$, and right across the gap, on the left plate, is a positive charge $+q$. You are locked in a tight electrostatic embrace. The electric field between the plates is strong, and your potential energy is minimized by staying exactly where you are.

Suddenly, the switch $S$ is closed! A new path opens up, leading to the left plate of capacitor $B$. The question is: do you take the path and flow to capacitor $B$?

At first glance, it might seem tempting. After all, electrons love to spread out and reduce their crowding. But physics is never that simple. Let's look at the bigger picture.

### The Power of Isolated Plates

To understand why charge won't flow, we must look at the outer boundaries of our circuit. Notice the left plate of capacitor $A$ and the right plate of capacitor $B$. They are completely isolated. They are not connected to a battery, a ground, or any closed loop.

Because of the fundamental law of **conservation of charge**, an isolated plate cannot magically gain or lose net charge. The left plate of $A$ is stuck with its $+q$ charge forever. Similarly, the right plate of $B$ is uncharged and must remain at $0$ charge.

### The Energy Trap

Now, what would happen if some negative charge $-q'$ actually decided to travel through the switch to the left plate of $B$?

If $-q'$ arrives at the left plate of $B$, it creates a strong electric field. Normally, in a working circuit, this would pull positive charge onto the right plate of $B$ to balance things out and keep the potential energy low. But remember, the right plate of $B$ is isolated! It cannot acquire a positive charge.

Without that balancing positive charge, the left plate of $B$ acts like a lone, isolated conductor in space. The capacitance of a single isolated plate (its self-capacitance) is incredibly tiny compared to a parallel-plate capacitor. Pushing even a microscopic amount of charge onto it would cause its electrical potential to skyrocket.

Nature is lazy; it always seeks the state of lowest potential energy. The energy required to push charge onto the left plate of $B$ without a balancing charge on the right plate is simply too massive.

### The Final Verdict

Meanwhile, back at capacitor $A$, the positive charge $+q$ on the isolated left plate is still exerting a massive attractive force on the negative charge. It refuses to let the electrons go.

Because the energy cost of moving to capacitor $B$ is astronomically high, and the electrostatic attraction at capacitor $A$ is so strong, the electrons stay exactly where they are.

**Absolutely no charge flows through the switch.** Capacitor $A$ retains its full charge $q$, and capacitor $B$ remains completely uncharged. The final charge on capacitor $B$ is exactly **zero**.

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

Select Answer:

Visualized Solution

$Q_{A} = q, Q_{B} = 0$

\text{Switch } S \text{ is closed}

\text{Isolated Plates Analysis}

\text{Electrostatic Attraction}

\text{Energy Constraint}

Q_B = 0

The Sigma Insight: Combination of Capacitors

The Illusion of the Closed Switch

The Power of Isolated Plates

The Energy Trap

The Final Verdict

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