A parallel plate capacitor of plate area A and plate separation d is charged to potential difference

Q: Why does the charge remain constant?

Because the battery is disconnected, the circuit is open. Electrons cannot jump across the gap, so the charge is trapped on the plates.

Q: Why does the capacitance increase with a dielectric?

The dielectric material gets polarized, creating an internal electric field that opposes the original field. This reduces the net electric field and potential difference, meaning the capacitor can store the same charge at a lower voltage, hence a higher capacitance.

Q: Why does the energy decrease?

The polarized dielectric is pulled into the capacitor by the non-uniform fringing electric field at the edges. The system does positive work to pull the slab in, which decreases its stored potential energy.

Q: What if the battery remained connected?

If the battery were kept connected, the potential difference $V$ would remain constant, and the charge $Q$ would increase to $K$ times its original value.

Q: Is the work done by the system or on the system?

The electrostatic force pulls the dielectric in, so the system does positive work. If an external agent slowly inserts the slab, it must do negative work to prevent the slab from accelerating. The magnitude of this work is the difference in potential energy.

Question

A parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to potential difference $V$ and then the battery is disconnected. A slab of dielectric constant $K$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q$, $E$ and $W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then

Options:

(A) $Q = \frac{\varepsilon_0 AV}{d}$

(B) $Q = \frac{\varepsilon_0 KAV}{d}$

(C) $E = \frac{V}{Kd}$

(D) $W = \frac{\varepsilon_0 AV^2}{2d} \left[1 - \frac{1}{K}\right]$

Sigma Percentile · Accepted Answer

The problem of a disconnected capacitor and a dielectric slab is a classic in electrostatics. It tests our understanding of what remains constant and how the internal properties of the system adapt to the new material. Let's dive into the physics of this process step by step!

### Analyzing the Setup

When a parallel plate capacitor is charged to a potential difference $V$, it stores a charge $Q$ on its plates. The relationship is given by the fundamental equation:
$$Q = CV$$

For a parallel plate capacitor with area $A$ and separation $d$, the capacitance in a vacuum (or air) is:
$$C = \frac{\varepsilon_0 A}{d}$$

Substituting this into our charge equation, we get the initial charge:
$$Q = \frac{\varepsilon_0 A V}{d}$$

**Crucial Insight:** The problem states that the battery is *disconnected* before the dielectric is inserted. This means the charge $Q$ is trapped on the plates. It has nowhere to flow. Therefore, throughout the entire process, the charge $Q$ remains absolutely constant. This immediately tells us that option (a) is correct and option (b) is incorrect.

### The Effect of the Dielectric

Now, we introduce a dielectric slab of constant $K$ that completely fills the space between the plates. A dielectric material gets polarized in the presence of an electric field, which effectively increases the capacitance of the system by a factor of $K$.

The new capacitance $C'$ is:
$$C' = KC = \frac{K \varepsilon_0 A}{d}$$

Since the charge $Q$ is constant, the new potential difference $V'$ across the plates must change to accommodate the new capacitance:
$$V' = \frac{Q}{C'} = \frac{Q}{KC} = \frac{V}{K}$$

The potential difference drops by a factor of $K$.

### The New Electric Field

The electric field $E$ between the plates of a parallel plate capacitor is directly related to the potential difference and the plate separation:
$$E = \frac{V'}{d}$$

Substituting our new potential difference $V'$, we find the new electric field:
$$E = \frac{V/K}{d} = \frac{V}{Kd}$$

This confirms that option (c) is also correct. The electric field is reduced by a factor of $K$ due to the opposing internal field created by the polarized dielectric.

### Energy and Work Done

Finally, let's look at the energy dynamics. The initial electrostatic potential energy $U_i$ stored in the capacitor is:
$$U_i = \frac{1}{2} CV^2 = \frac{\varepsilon_0 A V^2}{2d}$$

After the dielectric is inserted, the final energy $U_f$ is:
$$U_f = \frac{1}{2} C' V'^2 = \frac{1}{2} (KC) \left(\frac{V}{K}\right)^2 = \frac{1}{2} \frac{CV^2}{K} = \frac{\varepsilon_0 A V^2}{2Kd}$$

Notice that $U_f < U_i$. The system has lost potential energy. Where did this energy go?

When a dielectric is partially inserted into a charged capacitor, the non-uniform fringing electric field at the edges exerts an attractive force on the polarized dielectric, pulling it inward. The electric field does positive work on the slab.

The magnitude of the work done in this process is simply the difference between the initial and final energies:
$$|W| = U_i - U_f = \frac{\varepsilon_0 A V^2}{2d} - \frac{\varepsilon_0 A V^2}{2Kd}$$
$$|W| = \frac{\varepsilon_0 A V^2}{2d} \left[1 - \frac{1}{K}\right]$$

This perfectly matches option (d).

**A Note on Sign Convention:** If an external agent were holding the slab to insert it slowly without acceleration, the agent would have to pull *against* the electrostatic force, doing negative work. The problem asks for the "work done on the system", which technically would be negative. However, in multiple-choice questions of this type, it is standard convention to represent the magnitude of the work done (or the work done *by* the system), which is positive.

Select Answer:

Visualized Solution

$Initial Charge$

$New Capacitance$

$New Potential$

$New Electric Field$

$Initial Energy$

$Final Energy$

$Work Done$

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Effect of the Dielectric

The New Electric Field

Energy and Work Done

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Charge

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } New Capacitance

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Energy

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done