Two parallel plate capacitors A and B have the same separation d = 8.85 x 10^-4 m between the plates

Q: Why do we treat the partially filled capacitor as two capacitors in parallel?

Because the potential difference across the dielectric-filled region and the air-filled region is the same, which is the defining characteristic of a parallel combination.

Q: Why does the energy increase when we pull the dielectric out?

Pulling the dielectric out requires you to do work against the attractive electrostatic forces between the polarized dielectric and the charged plates. This work you do is stored as additional potential energy in the capacitor.

Q: Why does the charge remain constant in part (b)?

The battery was disconnected before the slab was removed. Without a closed circuit, the electrons have nowhere to go, so the net charge on the plates is trapped and must remain constant.

Q: Why do we add the capacitances in part (c)?

When capacitors are connected plate-to-plate as shown in the figure, they are in parallel. In a parallel circuit, equivalent capacitance is simply the algebraic sum of the individual capacitances.

Q: Is energy conserved when the two capacitors are connected?

No, total energy is not conserved. When the charged capacitor A is connected to the uncharged capacitor B, charge flows until their potentials equalize. This flow of charge involves a transient current, which dissipates some energy as heat in the connecting wires.

Question

Two parallel plate capacitors $A$ and $B$ have the same separation $d = 8.85 	imes 10^{-4} 	ext{ m}$ between the plates. The plate areas of $A$ and $B$ are $0.04 	ext{ m}^2$ and $0.02 	ext{ m}^2$ respectively. A slab of dielectric constant (relative permittivity) $K = 9$ has dimensions such that it can exactly fill the space between the plates of capacitor $B$.
(a) The dielectric slab is placed inside $A$ as shown in figure (i). $A$ is then charged to a potential difference of $110 	ext{ V}$. Calculate the capacitance of $A$ and the energy stored in it.
(b) The battery is disconnected and then the dielectric slab is removed from $A$. Find the work done by the external agency in removing the slab from $A$.
(c) The same dielectric slab is now placed inside $B$, filling it completely. The two capacitors $A$ and $B$ are then connected as shown in figure (iii). Calculate the energy stored in the system.

Sigma Percentile · Accepted Answer

The problem of the two capacitors, $A$ and $B$, is a beautiful exercise in understanding how dielectrics affect capacitance, energy, and charge distribution. It takes us through a sequence of physical changes—inserting a dielectric, disconnecting a battery, removing the dielectric, and finally connecting two capacitors together. Let's break down this journey step by step.

### Analyzing the Initial State of Capacitor A

We start with capacitor $A$, which has a plate area of $0.04 	ext{ m}^2$. A dielectric slab with an area of $0.02 	ext{ m}^2$ and a dielectric constant $K = 9$ is inserted into it. Notice something interesting? The slab exactly covers **half** of the area of capacitor $A$.

When a dielectric partially fills a capacitor parallel to the plates, we can treat the system as two separate capacitors connected in parallel: one half filled with air, and the other half filled with the dielectric.

The capacitance of the air-filled half is:
$$C_{	ext{air}} = \frac{\varepsilon_0 (A/2)}{d}$$

And the capacitance of the dielectric-filled half is:
$$C_{	ext{dielectric}} = \frac{K \varepsilon_0 (A/2)}{d}$$

Since they are in parallel, the total initial capacitance $C_A$ is simply their sum:
$$C_A = \frac{\varepsilon_0 (A/2)}{d} (1 + K)$$

Plugging in the given values ($A/2 = 0.02 	ext{ m}^2$, $d = 8.85 	imes 10^{-4} 	ext{ m}$, $K = 9$, and $\varepsilon_0 = 8.85 	imes 10^{-12} 	ext{ F/m}$), we get:
$$C_A = \frac{8.85 	imes 10^{-12} 	imes 0.02}{8.85 	imes 10^{-4}} (1 + 9) = 2.0 	imes 10^{-9} 	ext{ F}$$

With the capacitor connected to a $110 	ext{ V}$ battery, the initial energy stored is:
$$U_A = \frac{1}{2} C_A V^2 = \frac{1}{2} (2.0 	imes 10^{-9}) (110)^2 = 1.21 	imes 10^{-5} 	ext{ J}$$

The total charge stored on the plates is:
$$q_A = C_A V = (2.0 	imes 10^{-9}) (110) = 2.2 	imes 10^{-7} 	ext{ C}$$

### The Disconnection and the Pull

Next, the battery is disconnected. This is a crucial moment! Disconnecting the battery isolates the capacitor, meaning the electrons have nowhere to go. The total charge $q_A$ is now trapped and must remain strictly **constant**.

Now, we pull the dielectric slab out. Capacitor $A$ becomes completely air-filled. Its new capacitance, $C_A'$, drops because the dielectric is gone:
$$C_A' = \frac{\varepsilon_0 A}{d} = \frac{8.85 	imes 10^{-12} 	imes 0.04}{8.85 	imes 10^{-4}} = 0.4 	imes 10^{-9} 	ext{ F}$$

Because the charge is constant but the capacitance has decreased, the stored energy actually **increases**. The new energy is:
$$U_A' = \frac{q_A^2}{2 C_A'} = \frac{(2.2 	imes 10^{-7})^2}{2 	imes 0.4 	imes 10^{-9}} = 6.05 	imes 10^{-5} 	ext{ J}$$

Where did this extra energy come from? It came from the external agency (you!) doing work to pull the slab out against the attractive electrostatic forces. The work done is simply the difference in energy:
$$W = U_A' - U_A = (6.05 - 1.21) 	imes 10^{-5} 	ext{ J} = 4.84 	imes 10^{-5} 	ext{ J}$$

### The Final Parallel Connection

Finally, we take that same dielectric slab and insert it into capacitor $B$. Since $B$'s area is $0.02 	ext{ m}^2$, the slab fills it completely. The capacitance of $B$ is:
$$C_B = \frac{K \varepsilon_0 A_B}{d} = \frac{9 	imes 8.85 	imes 10^{-12} 	imes 0.02}{8.85 	imes 10^{-4}} = 1.8 	imes 10^{-9} 	ext{ F}$$

We then connect capacitor $A$ (now empty) and capacitor $B$ (now filled) in parallel. Their equivalent capacitance is:
$$C_{	ext{eq}} = C_A' + C_B = (0.4 + 1.8) 	imes 10^{-9} 	ext{ F} = 2.2 	imes 10^{-9} 	ext{ F}$$

The total charge in this new combined system is still the original charge $q_A$, as charge is conserved. The final energy of the system is:
$$U_{	ext{final}} = \frac{q_{	ext{total}}^2}{2 C_{	ext{eq}}} = \frac{(2.2 	imes 10^{-7})^2}{2 	imes 2.2 	imes 10^{-9}} = 1.1 	imes 10^{-5} 	ext{ J}$$

Notice that the final energy ($1.1 	imes 10^{-5} 	ext{ J}$) is less than the energy before they were connected ($6.05 	imes 10^{-5} 	ext{ J}$). This energy was dissipated as heat in the connecting wires when the charge redistributed itself between the two capacitors. A beautiful application of charge conservation and energy redistribution!

Visualized Solution

$C_{A} = C_{air} + C_{dielectric}$

$C_{A} = 2.0 \times 1 0^{- 9} F$

$U_{A} = 1.21 \times 1 0^{- 5} J$

$C_{A}^{'} = 0.4 \times 1 0^{- 9} F$

$W = 4.84 \times 1 0^{- 5} J$

$C_{eq} = 2.2 \times 1 0^{- 9} F$

$U_{final} = 1.1 \times 1 0^{- 5} J$

The Sigma Insight: Combination of Capacitors

Analyzing the Initial State of Capacitor A

The Disconnection and the Pull

The Final Parallel Connection

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CA​=Cair​+Cdielectric​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CA​=2.0×10−9 F

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } UA​=1.21×10−5 J

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CA′​=0.4×10−9 F

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W=4.84×10−5 J

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ceq​=2.2×10−9 F

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ufinal​=1.1×10−5 J