A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential differenc

Q: Why do we use $U = \frac{Q^2}{2C}$ instead of $U = \frac{1}{2}CV^2$ for the final state?

Because the battery is disconnected, the charge $Q$ is conserved, but the voltage $V$ changes. Using $U = \frac{Q^2}{2C}$ makes it much easier to see how energy changes purely with capacitance.

Q: Why does the energy decrease when the dielectric is inserted?

The dielectric is polarized by the electric field, and the fringing fields exert an attractive force on it, pulling it in. The capacitor does positive work on the slab, which comes at the expense of its stored potential energy.

Q: What if the battery remained connected while inserting the slab?

If the battery were connected, the voltage $V$ would remain constant. The energy would increase (using $U = \frac{1}{2}CV^2$), and the battery would do work to supply more charge to the plates.

Q: Is the work done by the capacitor positive or negative?

The capacitor does positive work on the slab because it pulls the slab inwards (force and displacement are in the same direction). Consequently, the capacitor's own potential energy decreases.

Q: How do we handle the units in this calculation?

It's best to keep the initial energy in pico-joules ($\text{pJ}$) throughout the calculation. Since $K$ is dimensionless, the final answer will naturally be in $\text{pJ}$, avoiding messy powers of 10.

Question

A parallel plate capacitor having capacitance $12 	ext{ pF}$ is charged by a battery to a potential difference of $10 	ext{ V}$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates. The work done by the capacitor on the slab is

Options:

(A) $560 	ext{ pJ}$

(B) $508 	ext{ pJ}$

(C) $692 	ext{ pJ}$

(D) $600 	ext{ pJ}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a parallel plate capacitor that has been fully charged by a battery. The moment the battery is disconnected, a critical physical constraint is established: the charge $Q$ on the plates is trapped. It has nowhere to go, meaning $Q$ remains strictly constant throughout the rest of the process.

This is the most important realization in the problem. Whenever a battery is disconnected, charge is conserved.

### The Initial Energy

Before we introduce the dielectric slab, let's determine how much electrostatic potential energy is stored in the capacitor. We can use the standard formula:

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

Given that the capacitance $C = 12 	ext{ pF}$ and the voltage $V = 10 	ext{ V}$, we can substitute these values directly:

$U_i = \frac{1}{2} 	imes 12 	imes 10^{-12} 	imes (10)^2$

$U_i = 600 	imes 10^{-12} 	ext{ J} = 600 	ext{ pJ}$

So, our capacitor starts with a stored energy of $600 	ext{ pJ}$.

### Inserting the Dielectric

Now, a porcelain slab with a dielectric constant $K = 6.5$ is slipped between the plates. The presence of the dielectric increases the capacitance of the system by a factor of $K$. So, the new capacitance is $C' = KC$.

Because the charge $Q$ is constant, it is mathematically elegant to use the energy formula that involves $Q$ and $C$:

$U_f = \frac{Q^2}{2C'}$

Substituting $C' = KC$, we get:

$U_f = \frac{Q^2}{2KC} = \frac{U_i}{K}$

This tells us that the stored energy decreases by a factor of $K$ when the dielectric is inserted.

### The Work Done

The question asks for the work done **by** the capacitor on the slab. As the slab approaches the capacitor, the fringing electric fields at the edges of the plates polarize the dielectric and exert an attractive force on it, pulling it inwards.

Because the capacitor does positive work on the slab, its own stored potential energy must decrease. By the work-energy theorem, the work done by the capacitor is exactly equal to this decrease in potential energy:

$W = U_i - U_f$

$W = U_i - \frac{U_i}{K} = U_i \left(1 - \frac{1}{K}ight)$

### Final Calculation

Now, we simply plug in our known values:

$W = 600 \left(1 - \frac{1}{6.5}ight)$

$W = 600 	imes \frac{5.5}{6.5}$

$W \approx 507.69 	ext{ pJ}$

Rounding to the nearest integer, we get $508 	ext{ pJ}$. This perfectly matches option (b).

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Visualized Solution

$C = 12 pF, V = 10 V$

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

$Calculating Initial Energy$

$Initial Energy Value$

$Inserting the Dielectric$

$U_{f} = \frac{Q ^{2}}{2 C ^{'}}$

$Work Done by Capacitor$

$Calculating Work Done$

$Final Substitution$

$Final Answer$

$The Way Forward$

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Initial Energy

Inserting the Dielectric

The Work Done

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C=12 pF,V=10 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ui​=21​CV2

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Uf​=2C′Q2​

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

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

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

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

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