A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capac

Q: Does it matter if the battery is kept connected or disconnected during the process?

No, it doesn't matter. If the battery is connected, the potential $V$ remains constant. If disconnected, the charge $Q$ remains constant. In either scenario, because the slab is reinserted, the final state is identical to the initial state, meaning the initial and final energies are exactly the same.

Q: What if the slab was only removed and not reinserted?

If the slab was only removed, the final state would be different from the initial state. The capacitance would change from $C$ to $C/K$, and the work done would be non-zero, depending on whether the battery was connected or not.

Q: Why is the work done equal to the change in potential energy?

According to the work-energy theorem for conservative forces, the work done by the system is equal to the negative change in potential energy. Since the process is cyclic, the change in potential energy $\Delta U$ is zero, making the net work done zero.

Q: Is there any heat dissipated in this process?

The problem states the slab is removed 'slowly'. This implies a quasi-static, reversible process where the system is always in equilibrium. Therefore, no kinetic energy is generated, and no heat is dissipated.

Q: Why do we call electrostatic force a conservative force?

A force is conservative if the work done by it in moving an object between two points is independent of the path taken. Equivalently, the work done in a closed loop is zero. Electrostatic force perfectly obeys this principle.

Question

A parallel plate condenser with a dielectric of dielectric constant $K$ between the plates has a capacity $C$ and is charged to a potential $V$ volts. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

Options:

(A) $\frac{1}{2}(K-1)CV^2$

(B) $CV^2(K-1)/K$

(C) $(K-1)CV^2$

(D) zero

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

When you first read this problem, your mind might immediately jump to complex formulas. You might start thinking about the force exerted by the capacitor plates on the dielectric slab, integrating that force over the distance it's pulled, and worrying about whether the battery is connected or disconnected.

But take a deep breath. This question is a classic trap designed to test your conceptual clarity rather than your mathematical stamina.

### The Power of State Functions

In physics, certain quantities are what we call **state functions**. This means their value depends strictly on the current state or configuration of the system, and absolutely not on the path taken to get there. Potential energy is a prime example of a state function.

Let's look at our initial state. We have a parallel plate capacitor with a dielectric slab of constant $K$ resting peacefully between its plates. The system has a capacitance $C$ and is charged to a potential $V$. The initial energy stored in this configuration is simply:

$$U_i = \frac{1}{2}CV^2$$

### The Cyclic Journey

Now, the problem states that the dielectric slab is slowly removed and then reinserted. Think about the geometry of the situation. We take the slab out, and then we put it exactly back where it was.

This is a **cyclic process**. We have taken the system on a journey, but the final destination is the exact same starting point. Because the physical configuration is identical to the initial state, the final capacitance is once again $C$, and the potential is $V$.

Consequently, the final energy of the system is:

$$U_f = \frac{1}{2}CV^2$$

### The Grand Conclusion

The electrostatic forces at play here are conservative. For any conservative force field, the net work done in a closed loop or cyclic process is always zero.

Mathematically, the net work done by the system is related to the change in its potential energy. Since the initial and final states are identical, the change in potential energy is zero:

$$\Delta U = U_f - U_i = \frac{1}{2}CV^2 - \frac{1}{2}CV^2 = 0$$

Therefore, the net work done by the system is **zero**. No complex integration required—just a solid understanding of conservative forces and cyclic processes!

Select Answer:

Visualized Solution

$U_{i} = \frac{1}{2} C V^{2}$

$Cyclic Process$

$U_{f} = \frac{1}{2} C V^{2}$

$W_{net} = 0$

The Sigma Insight: Capacitance and Capacitors

The Illusion of Complexity

The Power of State Functions

The Cyclic Journey

The Grand Conclusion

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