Animated Solution for Physics - Electrostatics: A parallel plate capacitor has plate of length $l$, width $w$ and separation of plates is $d$. It is connected to a battery of emf $V$. A dielectric slab of the same thickness $d$ and of dielectric constant $k=4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?

A Tale of Two Capacitances

Imagine a parallel plate capacitor connected to a battery of voltage $V$. The plates have length $l$ and width $w$, so the area is $l\times w$. The initial capacitance is simply given by the standard formula:

Since the battery remains connected, the voltage across the capacitor is constant at $V$. The initial energy stored in the capacitor is:

Now, we slide a dielectric slab of constant $k=4$ between the plates. Let's say it goes in up to a distance $x$. This changes the game completely. Look closely at the setup. We can treat this partially filled capacitor as two separate capacitors connected in parallel. One part has the dielectric, and the other part has just air. They are in parallel because they share the same top and bottom plates, meaning they have the same potential difference across them.

The capacitance of the dielectric part is $k{\epsilon }_0$ times its area, which is $x\times w$, divided by $d$. The air part has capacitance ${\epsilon }_0$ times its area, $(l-x)\times w$, divided by $d$. Adding them gives the total final capacitance:

The problem states that the final energy is twice the initial energy. Since the voltage $V$ is the same in both cases, this directly means the final capacitance must be exactly twice the initial capacitance.

Let's substitute the expressions for the capacitances. The common terms ${\epsilon }_0$, $w$, and $d$ cancel out beautifully. We are left with a simple linear equation:

Rearranging gives:

Finally, plug in the value of $k$, which is $4$. We get $3x=l$, which means $x$ is equal to $l/3$. So, the dielectric must be inserted one-third of the way in.

Final Answer: $x=\frac{l}{3}$

Here is a thought for you. What if we had disconnected the battery before inserting the dielectric? In that case, the charge $Q$ would remain constant, not the voltage. How would that change the energy equation? Think about it!