A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential diffe

Q: Why does the dielectric slab get pulled into the capacitor in the first place?

The fringing electric fields at the edges of the capacitor plates exert an attractive force on the induced dipoles within the dielectric material. This force pulls the slab inward, converting electrostatic potential energy into kinetic energy.

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

Because the battery was disconnected! This means the charge $Q$ has nowhere to go and remains strictly constant. The voltage $V$, however, drops when the dielectric is inserted. Using the formula with the constant variable ($Q$) makes the calculation straightforward and prevents errors.

Q: What would happen if the battery was kept connected while inserting the slab?

If the battery remained connected, the voltage $V$ would be forced to stay constant. The capacitance would still increase to $KC$, meaning the battery would have to pump more charge onto the plates. In that case, the stored energy would actually *increase*, and the battery would do work.

Q: Why does the slab oscillate instead of just stopping in the middle?

Because we are assuming no friction! As the slab is pulled in, it accelerates and gains kinetic energy. When it reaches the center, the net force is zero, but its velocity is maximum. It overshoots, gets pulled back by the electric field from the other side, and thus oscillates back and forth indefinitely.

Q: Is the mechanical energy of the slab constant throughout the oscillation?

Yes! In the absence of friction or air resistance, the system is conservative. The sum of the electrostatic potential energy and the mechanical kinetic energy of the slab remains perfectly constant at all times.

Question

A parallel plate capacitor whose capacitance $C$ is $14	ext{ pF}$ is charged by a battery to a potential difference $V = 12	ext{ V}$ between its plates. The charging battery is now disconnected and a porcelain plate with $K = 7$ is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of ......... $	ext{pJ}$. (Assume no friction)

Sigma Percentile · Accepted Answer

Welcome to a fascinating journey into the world of electrostatics and energy conservation! This problem is a classic example of how abstract electrical concepts manifest as tangible, mechanical motion. Let's break down the physics of a dielectric slab being pulled into a capacitor and understand exactly where the energy goes.

### The Setup: A Charged Capacitor

Imagine you have a parallel plate capacitor. We are given its initial capacitance, $C = 14	ext{ pF}$, and it is connected to a battery that provides a potential difference of $V = 12	ext{ V}$.

Before we do anything else, let's figure out how much electrostatic potential energy is stored in this system. The formula for the energy stored in a capacitor when the voltage is known is:

$$E_i = \frac{1}{2} C V^2$$

Substituting our given values into this equation:

$$E_i = \frac{1}{2} 	imes 14 	imes (12)^2 = 7 	imes 144 = 1008	ext{ pJ}$$

So, our capacitor is sitting there, fully charged, holding onto $1008	ext{ pJ}$ of energy.

### The Disconnection: Conservation of Charge

Now comes the critical step: **the battery is disconnected**.

Why is this so important? Because disconnecting the battery means the charge $Q$ trapped on the capacitor plates has nowhere to go. It is strictly conserved. Whatever charge was built up during the charging phase is now locked in place.

### The Dielectric Invasion: Energy Shift

Next, a porcelain dielectric slab with a dielectric constant $K = 7$ is inserted between the plates.

We know that inserting a dielectric increases the capacitance of the system by a factor of $K$. So, the new capacitance becomes $C' = KC$. But what happens to the energy? Since the charge $Q$ is constant, it is much safer and more intuitive to use the energy formula that involves $Q$ and $C$:

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

Substituting $C' = KC$, we get:

$$E_f = \frac{Q^2}{2(KC)} = \frac{1}{K} \left( \frac{Q^2}{2C} ight) = \frac{E_i}{K}$$

This is a profound result! The final energy is simply the initial energy divided by the dielectric constant. Let's calculate it:

$$E_f = \frac{1008}{7} = 144	ext{ pJ}$$

### The Missing Energy: Where Did It Go?

Our system started with $1008	ext{ pJ}$ of energy, and now it only has $144	ext{ pJ}$ of electrostatic energy. That's a massive drop! By the law of conservation of energy, this missing energy couldn't have just vanished.

So, where did it go?

As the dielectric slab approaches the edges of the capacitor, the fringing electric fields induce dipoles within the slab and exert a strong attractive force on it. The electric field literally pulls the slab inward, doing positive work on it. This work manifests as the **kinetic energy** of the slab.

Because we are assuming a frictionless environment, the slab will accelerate as it gets pulled in, reach maximum velocity at the center, overshoot due to inertia, and then get pulled back. It will oscillate back and forth forever! The constant mechanical energy of this oscillation is exactly equal to the electrostatic energy lost by the capacitor.

### The Final Calculation

To find this mechanical energy, we simply take the difference between the initial and final electrostatic energies:

$$\Delta E = E_i - E_f$$

$$\Delta E = 1008 - 144 = 864	ext{ pJ}$$

And there we have it! The mechanical energy of the oscillating porcelain plate is $864	ext{ pJ}$. This beautiful interplay between electricity and mechanics is what makes physics so incredibly elegant.

Enter Numerical Value:

Visualized Solution

Initial State

Initial Energy Formula

Substitution

Initial Energy Calculation

Battery Disconnected

Final Energy Formula

Final Energy Calculation

Mechanical Energy

Final Answer

The Way Forward

The Sigma Insight: Capacitors and Capacitance

The Setup

The Disconnection

The Dielectric Invasion

The Missing Energy

The Final Calculation