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?

We start with a parallel-plate capacitor whose plates have length $$l$$, width $$w$$ and are separated by a distance $$d$$. The total plate area is therefore $$A = lw$$. While the capacitor is connected to a battery of constant emf $$V$$, its voltage remains fixed at $$V$$ throughout the entire process.

For any capacitor of capacitance $$C$$ kept at a fixed potential difference $$V$$, the electrostatic energy stored is given by the basic formula

$$U = \dfrac12\,C\,V^2.$$

Initial situation (no dielectric)
The initial capacitance is that of an empty parallel-plate capacitor,

$$C_i \;=\; \dfrac{\varepsilon_0 A}{d} \;=\; \dfrac{\varepsilon_0\,lw}{d}.$$

Therefore the initial energy is

$$U_i \;=\; \dfrac12\,C_i\,V^2 \;=\; \dfrac12\left(\dfrac{\varepsilon_0\,lw}{d}\right)V^2.$$

Situation after inserting the slab
A dielectric slab of the same thickness $$d$$ and dielectric constant $$K = 4$$ is now pushed in from one side. Let the length of slab already inside be $$x$$ (so the slab covers an area $$A_d = w\,x$$), while the remaining portion of the plates, having length $$l - x$$, is still filled with air. Because both regions share the same pair of plates and the same voltage $$V$$, the two regions act like two capacitors connected in parallel.

• Capacitance of the dielectric-filled region:

$$C_{\text{dielectric}} \;=\; \dfrac{K\,\varepsilon_0 A_d}{d} \;=\; \dfrac{4\,\varepsilon_0\,w\,x}{d}.$$

• Capacitance of the air-filled region:

$$C_{\text{air}} \;=\; \dfrac{\varepsilon_0 A_{\text{air}}}{d} \;=\; \dfrac{\varepsilon_0\,w\,(l - x)}{d}.$$

Because they are in parallel, their capacitances simply add:

$$C_f \;=\; C_{\text{air}} + C_{\text{dielectric}} \;=\; \dfrac{\varepsilon_0 w}{d}\,\bigl[(l - x) + 4x\bigr] \;=\; \dfrac{\varepsilon_0 w}{d}\,\bigl[l + 3x\bigr].$$

The new energy stored is therefore

$$U_f \;=\; \dfrac12\,C_f\,V^2 \;=\; \dfrac12\left[\dfrac{\varepsilon_0 w}{d}(l + 3x)\right]V^2.$$

Condition for doubling the energy
The problem asks for the slab length $$x$$ that makes the stored energy twice the initial energy, i.e.

$$U_f \;=\; 2\,U_i.$$

Substituting the expressions for $$U_f$$ and $$U_i$$ we have

$$\dfrac12\left[\dfrac{\varepsilon_0 w}{d}(l + 3x)\right]V^2 \;=\; 2\left[\dfrac12\left(\dfrac{\varepsilon_0\,lw}{d}\right)V^2\right].$$

The common factors $$\dfrac12,\,\varepsilon_0,\,w/d,\,V^2$$ cancel out immediately, leaving

$$l + 3x \;=\; 2l.$$

Now we isolate $$x$$ step by step:

$$3x \;=\; 2l - l \;=\; l,$$

$$x \;=\; \dfrac{l}{3}.$$

Thus the dielectric slab must be inserted to a length $$\dfrac{l}{3}$$ inside the plates in order for the energy stored in the capacitor to become double its original value.

Hence, the correct answer is Option B.